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23-Chem-B2 Environmental Engineering · May 2015

Question 4 of 7: pH control, ion exchange and reverse osmosis design principles, and activated-sludge process design

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. EGBC 04-Chem-B2 Environmental Engineering, May 2015, 3 hours, closed-book with a candidate-prepared double-sided 8½×11-inch aid sheet. Seven problems, each worth 20 marks; candidates attempt any five, and only the first five answers in the workbook are marked. All seven problems are solved below as a complete study resource.

Reference texts: G. Tchobanoglous, F. L. Burton & H. D. Stensel (Metcalf & Eddy), Wastewater Engineering: Treatment and Reuse (4th ed., McGraw-Hill) — BOD kinetics, dissolved air flotation, activated-sludge design, nutrient removal; M. L. Davis & D. A. Cornwell, Introduction to Environmental Engineering (5th ed., McGraw-Hill) — drinking-water treatment, air pollution control, ion exchange, reverse osmosis, soil remediation; C. D. Cooper & F. C. Alley, Air Pollution Control: A Design Approach — fabric filtration, thermal oxidation, adsorption, odour control; S. P. Turner, Workbook of Atmospheric Dispersion Estimates (2nd ed., CRC Press) — the Gaussian plume model and Pasquill–Gifford stability classes. Canadian context follows the Canadian Environmental Protection Act (CEPA 1999), the Canadian Council of Ministers of the Environment (CCME) Municipal Wastewater Effluent and Drinking Water Quality guidelines, and provincial air/water permitting practice (e.g. BC Environmental Management Act, Metro Vancouver air-quality bylaws).

Question 4: pH control, ion exchange and reverse osmosis 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 principle and O&M parameter per technology

TechnologyKey design principleImportant O&M parameter
(a) pH controlThe reagent dose (lime, caustic, or acid) must be sized against the influent's actual titration/buffering curve, not a naive stoichiometric target, since a poorly buffered stream can swing far past a pH set-point on a small overdose.Continuous in-line pH monitoring feeding a closed-loop dosing-pump control (not a static, un-monitored dose), since influent pH/flow can vary faster than a manual adjustment can track.
(b) Ion exchangeResin exchange capacity and selectivity must be matched to the target ion against the competing-ion background (e.g. strong-acid cation resin affinity Ca²⁺>Mg²⁺>Na⁺), since a high concentration of a competing higher-affinity ion prematurely exhausts capacity for the target.Defined breakthrough monitoring (effluent hardness/target-ion trigger) governing the regeneration cycle (brine dose, frequency), since exchange efficiency falls sharply once breakthrough begins.
(c) Reverse osmosisApplied pressure must exceed the feed osmotic pressure by a sufficient net driving pressure for the target flux/recovery; higher-TDS feed requires correspondingly higher operating pressure and membrane/vessel rating.Pretreatment performance monitoring (fine filtration, antiscalant dose) and periodic membrane cleaning-in-place, since fouling/scaling of the concentrating reject stream is the dominant driver of RO operating cost and membrane life.

Given.

QuantitySymbolValue
Influent flow$Q_0$$500{,}000\ \text{m}^3/\text{d}$
Influent BOD₅/TSS$S_0$$200\ \text{mg/L}$
Effluent BOD₅/TSS$S$$2\ \text{mg/L}$
Yield coefficient$Y$$0.7$
Decay rate$k_d$$0.04\ \text{d}^{-1}$
Mean cell residence time$\theta_c$$20\ \text{d}$
Mixed-liquor suspended solids$X$$4{,}000\ \text{mg/L}$
Waste (recycle) MLSS$X_w$$12{,}000\ \text{mg/L}$

Find. (a) Aeration tank volume $V$ and HRT $\tau$. (b) Daily sludge mass wasted $Q_w$ (kg/d). (c) Recycle ratio $Q_r/Q_0$.

Approach. The Lawrence–McCarty SRT-based design equation sizes the aeration tank directly from $\theta_c$, $Y$, $k_d$ and the substrate removed; sludge production follows from the same substrate balance, and the recycle ratio comes from a pure solids balance around the clarifier, independent of the biological kinetics.

  1. Aeration tank volume. $$V=\frac{\theta_c\,Y\,Q_0(S_0-S)}{X(1+k_d\theta_c)}=\frac{20\times0.7\times500{,}000\times(200-2)}{4{,}000\times(1+0.04\times20)}=192{,}500\ \text{m}^3$$ ==**V = 192,500 m³**==
  2. Hydraulic retention time. $$\tau=\frac{V}{Q_0}\times24=\frac{192{,}500}{500{,}000}\times24=9.24\ \text{h}$$ ==**τ = 9.24 h**==
  3. Daily sludge mass wasted. $$Q_w=P_x=\frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c}=\frac{0.7\times500{,}000\times198}{1.8}=38{,}500\ \text{kg/d}$$ ==**Qw = 38,500 kg/d**==
  4. Cross-check via the SRT identity. Independently, $Q_w=VX/\theta_c=(192{,}500\times4{,}000\times10^{-3})/20=38{,}500\ \text{kg/d}$ — the two routes agree exactly, confirming $V$ and $Q_w$ are internally consistent.
  5. Sludge recycle ratio. A solids balance around the clarifier alone (no biology, no $\theta_c$) gives $$\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{4{,}000}{12{,}000-4{,}000}=0.5$$ ==**Qr/Q0 = 0.5**==
QuantityResult
Aeration tank volume192,500 m³
Hydraulic retention time9.24 h
Sludge wasted daily38,500 kg/d
Recycle ratio $Q_r/Q_0$0.5
Aeration TankSecondary ClarifierQ0=500,000 m3/dBOD5=200 mg/Lmixed liquorX=4,000 mg/LEffluentBOD5=2 mg/LRAS QrWAS Xw=12,000 mg/L
Fig. 3: Conventional activated-sludge process flow — aeration tank → secondary clarifier, with return activated sludge (RAS) recycled to the aeration tank inlet and waste activated sludge (WAS) removed from the RAS line to solids handling.
Check — recycle-ratio independence from θ𝑐

The recycle ratio $Q_r/Q_0=X/(X_w-X)$ follows purely from a steady-state solids mass balance across the clarifier/recycle split and does not depend on $\theta_c$, $Y$ or $k_d$ — a useful self-check, since an answer that appears to need the biokinetic parameters for part (c) signals an error.