Question

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. The 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 equilibrium, the flowrate (in m³/day) of sludge that is to be wasted from the system is ______ (in integer).

Hint:

Find the return sludge concentration \(X_r\) from a mass balance at the aeration tank inlet, then use \(\theta_c=VX/(Q_wX_r)\) to solve for \(Q_w\).

Answer 1

Correct Answer: 231

Step 1: Recall the definition of Biological Sludge Retention Time (BSRT / SRT).
SRT (also called mean cell residence time, \(\theta_c\)) is the average time a unit of biomass (MLSS) stays in the whole biological system before being removed. It equals the total mass of biomass held in the aeration tank divided by the rate at which biomass leaves the system:
\[ \theta_c = \frac{V X}{Q_w X_r + Q_e X_e} \]
where \(V\) = aeration tank volume, \(X\) = MLSS in the tank, \(Q_w\) = wasted sludge flowrate, \(X_r\) = biomass concentration in the wasted (return/RAS) line, \(Q_e\) = effluent flow, and \(X_e\) = effluent biomass concentration. Since effluent biomass is negligible, \(Q_e X_e \approx 0\), so
\[ \theta_c = \frac{V X}{Q_w X_r} \]

Step 2: Find \(X_r\), the concentration in the recycle (return sludge) line, from a mass balance around the aeration tank.
At the tank inlet, the incoming wastewater (flow \(Q=20000\) m³/day) carries negligible biomass, and the return sludge (flow \(Q_r=6000\) m³/day at concentration \(X_r\)) supplies essentially all the biomass entering the tank. At steady state this mixes to the tank's MLSS \(X=3000\) mg/l over the combined flow \((Q+Q_r)\):
\[ (Q+Q_r)X = Q_r X_r \implies X_r = \frac{(Q+Q_r)X}{Q_r} = \frac{(20000+6000)\times3000}{6000} \]

Step 3: Evaluate \(X_r\).
\[ X_r = \frac{26000\times3000}{6000}=\frac{78000000}{6000}=13000 \text{ mg/l} \]
This is higher than the tank MLSS because the return line carries the settled, concentrated sludge from the bottom of the secondary sedimentation tank.

Step 4: Substitute into the SRT equation and solve for \(Q_w\).
\[ 6 = \frac{6000\times3000}{Q_w\times13000} \implies Q_w = \frac{6000\times3000}{13000\times6}=\frac{18000000}{78000} \]

Step 5: Compute \(Q_w\).
\[ Q_w = 230.8 \text{ m}^3/\text{day} \]

Final Answer:
Wasting the sludge from the concentrated return line (at \(X_r=13000\) mg/l, not at the dilute tank MLSS of 3000 mg/l) needs a much smaller flowrate to remove the same mass of solids and hold the 6-day SRT.
\[ \boxed{Q_w \approx 231 \text{ m}^3/\text{day}} \]