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23-Chem-B2 Environmental Engineering · December 2013

Question 4 of 7: pH control, ion exchange, reverse osmosis applications, 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 2013, 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 — membrane/condensation/adsorption control technologies, thermal 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 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), which govern effluent/emission limits and treatment-technology selection referenced throughout.

Question 4: pH control, ion exchange, reverse osmosis applications, 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) Application examples

(a) pH control. Neutralization of an acidic metal-finishing (electroplating) rinse-water discharge with lime (Ca(OH)₂) or caustic (NaOH) ahead of a settling basin, both to bring the effluent pH within the sewer-use bylaw range and to precipitate dissolved heavy metals as hydroxides for removal.

(b) Ion exchange. Cation-exchange (sodium-cycle) water softening of a municipal or industrial boiler-feed supply, exchanging influent Ca²⁺/Mg²⁺ hardness ions for Na⁺ on the resin to prevent scale formation in downstream boiler tubes and heat exchangers.

(c) Reverse osmosis. Desalination of brackish groundwater (or seawater) for potable water supply, or polishing of tertiary municipal effluent to produce high-purity reclaimed water for industrial reuse (e.g. semiconductor manufacturing) where conventional treatment cannot meet the required total-dissolved-solids specification.

Given.

QuantitySymbolValue
Plant flow$Q_0$$200{,}000\ \text{m}^3/\text{d}$
Influent BOD5/TSS$S_0$$250\ \text{mg/L}$
Effluent BOD5/TSS$S$$5\ \text{mg/L}$
Yield coefficient$Y$$0.8$
Endogenous decay rate$k_d$$0.05\ \text{d}^{-1}$
Average MLSS$X$$3{,}000\ \text{mg/L}$
Waste (underflow) MLSS$X_w$$10{,}000\ \text{mg/L}$
Mean cell residence time$\theta_c$$15\ \text{d}$

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

Aeration TankV, X=3000 mg/LSecondaryClarifierQ0=200,000 m3/dBOD5,0=250 mg/Lmixed liquorX=3000 mg/LQeBOD5=5 mg/LTSS=5 mg/LQr (RAS), Xw=10,000 mg/LQw (WAS)Xw=10,000 mg/L
Fig. 4: Conventional activated sludge process. Return activated sludge (RAS) recycles clarifier underflow (concentration $X_w$) back to the aeration tank inlet to maintain MLSS $X$; waste activated sludge (WAS) is removed from the same underflow to control the mean cell residence time $\theta_c$.

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 gives the daily solids production, equal to the mass that must leave via wasting; dividing by the wasting concentration ($X_w$) gives the wasting rate, and a clarifier solids balance gives the recycle ratio.

  1. Aeration tank volume. The $\theta_c$ design equation $1/\theta_c=\dfrac{YQ_0(S_0-S)}{VX}-k_d$ rearranges to $$V=\frac{\theta_c\,Y\,Q_0(S_0-S)}{X(1+k_d\theta_c)}=\frac{15\times0.8\times200{,}000\times245}{3{,}000\times(1+0.05\times15)}=\frac{5.88\times10^8}{5{,}250}$$ $$V\approx112{,}000\ \text{m}^3$$ ==**Aeration tank volume V ≈ 112,000 m³**==
  2. Hydraulic retention time. $$\tau=\frac{V}{Q_0}=\frac{112{,}000}{200{,}000}=0.56\ \text{d}=13.44\ \text{h}$$ ==**τ = 13.44 hours (0.56 d)**==
  3. Daily sludge production (mass wasted). By definition $\theta_c=VX/P_x$, so $$P_x=\frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c}=\frac{0.8\times200{,}000\times245}{1.75}=\frac{3.92\times10^7\ \text{g/d}}{1.75}=22{,}400{,}000\ \text{g/d}$$ ==**Sludge wasted Px ≈ 22,400 kg/d (≈ 2,240 m³/d at Xw = 10,000 mg/L)**==
  4. Sludge recycle ratio. A steady-state solids balance around the aeration tank/clarifier (RAS and WAS drawn from the same underflow, both at $X_w$) gives $Q_r X_w=(Q_0+Q_r)X$, so $$\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{3{,}000}{10{,}000-3{,}000}=0.4286$$ ==**Recycle ratio Qr/Q0 ≈ 0.429 (i.e. Qr ≈ 85,700 m³/d)**==
QuantityResult
Aeration tank volumeV ≈ 112,000 m³
Hydraulic retention timeτ = 13.44 h (0.56 d)
Sludge wasted daily≈ 22,400 kg/d (≈2,240 m³/d at Xw)
Recycle ratioQr/Q0 ≈ 0.429
Check — effluent solids in the SRT balance

The wasting and recycle-ratio equations above neglect the small mass of solids leaving in the clarified effluent (5 mg/L TSS × 200,000 m³/d = 1,000 kg/d, about 4.5% of the wasted mass) — a standard simplifying assumption when RAS/WAS share the same underflow concentration $X_w$, since $X_w$ then cancels out of the mass-wasted figure. A rigorous balance would use $\theta_c=VX/(Q_wX_w+Q_eX_e)$, marginally raising the required $Q_w$ to compensate for solids already leaving with the effluent.