NivaarExam PrepOfficial exam papers ↗

23-Chem-B2 Environmental Engineering · May 2018

Question 5 of 7: pH Control / Ion Exchange / RO 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 — May 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: pH Control / Ion Exchange / RO 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: function and operational issues

Selecting the low-pH ground-aquifer drinking-water example (acidic, corrosive groundwater, e.g. pH 5.5–6.0):

TechnologyMain functionOperational issue
(a) pH controlRaises the finished-water pH into the non-corrosive/non-scaling Langelier range (typically pH 7.5–8.5) by feeding a base (lime, soda ash, or caustic) into the raw or finished water, protecting the distribution system from corrosion (lead/copper leaching) and reducing coagulant/disinfectant demand shifts caused by low pH.Overdosing risks excessive scaling (calcium carbonate deposition) and turbidity from precipitated hydroxides; the dosing system must be flow-paced and continuously monitored since demand tracks raw-water alkalinity, which can vary seasonally.
(b) Ion exchangeExchanges the aquifer's hardness/metal ions (Ca²⁺, Mg²⁺, Fe/Mn) for Na⁺ on a resin bed, indirectly improving the finished water's aesthetic and corrosivity properties without necessarily changing pH directly.Resin fouling by iron/manganese precipitates or organic matter progressively reduces exchange capacity, requiring periodic backwash and acid/brine regeneration and generating a saline regenerant waste stream that needs disposal.
(c) Reverse osmosisRemoves dissolved ions (hardness, metals, TDS) across a semi-permeable membrane under applied pressure, producing a high-purity permeate. Dissolved CO2 (the usual cause of low groundwater pH) passes through the membrane, so the permeate stays acidic and has almost no alkalinity; it must be degassed and re-mineralized/pH-adjusted afterward for distribution stability. Membrane scaling/fouling from the concentrated reject stream requires antiscalant dosing and periodic chemical cleaning, and the reject (concentrate) brine stream requires its own disposal — RO is energy-intensive relative to pH control or ion exchange alone.

(ii) Conventional activated sludge preliminary design (Lawrence–McCarty)

Given.

QuantityValue
Influent flow, Q0100,000 m³/d
Influent BOD5/TSS, S0300 mg/L
Effluent BOD5/TSS, S40 mg/L
Yield coefficient, Y0.5 kg VSS/kg BOD5
Decay rate, kd0.04 d−1
Aeration-tank MLSS, X5,000 mg/L
Waste (clarifier underflow) MLSS, Xw8,000 mg/L
Mean cell residence time, θc8 days

Find. (a) Aeration tank volume V and hydraulic retention time θ; (b) daily sludge wasting rate Px (kg/d); (c) sludge recycle ratio Qr/Q0.

Approach. Apply the standard Lawrence–McCarty design equations for a completely-mixed activated-sludge system relating θc (solids retention time) to reactor volume and biomass production, then close a steady-state solids balance around the aeration tank/clarifier for the recycle ratio.

  1. (a) Aeration tank volume, from the θc design equation. $$V = \frac{Y\,\theta_c\,Q_0(S_0-S)}{X(1+k_d\theta_c)} = \frac{0.5\times8\times100{,}000\times(300-40)}{5{,}000\times(1+0.04\times8)}$$ $$V = \frac{104{,}000{,}000}{6{,}600} = \boxed{15{,}758\ \text{m}^3}$$
  2. Hydraulic retention time. $$\theta = \frac{V}{Q_0} = \frac{15{,}758}{100{,}000} = 0.1576\ \text{d} = \boxed{3.78\ \text{h}}$$
  3. (b) Daily sludge wasting rate. $$P_x = \frac{Y\,Q_0(S_0-S)}{1+k_d\theta_c} = \frac{0.5\times100{,}000\times260}{1.32} = \boxed{9{,}848\ \text{kg/d}}$$ This is the net solids production that must leave the system each day (Px = VX/θc), the standard preliminary-design answer when effluent solids are neglected. As a refinement, if the stated effluent TSS of 40 mg/L is counted as leaving over the weir (about (Q0−Qw) × 0.040 ≈ 4,000 kg/d), the balance QwXw + (Q0−Qw)Xe = 9,848 kg/d gives Qw ≈ 735 m³/d, so the sludge actually wasted from the clarifier underflow is about 5,880 kg/d.
  4. (c) Sludge recycle ratio, from a steady-state solids balance around the aeration tank. Combining the influent (negligible X0) and recycle streams to give the aeration-tank MLSS X: $$Q_r X_w = (Q_0+Q_r)X \ \Rightarrow\ \alpha=\frac{Q_r}{Q_0}=\frac{X}{X_w-X} = \frac{5{,}000}{8{,}000-5{,}000} = \boxed{1.67}$$
QuantityValue
(a) Aeration tank volume, V15,758 m³
(a) Hydraulic retention time, θ3.78 h
(b) Daily sludge wasted, Px9,848 kg/d
(c) Recycle ratio, Qr/Q01.67