Question 7 of 7: pH control / ion exchange / RO design principles, and activated sludge process design

Question 7: pH control / ion exchange / RO design principles, and activated sludge process design (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Design principles — (a) pH control and (b) ion exchange / reverse osmosis

(a) pH control — three key design principles. (1) Reagent selection matched to the buffering capacity and endpoint chemistry of the stream: a strongly buffered stream (e.g. carbonate-alkalinity-rich water) needs a stronger acid/base and more careful dose-response characterization than a weakly buffered one, and the titration curve (not a linear dose assumption) must be used near the equivalence point where pH is most sensitive to small dose changes. (2) Residence time and mixing sized for the reaction kinetics, not just flow: the reactor/basin must provide enough hydraulic residence time and complete mixing for the neutralization reaction and any associated precipitation to go to completion before discharge, with multi-stage (two- or three-tank) systems often used to flatten the titration curve's steep region across more than one control loop. (3) Fast, closed-loop instrumentation: because the titration curve is highly non-linear near neutral pH, a single feedback loop with a fast-responding pH probe (and, for large or variable flows, feed-forward flow-paced dosing ahead of the feedback trim) is required to avoid dosing overshoot/oscillation.

(b) Ion exchange or reverse osmosis — three key design principles. (1) Pretreatment to protect the exchange resin / membrane: suspended solids, oxidants (free chlorine attacks RO membranes and some resins), and scale-forming ions (Ca, Ba, Sr with sulfate/carbonate) must be removed or sequestered (antiscalant dosing, cartridge filtration, dechlorination) upstream, since fouling/scaling is the dominant cause of premature performance loss in both technologies. (2) Capacity/recovery sized against the specific feed-water chemistry: for ion exchange, resin exchange capacity (eq/L resin) is sized against the specific ionic load and required run length between regenerations; for RO, the design recovery (permeate/feed ratio) is limited by the concentration factor at which the least-soluble scale-forming salt reaches saturation in the reject stream, not by membrane hydraulics alone. (3) Regeneration/reject-stream management: ion exchange requires a regeneration cycle (acid/caustic or brine) producing a concentrated waste stream, and RO concentrates the rejected salts into a smaller reject flow — both require a designed disposal or further-treatment path for that concentrated stream, which is frequently the limiting design/permitting constraint rather than the primary treatment step itself.

Given.

Find. (a) Aeration tank volume $V$ and hydraulic retention time $\tau$. (b) Mass of sludge wasted daily. (c) Sludge recycle ratio $Q_r/Q_0$.

Approach. The completely-mixed activated-sludge design equation links $\theta_c$, the biomass yield/decay kinetics, and the aeration tank's MLSS inventory; solving it for $V$ gives the tank size and HRT. The same $\theta_c$ definition (mass of solids in the system divided by mass wasted per day) gives the daily solids production, which equals the mass that must leave via wasting; dividing by the concentration at which it is wasted ($X_w$) gives the wasting rate. The recycle ratio then follows from a clarifier solids balance at steady state.

1. Aeration tank volume. The standard $\theta_c$ design equation, $1/\theta_c=\dfrac{YQ_0(S_0-S)}{VX}-k_d$, rearranges to $$V=\frac{\theta_c\,Y\,Q_0(S_0-S)}{X(1+k_d\theta_c)}=\frac{25\times0.7\times100{,}000\times275}{5{,}000\times(1+0.05\times25)}=\frac{4.8125\times10^8}{11{,}250}$$ $$V\approx42{,}778\ \text{m}^3$$ ==**Aeration tank volume V ≈ 42,800 m³**==
2. Hydraulic retention time. $$\tau=\frac{V}{Q_0}=\frac{42{,}778}{100{,}000}=0.4278\ \text{d}=10.27\ \text{h}$$ ==**τ ≈ 10.3 hours**==
3. Daily sludge production (mass wasted). By definition $\theta_c=VX/P_x$, so the net biomass (VSS) produced — and hence wasted — per day is $$P_x=\frac{VX}{\theta_c}=\frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c}=\frac{0.7\times100{,}000\times275}{1+1.25}=\frac{1.925\times10^7\ \text{g/d}}{2.25}\approx8{,}556\ \text{kg/d}$$ ==**Sludge wasted P_x ≈ 8,560 kg/d (≈ 855.6 m³/d at Xw = 10,000 mg/L)**==
4. Sludge recycle ratio. A steady-state solids balance around the aeration tank/clarifier (influent VSS negligible, RAS drawn from the same underflow as WAS so both are at $X_w$) gives $Q_r X_w=(Q_0+Q_r)X$, so $$\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{5{,}000}{10{,}000-5{,}000}=1.0$$ ==**Recycle ratio Qr/Q0 = 1.0 (i.e. Qr = Q0 = 100,000 m³/d)**==