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: Define the recycle ratio.
The recycle (return sludge) ratio is $R = Q_r/Q = 6000/20000 = 0.3$. This means the return flow is 30% of the influent flow.

Step 2: Relate the MLSS concentration to the return sludge concentration through $R$.
At the aeration tank inlet, a solids mass balance (with negligible solids entering with the influent) gives, per unit of influent flow $Q$:
$R\,Q\,X_r = (1+R)Q\,X$
Dividing through by $Q$: $R X_r = (1+R)X$, so
$X_r = X\times\dfrac{1+R}{R} = 3000\times\dfrac{1.3}{0.3}$

Step 3: Evaluate $X_r$.
$X_r = 3000\times4.333 = 13000$ mg/l, the same return-sludge concentration found through the direct mass balance, confirming the recycle-ratio shortcut is consistent.

Step 4: Use the solids-inventory form of the SRT definition.
The mass of biomass held in the tank is $M = V\times X = 6000\times3000 = 18{,}000{,}000$ (mg/l times m$^3$ units). At steady state, this stock must be replaced (wasted) once every $\theta_c=6$ days, so the daily solids wasting rate needed is
$\dfrac{M}{\theta_c} = \dfrac{18{,}000{,}000}{6}=3{,}000{,}000$ (same units per day)

Step 5: Since sludge is wasted from the return line at concentration $X_r$, find the flowrate that removes exactly this much solid mass per day.
$Q_w\times X_r = 3{,}000{,}000 \implies Q_w = \dfrac{3{,}000{,}000}{13000}=230.8$ m$^3$/day
Rounding, the wasting flowrate needed is about 231 m$^3$/day, matching the direct substitution method exactly.
\[ \boxed{Q_w \approx 231 \text{ m}^3/\text{day}} \]