pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge: 23-Chem-B2 Environmental Engineering, May 2016, Q4

Problem 4: pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge (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 / operation / maintenance table

Technology	Key design principle	Important operational issue	Critical maintenance condition
(a) pH control	Reagent (acid/base) selection and feed-control strategy (feedback trim with feed-forward on flow) sized for the buffering capacity and flow variability of the stream, with adequate mixing/contact time before the sensor.	pH-probe response is highly non-linear near neutral and very sensitive near the buffering capacity of the stream; controller tuning must avoid overshoot/hunting from a fast-responding but noisy signal.	Routine probe cleaning/re-calibration — electrode fouling (scale, oil, biofilm) is the single most common cause of pH control failure.
(b) Ion exchange	Resin type (cation/anion) and bed volume are sized from the target ion's influent concentration and the resin's exchange capacity (meq/L) to give an acceptable service run before breakthrough.	Continuously (or frequently) monitor effluent quality for early breakthrough, since capacity is consumed progressively and failure is a gradual quality slip rather than a sudden event.	Periodic regeneration (acid/caustic or brine) on a schedule tied to measured capacity, plus backwashing to prevent channeling and resin fouling/fracture from competing ions or particulates.
(c) Reverse osmosis	Pretreatment (cartridge filtration, antiscalant dosing) is designed around the feed's Silt Density Index and scaling potential so the membrane operates within its fouling/scaling limits at the chosen recovery ratio.	Track normalized permeate flow and salt rejection against baseline (not raw values) since feed temperature/pressure changes mask early fouling if compared to raw readings.	Scheduled clean-in-place (CIP) chemical cleaning when normalized flux/rejection drifts, and cartridge pre-filter replacement, to prevent irreversible membrane fouling/scaling.

(ii) Activated sludge aeration-tank design

This is the classic Lawrence–McCarty design procedure: the mean cell residence time (SRT, θc) is the design lever that, together with the kinetic coefficients Y and kd, fixes the required biomass inventory (V·X) for the given substrate removal, from which the tank volume, hydraulic retention time, sludge wasting rate and recycle ratio all follow.

Fig. 4 — Conventional activated-sludge process with RAS/WAS.

Given.

Quantity	Symbol	Value
Flow	Q0	100,000 m³/d
Influent BOD5	S0	250 mg/L
Effluent BOD5	S	5 mg/L
Yield coefficient	Y	0.6 kg VSS/kg BOD5
Decay rate	kd	0.03 d^-1
Aeration-tank MLSS	X	5,000 mg/L
Waste (RAS) MLSS	Xw	10,000 mg/L
Mean cell residence time	θc	15 d

Find. Aeration volume V, HRT θ, daily sludge wasting rate, and recycle ratio Qr/Q0.

Approach. Use the SRT design equation for V, divide by Q0 for θ, use the biomass production equation for the wasting rate, and close a solids balance across the aeration tank/clarifier for the recycle ratio.

(a) Aeration tank volume. $$V=\frac{Y\,\theta_c\,Q_0\,(S_0-S)}{X\,(1+k_d\theta_c)}=\frac{0.6(15)(100{,}000)(250-5)}{5000\,[1+0.03(15)]}=\frac{220{,}500{,}000}{7250}=\boxed{30{,}414\ \text{m}^3}$$
(b) Hydraulic retention time. $$\theta=\frac{V}{Q_0}=\frac{30{,}414}{100{,}000}=0.304\ \text{d} \times 24 = \boxed{7.30\ \text{h}}$$
(c) Daily sludge wasting rate. The net biomass (VSS) produced per day — which must equal what is wasted at steady state — follows directly from Y, the substrate removed, and the decay-corrected yield: $$P_x=\frac{Y\,Q_0\,(S_0-S)}{1+k_d\theta_c}=\frac{0.6(100{,}000)(245)}{1.45}=10{,}137{,}931\ \text{g/d}=\boxed{10{,}138\ \text{kg/d}}$$ The same figure follows from the SRT definition, VX/θc = 30,414 × 5.0 kg/m³ / 15 d = 10,138 kg/d. If the 5 mg/L effluent TSS lost over the clarifier weir is credited against it, Q_wX_w + (Q₀−Q_w)X_e = VX/θc gives Q_w ≈ 964 m³/d of waste sludge at 10,000 mg/L, i.e. about 9,643 kg/d through the WAS line and about 495 kg/d leaving in the effluent.
(d) Sludge recycle ratio. A steady-state solids balance around the aeration-tank inlet (negligible influent solids, mixing X with the returned RAS at Xw) gives $$\alpha=\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{5000}{10{,}000-5000}=\boxed{1.00}$$ i.e. the return activated-sludge flow equals the plant influent flow.

Quantity	Value
Aeration tank volume, V	30,414 m³
Hydraulic retention time, θ	7.30 h (0.304 d)
Sludge wasting rate, Px	10,138 kg VSS/d
Recycle ratio, Qr/Q0	1.00

Topic: pH/ion-exchange/RO operability, and Lawrence–McCarty activated-sludge design.

Key relations: $V=\dfrac{Y\theta_c Q_0(S_0-S)}{X(1+k_d\theta_c)}$, $P_x=\dfrac{YQ_0(S_0-S)}{1+k_d\theta_c}$, $\alpha=Q_r/Q_0=\dfrac{X}{X_w-X}$.

Why this works: Fixing θc (not θ) is what makes activated-sludge design different from a simple CSTR sizing problem: SRT decouples the solids (biomass) residence time from the liquid residence time via recycle, so a small tank (short θ) can still hold a old, well-acclimated sludge (long θc) — that decoupling is exactly what the recycle ratio calculation exposes.

Common pitfall: Using θc in place of θ in the V/Q0 step, or forgetting the (1+kdθc) decay correction in both the volume and the Px equations — omitting it overestimates both by ∼45% here.

Source: Metcalf & Eddy, Wastewater Engineering, 4th ed., Ch. 8 (activated sludge kinetics/design); Davis & Cornwell, Introduction to Environmental Engineering, 5th ed., Ch. 7 (ion exchange, RO, pH control).

Question 4 of 7: pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge

Problem 4: pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge (20 marks)

(i) Design / operation / maintenance table

(ii) Activated sludge aeration-tank design