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23-Chem-B2 Environmental Engineering · December 2015

Question 5 of 7: pH control, ion exchange and reverse osmosis mechanisms, 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, December 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; 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 — cyclones, scrubbers, fabric filtration, electrostatic precipitation, 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 5: pH control, ion exchange and reverse osmosis mechanisms, 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) How pH control, ion exchange and reverse osmosis work

TechnologyHow it works
(a) pH controlAn acid (e.g. H₂SO₄) or base (e.g. lime, NaOH) reagent is metered into the flow to shift the hydrogen-ion activity toward a target set-point, following the stream's titration/buffering curve rather than a simple stoichiometric neutralization — effective dosing therefore requires knowing how strongly buffered the specific stream is, since a poorly buffered stream can swing rapidly past the set-point on a small dose change.
(b) Ion exchangeWater is passed through a bed of resin beads carrying fixed, exchangeable ionic sites (e.g. Na⁺ on a strong-acid cation resin); target ions in solution (e.g. Ca²⁺, Mg²⁺) exchange onto the resin in preference to the resin's mobile ion according to the resin's selectivity series, removing them from the water stream until the resin's exchange capacity is exhausted and must be regenerated with a concentrated brine solution.
(c) Reverse osmosisFeed water is pressurized above its osmotic pressure and forced through a semi-permeable membrane; water molecules pass through while dissolved ions/larger molecules are rejected, splitting the feed into a low-TDS permeate and a concentrated reject (brine) stream — the applied pressure must exceed the feed osmotic pressure by a sufficient net driving pressure for the target flux and recovery.

Given.

QuantitySymbolValue
Influent flow$Q_0$$100{,}000\ \text{m}^3/\text{d}$
Influent BOD₅/TSS$S_0$$300\ \text{mg/L}$
Effluent BOD₅/TSS$S$$10\ \text{mg/L}$
Yield coefficient$Y$$0.5$
Decay rate$k_d$$0.04\ \text{d}^{-1}$
Mean cell residence time$\theta_c$$25\ \text{d}$
Mixed-liquor suspended solids$X$$3{,}000\ \text{mg/L}$
Waste (recycle) MLSS$X_w$$9{,}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. $$V=\frac{\theta_c\,Y\,Q_0(S_0-S)}{X(1+k_d\theta_c)}=\frac{25\times0.5\times100{,}000\times(300-10)}{3{,}000\times(1+0.04\times25)}=60{,}416.7\ \text{m}^3$$ ==**V ≈ 60,417 m³**==
  2. Hydraulic retention time. $$\tau=\frac{V}{Q_0}\times24=\frac{60{,}416.7}{100{,}000}\times24=14.5\ \text{h}$$ ==**τ = 14.5 h**==
  3. Daily sludge mass wasted. $$Q_w=P_x=\frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c}=\frac{0.5\times100{,}000\times290}{2.0}=7{,}250\ \text{kg/d}$$ ==**Qw = 7,250 kg/d**==
  4. Cross-check via the SRT identity. Independently, $Q_w=VX/\theta_c=(60{,}416.7\times3{,}000\times10^{-3})/25=7{,}250\ \text{kg/d}$ — 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 $$\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{3{,}000}{9{,}000-3{,}000}=0.5$$ ==**Qr/Q0 = 0.5**==
QuantityResult
Aeration tank volume60,417 m³
Hydraulic retention time14.5 h
Sludge wasted daily7,250 kg/d
Recycle ratio $Q_r/Q_0$0.5
Aeration TankSecondary ClarifierQ0=100,000 m3/dBOD5=300 mg/Lmixed liquorX=3,000 mg/LEffluentBOD5=10 mg/LRAS QrWAS Xw=9,000 mg/L
Fig. 4: 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.