An Activated Sludge Process (ASP) has an inlet wastewater flowrate of 20000 m³/day with a Biochemical Oxygen Demand (BOD) concentration of 250 mg/l. It produces treated wastewater containing 20 mg/l BOD. The aeration tank has a working volume of 6000 m³ and a biomass concentration of 3000 mg/l. Biological Sludge Residence Time (BSRT) of the system is 6 days. The influent wastewater and the treated effluent from the system have negligible concentrations of biomass. The sludge recycle line from the bottom of the Secondary Sedimentation Tank (SST) to the inlet of the aeration tank has a flowrate of 6000 m³/day. To maintain equi librium, the flowrate (in m³/day) of sludge that is to be wasted from the system is________ (in integer).

The correct answer is 231.

1. Identify the given parameter values:

Influent flowrate, $Q=20000{\text{m}}^3\text{/day}$
Aeration tank volume, $V=6000{\text{m}}^3$
Biomass concentration in aeration tank (MLSS), $X=3000\text{mg/L}$
Biological Sludge Residence Time (BSRT), ${\theta }_c=6\text{days}$
Recycle sludge flowrate, $Q_r=6000{\text{m}}^3\text{/day}$
Effluent biomass concentration is negligible, $X_e\approx 0$

2. Mass Balance at the Secondary Sedimentation Tank (SST):

Assuming no biomass in the effluent and no biological growth occurring within the settling tank, we write a mass balance for biomass entering and leaving the SST:

$(Q+Q_r)X=Q_r{X}_w$

where $X_w$ is the concentration of biomass in the underflow/recycle line. Substituting the given values:

$(20000+6000)\times 3000=6000\times X_w$

$26000\times 3000=6000\times X_w$

Solving for $X_w$:

$X_w=\frac{26000\times 3000}{6000}=13000\text{mg/L}$

3. Calculate the Sludge Wastage Flowrate ($Q_w$):

The definition of BSRT (or Mean Cell Residence Time, ${\theta }_c$) is given by the ratio of the total biomass in the system to the biomass wasted per day:

${\theta }_c=\frac{VX}{Q_w{X}_w+Q_e{X}_e}$

Since the effluent biomass concentration ($X_e$) is negligible, the equation simplifies to:

${\theta }_c=\frac{VX}{Q_w{X}_w}$

Rearranging the equation to solve for the wastage flowrate ($Q_w$):

$Q_w=\frac{VX}{{\theta }_c{X}_w}$

Substitute the known values:

$Q_w=\frac{6000\times 3000}{6\times 13000}$

$Q_w=\frac{3000}{13}\approx 230.77{\text{m}}^3\text{/day}$

Rounding to the nearest integer, we find the flowrate of sludge that is to be wasted is 231 ${\text{m}}^3\text{/day}$.