Question

A community generates 1 million litres/day (MLD) of wastewater. This wastewater is treated using activated sludge process (ASP). The working volume of the aeration tank of the ASP is 250 m\(^3\), and the biomass concentration in the tank is 3000 mg/L. Analyses results showed that a biomass concentration of 10 mg/L is present in the treated effluent from the secondary sedimentation tank of the ASP. Sludge wastage from the system is at a rate of 5000 L/day with a biomass concentration of 10000 mg/L. The system is in steady state condition.
The biological sludge residence time (BSRT) of the system (in days) is ......... (round off to one decimal place).

Hint:

The biological sludge residence time (BSRT) is a key indicator of the performance of activated sludge systems. It can be calculated by considering the biomass concentration, flow rates of influent and effluent, and the amount of sludge being wasted.

Answer 1

Given: - \( Q_0 = 1 \, {MLD} = 1 \times 10^6 \, {L/day} \) (wastewater flow rate)
- \( V = 250 \, {m}^3 \) (volume of aeration tank)
- \( X = 3000 \, {mg/L} \) (biomass concentration in aeration tank)
- \( X_e = 10 \, {mg/L} \) (biomass concentration in treated effluent)
- \( Q_w = 5000 \, {L/day} \) (sludge wastage flow rate)
- \( X_u = 10000 \, {mg/L} \) (biomass concentration in sludge wastage) The biological sludge residence time (BSRT) is given by the formula: \[ \theta_s = \frac{V \times X}{Q_w \times X_u + (Q_0 - Q_w) \times X_e} \] Substitute the given values into the formula: \[ \theta_s = \frac{250 \times 3000 \times 10^3}{5000 \times 10000 + (10^6 - 5000) \times 10} \] \[ \theta_s = \frac{250 \times 3000 \times 10^3}{5000 \times 10000 + (10^6 - 5000) \times 10} = 12.5 \, {days} \] Thus, the biological sludge residence time (BSRT) is \( 12.5 \, {days} \).