NivaarExam PrepOfficial exam papers ↗

23-Chem-B2 Environmental Engineering · December 2016

Question 5 of 7: pH Control, Ion Exchange, RO and Extended-Aeration Activated Sludge

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

Notes on this paper

National Exam 04-Chem-B2, Environmental Engineering — December 2016. 3 hours, Closed-Book Exam with a candidate-prepared 8½×11" double-sided aid sheet. Any five (5) of the seven questions constitute a complete paper (100 marks); all seven are solved below for completeness.

Reference texts: Metcalf & Eddy (Tchobanoglous, Burton, Stensel), Wastewater Engineering: Treatment and Reuse, 4th ed.; Davis & Cornwell, Introduction to Environmental Engineering, 5th ed.; Turner, Workbook of Atmospheric Dispersion Estimates, 2nd ed.; Cooper & Alley, Air Pollution Control: A Design Approach, 4th ed.

Problem 5: pH Control, Ion Exchange, RO and Extended-Aeration 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) pH control, ion exchange and reverse osmosis — application examples

TechnologyApplication example
(a) pH control Lime or soda-ash addition raises the pH of a soft, corrosive surface-water supply ahead of distribution (corrosion-control/Langelier balancing) in drinking-water treatment; in wastewater, caustic dosing neutralizes an acidic industrial effluent before biological treatment, since nitrifiers are pH-sensitive below ∼6.5.
(b) Ion exchange A strong-acid cation resin in the sodium cycle softens hard surface water (exchanging Ca2+/Mg2+ for Na+) ahead of distribution; in wastewater, a selective resin removes trace heavy metals or nitrate from an industrial or agricultural discharge to meet a discharge limit.
(c) Reverse osmosis RO polishes a surface-water supply for a high-purity drinking-water application (or desalinates a brackish source) by rejecting dissolved salts across a semi-permeable membrane; in wastewater, RO is the final polishing step in an indirect/direct potable-reuse train, rejecting dissolved organics and salts from secondary/tertiary effluent.

(ii) Extended-aeration activated sludge 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. Extended aeration simply designs to a much longer θc (here 30 d, vs. 5–15 d for conventional activated sludge) to minimize net sludge production.

Aeration tankV, X=4,000 mg/LSecondaryclarifierQ0=200,000 m3/dS0=400 mg/LEffluentS=30 mg/LRAS: Qr, Xr=10,000 mg/LWAS: QwPx (kg/d)
Fig. 5 — Extended-aeration activated-sludge process with RAS/WAS.

Given.

QuantitySymbolValue
FlowQ0200,000 m³/d
Influent BOD5S0400 mg/L
Effluent BOD5S30 mg/L
Yield coefficientY0.4 kg VSS/kg BOD5
Decay ratekd0.05 d-1
Aeration-tank MLSSX4,000 mg/L
Waste (RAS) MLSSXw10,000 mg/L
Mean cell residence timeθc30 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.

  1. (a) Aeration tank volume and HRT. $$V=\frac{Y\,\theta_c\,Q_0\,(S_0-S)}{X\,(1+k_d\theta_c)}=\frac{0.4(30)(200{,}000)(400-30)}{4000\,[1+0.05(30)]}=\frac{888{,}000{,}000}{10{,}000}=\boxed{88{,}800\ \text{m}^3}$$ $$\theta=\frac{V}{Q_0}=\frac{88{,}800}{200{,}000}=0.444\ \text{d} \times 24 = \boxed{10.66\ \text{h}}$$
  2. (b) 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: $$P_x=\frac{Y\,Q_0\,(S_0-S)}{1+k_d\theta_c}=\frac{0.4(200{,}000)(370)}{2.5}=11{,}840{,}000\ \text{g/d}=\boxed{11{,}840\ \text{kg/d}}$$
  3. (c) 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 $$\alpha=\frac{Q_r}{Q_0}=\frac{X}{X_w-X}=\frac{4000}{10{,}000-4000}=\boxed{0.67}$$
QuantityValue
Aeration tank volume, V88,800 m³
Hydraulic retention time, θ10.66 h (0.444 d)
Sludge wasting rate, Px11,840 kg VSS/d
Recycle ratio, Qr/Q00.67