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23-Chem-B2 Environmental Engineering · December 2018

Question 5 of 7: Flotation, pH Control, Ion Exchange 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 2018. 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: Flotation, pH Control, Ion Exchange 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) Flotation, pH control and ion exchange — function and operational issue

TechnologyMain functionOperational issue
(a) Dissolved-air flotation (DAF) Removes low-density colloidal and suspended solids (algae, light floc) by attaching fine air bubbles (released from a pressurized-then-depressurized recycle stream) to the particles so they float to the surface as a skimmable froth — effective where the raw water's solids are too light or slow-settling for conventional sedimentation. Recycle ratio and saturator pressure must be tuned to produce a consistent fine bubble size (too coarse a bubble reduces attachment efficiency); over- or under-dosed coagulant ahead of the DAF leaves floc either too large (settles instead of floating) or too fragile (shears apart under the bubble contact).
(b) pH control (neutralization) Raises the low-pH raw water into the range where coagulants hydrolyze effectively and finished-water corrosivity/Langelier saturation index targets are met — typically lime, soda ash or caustic soda dosed ahead of (or with) the coagulant. Reagent feed-rate control against a fluctuating raw-water alkalinity/pH (feed-forward or feedback pH control loop) to avoid overshoot; overdosing lime can itself add hardness/turbidity that must then be removed downstream.
(c) Ion exchange Removes the ions responsible for high conductivity/hardness (Ca₂⁺, Mg₂⁺, and other dissolved cations) by exchanging them for Na⁺ (softening resin) or H⁺/OH⁻ (demineralization resin) on a fixed-bed resin. Resin exhaustion (breakthrough) must be tracked against the bed's exchange capacity and the raw water's total ionic load, and the resin regenerated (brine for softening; acid/caustic for demineralization) on a schedule that avoids either premature regeneration (wasted chemical) or breakthrough of untreated hard water.

(ii) Conventional activated sludge preliminary process design

Given.

QuantityValue
Flow, Q₀200,000 m³/d
Influent BOD₅, S₀240 mg/L
Effluent BOD₅, S25 mg/L
Yield coefficient, Y0.4
Decay rate, kd0.05 d⁻¹
MLSS, X6,000 mg/L
Waste MLSS, Xw9,000 mg/L
Mean cell residence time, θc12 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. This is the classic Lawrence–McCarty design-SRT formulation: the biomass mass balance over the whole system (aeration tank + clarifier, at steady state and a fixed target SRT) directly gives the required tank volume, the net biomass production sets the daily waste-sludge mass, and a solids mass balance around the aeration-tank mixing point gives the return/recycle ratio.

  1. Aeration tank volume from the design-SRT relation.
    $$ V = \frac{Y\,\theta_c\,Q_0\,(S_0-S)}{X\,(1+k_d\,\theta_c)} = \frac{0.4\times12\times200{,}000\times(240-25)}{6{,}000\times(1+0.05\times12)} $$
    $$ V = \frac{206{,}400{,}000}{9{,}600} = \boxed{21{,}500\ \text{m}^3} $$
  2. Hydraulic retention time.
    $$ \theta = \frac{V}{Q_0} = \frac{21{,}500}{200{,}000} = 0.1075\ \text{d} = \boxed{2.58\ \text{h}} $$
  3. Sludge wasted daily from the net biomass yield.
    $$ P_x = \frac{Y\,Q_0\,(S_0-S)}{1+k_d\,\theta_c} = \frac{0.4\times200{,}000\times(240-25)}{1.6\times1000} = \boxed{10{,}750\ \text{kg/d}} $$

    This uses the usual preliminary-design convention that effluent suspended solids are negligible, so the whole net biomass production (Px = VX/θc = 21,500×6,000/12 g/d = 10,750 kg/d) leaves as waste sludge — a waste flow of about 10,750/9.0 ≈ 1,194 m³/d at Xw = 9,000 mg/L. If the stated 25 mg/L effluent TSS is instead counted as solids leaving the system, θc = VX/(QwXw + QeXe) and the waste stream carries 10,750 − 200,000×25/1000 ≈ 5,750 kg/d (state whichever convention is used).

  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:
    $$ \alpha = \frac{Q_r}{Q_0} = \frac{X}{X_w-X} = \frac{6{,}000}{9{,}000-6{,}000} = \boxed{2.00} $$
QuantityValue
Aeration tank volume, V21,500 m³
Hydraulic retention time, θ2.58 h
Sludge wasted daily, Px10,750 kg/d
Recycle ratio, Qr/Q₀2.00