Find. The solids retention time (SRT, $\theta_c$); if short, how it can be increased.
Approach. $\theta_c$ is the mass of solids held in the reactor divided by the rate at which solids LEAVE the system (wasting + clarifier overflow). The waste flow is drawn from the RAS line, whose concentration $X_r$ is not given directly — recover it first from an overall flow balance and a clarifier solids mass balance.
Recover the final-effluent flow, $Q_e$. The whole system (reactor + clarifier) receives only the 4000 L/d influent and loses it only as final effluent overflow plus wasted sludge, so by an overall steady-state flow balance:
$$Q_e=Q-Q_w=4000-1200=\boxed{2800\ \text{L/d}}$$
Recover the RAS/WAS underflow concentration, $X_r$. The recycle and waste streams are drawn from the same clarifier underflow line, so they share one concentration $X_r$. A steady-state solids balance around the clarifier (solids entering with the mixed-liquor flow $Q+Q_r$ at concentration $X$ = solids leaving as effluent overflow + solids leaving as underflow at $X_r$) gives:
$$(Q+Q_r)X=Q_eX_e+(Q_r+Q_w)X_r$$
$$X_r=\frac{(Q+Q_r)X-Q_eX_e}{Q_r+Q_w}=\frac{(6000)(2000)-(2800)(40)}{3200}=\boxed{3715\ \text{mg/L}}$$
This is a physically reasonable (if somewhat dilute) return-sludge concentration, consistent with the clarifier providing only modest thickening here.
Mass of solids in the reactor.
$$M=V\times X=2000\ \text{L}\times2000\ \text{mg/L}=4{,}000{,}000\ \text{mg}=4.0\ \text{kg}$$
Solids retention time. Divide the reactor's solids inventory by the total rate at which solids leave the system (wasting at $X_r$ plus overflow at $X_e$):
$$\theta_c=\frac{VX}{Q_wX_r+Q_eX_e}=\frac{4{,}000{,}000}{(1200)(3715)+(2800)(40)}=\frac{4{,}000{,}000}{4{,}570{,}000}=\boxed{0.875\ \text{d}\ (\approx21\ \text{hr})}$$
Final results
Quantity
Value
Final effluent flow, $Q_e$
2800 L/d
RAS/WAS concentration, $X_r$
3715 mg/L
Solids retention time, $\theta_c$
0.875 d (≈21 hr)
An SRT of under one day is very short for an activated-sludge process — conventional BOD/TOC removal typically targets $\theta_c\approx3$–15 days, and any degree of nitrification requires 8–20 days or more, so this reactor is almost certainly under-performing (poor floc formation, low biomass yield, effluent quality at risk) at its current wasting rate.
How to increase the SRT. Since $\theta_c=VX/(Q_wX_r+Q_eX_e)$, and the effluent-loss term $Q_eX_e$ is fixed by the required treated-flow rate and achievable clarifier performance, the SRT is increased primarily by REDUCING the wasting term $Q_wX_r$: (i) reduce the wasted volume $Q_w$ directly (waste less frequently or a smaller volume per event) — the single most direct lever, since $\theta_c$ is inversely related to $Q_w$ at fixed $X_r$; (ii) increase the reactor's solids inventory $VX$ by raising the operating MLVSS setpoint (more aeration-tank biomass for the same wasting rate) or, less practically, increasing reactor volume; (iii) improve clarifier thickening (a higher achievable $X_r$ lets the SAME solids mass be wasted in a smaller volume $Q_w$, which only helps if $Q_w$ is then reduced accordingly — thickening alone, with $Q_w$ unchanged, does not change $\theta_c$ since the same total mass $Q_wX_r$ still leaves).
Check: $X_r$ (RAS/WAS concentration) is not stated in the source data and is recovered here from a clarifier solids mass balance under the standard steady-state, no-accumulation assumption — a legitimate open-book-exam technique when the underflow concentration is not tabulated directly. The short SRT the mass balance returns (<1 day) is internally consistent with the question's own prompt ("if short, how can it be increased"), supporting the approach.
