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23-Chem-B2 Environmental Engineering · December 2019

Question 4 of 7: pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge Design

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

Notes on this paper

National Exam 16-Chem-B2, Environmental Engineering — December 2019. 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 4: pH Control, Ion Exchange, Reverse Osmosis and Activated Sludge 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) pH control, ion exchange and reverse osmosis — example and design parameter

TechnologyExample applicationKey process design parameter
(a) pH controlNeutralizing low-pH industrial (e.g. metal-finishing) wastewater with lime or caustic soda before biological treatment. Reagent dose set from the wastewater's titration (buffering) curve, not just its raw pH, because alkalinity/acidity determines how much reagent is actually needed to shift the pH into the target range.
(b) Ion exchangeSoftening a hard groundwater supply by exchanging Ca₂⁺/Mg₂⁺ for Na⁺ on a strong-acid cation resin. Resin exchange capacity (equivalents of hardness removable per unit resin volume before breakthrough), which sets the bed volume and regeneration (brine) frequency for a given raw-water hardness and flow.
(c) Reverse osmosisDesalinating brackish groundwater or polishing tertiary wastewater effluent for potable reuse. Operating pressure relative to the feedwater's osmotic pressure — the applied pressure must exceed the osmotic pressure of the (increasingly concentrated) reject stream by enough net driving pressure to sustain the target permeate flux across the membrane.

(ii) Activated sludge preliminary process design

Given.

QuantityValue
Flow, Q₀200,000 m³/d
Influent BOD₅, S₀250 mg/L
Effluent BOD₅, S10 mg/L
Yield coefficient, Y0.4
Decay rate, kd0.05 d⁻¹
MLSS, X5,000 mg/L
Waste MLSS, Xw12,000 mg/L
Mean cell residence time, θc8 d

Find. (a) Aeration tank volume V (m³) and HRT θ (h). (b) Sludge wasted daily Qw (kg/d). (c) Recycle ratio Qr/Q₀.

Approach. Classic Lawrence–McCarty design-SRT formulation: a biomass steady-state mass balance over the aeration tank + clarifier at the fixed target SRT gives the tank volume directly, the net biomass production sets the daily wasted-sludge mass, and a solids balance at the aeration-tank inlet gives the recycle ratio.

  1. Aeration tank volume from the design-SRT relation.
    $$ V = \frac{Y\,\theta_c\,Q_0\,(S_0-S)}{X\,(1+k_d\,\theta_c)} = \frac{0.4\times8\times200{,}000\times(250-10)}{5{,}000\times(1+0.05\times8)} $$
    $$ V = \frac{153{,}600{,}000}{7{,}000} = \boxed{21{,}943\ \text{m}^3} $$
  2. Hydraulic retention time.
    $$ \theta = \frac{V}{Q_0} = \frac{21{,}943}{200{,}000} = 0.1097\ \text{d} = \boxed{2.63\ \text{h}} $$
  3. Sludge wasted daily from the net biomass yield.
    $$ P_x = \frac{Y\,Q_0\,(S_0-S)}{1+k_d\,\theta_c} = \frac{0.4\times200{,}000\times(250-10)}{1.4\times1000} = \boxed{13{,}714\ \text{kg/d}} $$
  4. Recycle ratio from the aeration-tank solids balance. A steady-state solids balance around the aeration-tank inlet (Q₀X₀≈0, return-line concentration Xr=Xw) gives (Q₀+Qr)X = QrXw, which rearranges to:
    $$ \alpha = \frac{Q_r}{Q_0} = \frac{X}{X_w-X} = \frac{5{,}000}{12{,}000-5{,}000} = \boxed{0.71} $$
QuantityValue
Aeration tank volume, V21,943 m³
Hydraulic retention time, θ2.63 h
Sludge wasted daily, Px13,714 kg/d
Recycle ratio, Qr/Q₀0.71