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23-Chem-B2 Environmental Engineering · May 2014

Question 4 of 7: pH control, ion exchange, 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 2014, 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, nutrient removal, activated-sludge design, sedimentation 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, air quality modelling; C. D. Cooper & F. C. Alley, Air Pollution Control: A Design Approach — particulate/gas/vapour control, thermal/catalytic oxidation, 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 Guidelines for Canadian Drinking Water Quality (Health Canada), the Canadian Council of Ministers of the Environment (CCME) Municipal Wastewater Effluent guidelines, and provincial air/water permitting practice (e.g. BC Environmental Management Act and Metro Vancouver air-quality bylaws).

Question 4: pH control, ion exchange, 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 for pH control, ion exchange, and reverse osmosis

(a) pH control. Key design principle: reagent selection and dose sizing must follow the stream's actual titration curve, not a linear dose assumption, because pH is a logarithmic (non-linear) response to added acid/base that is steepest near the neutral endpoint — multi-stage dosing/mixing is often used to flatten that steep region across more than one control loop. O&M parameter: continuous online pH-probe calibration/verification, since a drifted or fouled probe silently defeats the control loop (the process appears “in control” on the readout while actually over- or under-dosing).

(b) Ion exchange. Key design principle: resin exchange capacity (eq/L resin) is sized against the specific ionic load (hardness, target ion) and the desired service-run length between regenerations, since undersizing the resin bed relative to the influent ionic strength causes premature breakthrough. O&M parameter: regeneration frequency and reagent (acid/caustic/brine) dose, tracked against effluent quality (e.g. hardness or target-ion breakthrough monitoring) so regeneration is triggered by actual exhaustion rather than a fixed calendar schedule that can either waste reagent or risk breakthrough.

(c) Reverse osmosis. Key design principle: the design recovery (permeate/feed ratio) is limited by the concentration factor at which the least-soluble scale-forming salt (commonly CaCO₃ or CaSO₄) reaches saturation in the reject stream, not by membrane hydraulics alone — antiscalant dosing and recovery must be co-designed. O&M parameter: trans-membrane pressure and normalized permeate flux, trended over time to detect fouling/scaling before it causes irreversible membrane damage, triggering a cleaning-in-place (CIP) cycle while it is still reversible.

Given.

QuantitySymbolValue
Plant 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$
Endogenous decay rate$k_d$$0.04\ \text{d}^{-1}$
Average MLSS$X$$4{,}000\ \text{mg/L}$
Waste (return-line) MLSS$X_w$$12{,}000\ \text{mg/L}$
Mean cell residence time$\theta_c$$20\ \text{d}$

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

Aeration TankV, X=4,000 mg/LSecondaryClarifierQ0=500,000 m3/dBOD5,0=200 mg/Lmixed liquorX=4,000 mg/LQeBOD5=2 mg/LTSS=2 mg/LQw (WAS)Xw=12,000 mg/LQr (RAS), Xw=12,000 mg/L
Fig. 4: Conventional activated sludge process for the 500,000 m³/d plant. Return activated sludge (RAS) recycles clarifier underflow (concentration $X_w$) to the aeration tank inlet to maintain MLSS $X$; waste activated sludge (WAS) is removed from the same underflow at $X_w$ to control the mean cell residence time $\theta_c$.

Approach. The $\theta_c$ design equation links the biomass yield/decay kinetics and the aeration tank's MLSS inventory to give the required tank volume; the same $\theta_c$ definition (solids inventory divided by mass wasted per day) gives the daily sludge production, which is wasted at concentration $X_w$; the recycle ratio then follows from a steady-state clarifier solids balance.

  1. Aeration tank volume. The mean-cell-residence-time design equation, $1/\theta_c=\dfrac{YQ_0(S_0-S)}{VX}-k_d$, rearranges to $$V=\frac{\theta_c\,Y\,Q_0(S_0-S)}{X(1+k_d\theta_c)}=\frac{20\times0.7\times500{,}000\times198}{4{,}000\times(1+0.04\times20)}=\frac{1.386\times10^9}{7{,}200}$$ $$V=192{,}500\ \text{m}^3$$ ==**Aeration tank volume V = 192,500 m³**==
  2. Hydraulic retention time. $$\tau=\frac{V}{Q_0}=\frac{192{,}500}{500{,}000}=0.385\ \text{d}=9.24\ \text{h}$$ ==**τ = 9.24 hours**==
  3. Daily sludge wasted. By definition $\theta_c=VX/P_x$, so $$P_x=\frac{VX}{\theta_c}=\frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c}=\frac{0.7\times500{,}000\times198}{1+0.8}=\frac{6.93\times10^7\ \text{g/d}}{1.8}=38{,}500{,}000\ \text{g/d}$$ $$P_x=38{,}500\ \text{kg/d}$$ ==**Sludge wasted Qw ≈ 38,500 kg/d (≈ 3,208 m³/d at Xw = 12,000 mg/L)**==
  4. Sludge recycle ratio. A steady-state solids balance around the aeration tank/clarifier (RAS and WAS both drawn from the same underflow at $X_w$, influent VSS negligible) gives $Q_rX_w=(Q_0+Q_r)X$, so $$\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{4{,}000}{12{,}000-4{,}000}=0.5$$ ==**Recycle ratio Qr/Q0 = 0.5 (Qr = 250,000 m³/d)**==
QuantityResult
Aeration tank volumeV = 192,500 m³
Hydraulic retention timeτ = 9.24 h (0.385 d)
Sludge wasted dailyQw ≈ 38,500 kg/d (≈3,208 m³/d at Xw)
Recycle ratioQr/Q0 = 0.5
Check — effluent solids in the SRT balance

The Steps above neglect the small mass of solids leaving in the 2 mg/L TSS effluent (2 mg/L × 500,000 m³/d = 1,000 kg/d, about 2.6% of the wasted mass) — a standard simplification when RAS/WAS share the concentrated underflow at $X_w$ (which cancels out of the mass-wasted figure regardless). A rigorous balance, $\theta_c=VX/(Q_wX_w+Q_eX_e)$, would raise the required wasting rate marginally to compensate for solids already leaving with the effluent.