16-Civ-A3 Elementary Environmental Engineering · December 2014

Question 2 of 7: Particle Characteristics, Solution Chemistry and Gases

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

Notes on this paper

Paper format. National Exams, December 2014 — 98-Civ-A3 Environmental Engineering. Three hours, closed book with one candidate-prepared double-sided aid sheet. Seven problems, each worth 20 marks; any five constitute a complete paper (maximum 100 marks), and only the first five answers in the work book are marked. All seven problems are solved below, because the set is a study resource rather than an exam attempt.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering; Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design; Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery; Crittenden et al. (MWH), Water Treatment: Principles and Design; CCME Canadian Environmental Quality Guidelines; Impact Assessment Agency of Canada, Impact Assessment Act guidance.

Check: two source inconsistencies are carried through deliberately. (1) Problem 1(i) prints the dipropylene glycol formula as C₆H₁₄O₂ (118.2 g/mol); the actual compound is C₆H₁₄O₃ (134.2 g/mol). (2) The same sentence states the dose as “76 kg (1000 mol)”, which implies a molar mass of 76 g/mol and matches neither formula — 1000 mol of the real compound is 134 kg. The mole quantity is the load-bearing datum for a closed-system balance, so 1000 mol is adopted and both molar masses are reported where a mass concentration is asked for. NOTE 1 on page 1 expressly invites this kind of stated assumption.

Question 2: Particle Characteristics, Solution Chemistry and Gases (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.

2 (i) — Removing colloids, and why particle counters help (9 marks)

Colloids are the particles between roughly 0.001 and 1 µm that plain sedimentation cannot capture. Stokes' law explains why: settling velocity scales with the square of diameter, so a 1 µm clay platelet settles at a few centimetres per day and would need a clarifier the size of a small lake. Worse, these particles carry a net negative surface charge in natural waters, which produces a repulsive electrical double layer that keeps them permanently dispersed. Two engineering principles are therefore applied in series.

Principle 1 — destabilisation by coagulation (charge neutralisation and sweep floc). A hydrolysing metal salt, typically alum or a ferric or polyaluminium chloride coagulant, is dosed into a high-intensity rapid-mix zone with a velocity gradient of order 600–1000 s⁻¹ and a detention time of seconds. The hydrolysis products are positively charged and adsorb onto the colloid surface, compressing the double layer and driving the zeta potential toward zero so that van der Waals attraction can dominate at close approach. At higher doses the metal hydroxide precipitates as an amorphous floc that physically enmeshes colloids — the sweep-floc mechanism that most Canadian surface-water plants actually operate in. The engineering variables are dose, mixing intensity and pH, which must be held in the narrow window where the hydroxide is least soluble (about pH 6.0–6.8 for alum).

Principle 2 — aggregation by flocculation, and transport-limited capture in the filter. Destabilised particles must still be brought into contact. Gentle tapered mixing at $G \approx 70$ down to 10 s⁻¹ over 20–30 minutes promotes orthokinetic flocculation, in which the collision rate scales with $G$ and with the floc volume fraction, growing particles until they are large and dense enough to settle. Whatever survives the clarifier is then captured in a granular filter by depth filtration — interception, sedimentation onto the grain surface and Brownian diffusion for the finest fraction — where attachment is only possible because the coagulation step already destabilised the particles. This is why filters immediately lose performance if coagulant feed fails, even though the media are unchanged.

Why particle counters improve on turbidity. A turbidimeter measures scattered light and reports one aggregate number in NTU. Because scattering intensity depends strongly on particle size, a turbidity reading is dominated by the abundant particles near 1 µm and is nearly blind to the small number of large particles that matter most for public health. A filtered water can sit comfortably at 0.05 NTU while passing tens of Cryptosporidium-sized (4–6 µm) oocysts per litre during a filter-ripening or end-of-run breakthrough event. An online particle counter reports the actual count in discrete size channels, so it resolves the 3–10 µm fraction directly, detects breakthrough one to two orders of magnitude more sensitively than turbidity, and gives operators an early warning while the turbidity trace is still flat. That sensitivity is the reason particle counting is used to demonstrate log-removal credit and to optimise filter-to-waste duration.

2 (ii) — Hardness as CaCO₃ (6 marks)

Given. Average divalent cation concentrations in a lake near a rock quarry, together with the atomic weights the question fixes.

Given data — Problem 2(ii)

Ion	Concentration	Atomic weight	Equivalent weight
Ca2+	100 mg/L	40	20
Mg2+	80 mg/L	24	12
Fe2+	30 mg/L	56	28
CaCO3 reference	M = 40 + 12 + 3(16) = 100		50

Find. Total hardness expressed in mg/L as CaCO₃, and the resulting classification of the water.

Approach. Convert each divalent cation to the common calcium-carbonate basis through equivalent weights, sum the contributions, and compare the total against the conventional hardness bands.

1. Establish the calcium carbonate reference. Hardness is conventionally reported as an equivalent CaCO₃ concentration so that different cations can be added. From the given atomic weights $M_{\mathrm{CaCO_3}} = 100$ and, since carbonate is divalent, $$EW_{\mathrm{CaCO_3}} = \frac{100}{2} = 50\ \text{mg/meq}$$
2. Write the conversion on an equivalents basis. Equivalents, not moles, are what must be conserved when re-expressing one species as another: $$C_{\text{as CaCO}_3} = C_{\text{ion}} \times \frac{EW_{\mathrm{CaCO_3}}}{EW_{\text{ion}}}, \qquad EW_{\text{ion}} = \frac{\text{atomic weight}}{\text{charge}}$$
3. Convert calcium. With $EW_{\mathrm{Ca}} = 40/2 = 20$, $$C_{\mathrm{Ca}} = 100 \times \frac{50}{20} = 250\ \text{mg/L as CaCO}_3$$
4. Convert magnesium. With $EW_{\mathrm{Mg}} = 24/2 = 12$, $$C_{\mathrm{Mg}} = 80 \times \frac{50}{12} = 333.3\ \text{mg/L as CaCO}_3$$ Note that magnesium out-contributes calcium here despite the lower reported mg/L, because its equivalent weight is much smaller. Reading the raw mg/L figures and assuming calcium dominates is the standard error on this question.
5. Convert ferrous iron. With $EW_{\mathrm{Fe}} = 56/2 = 28$, $$C_{\mathrm{Fe}} = 30 \times \frac{50}{28} = 53.6\ \text{mg/L as CaCO}_3$$
6. Sum and report both the conventional and the strict total. Hardness is conventionally taken as calcium plus magnesium, since these dominate almost every natural water: $$\mathrm{TH_{Ca+Mg}} = 250 + 333.3 = \boxed{583\ \text{mg/L as CaCO}_3}$$ Ferrous iron is a genuine divalent cation and strictly does contribute, giving $$\mathrm{TH_{total}} = 583.3 + 53.6 = 637\ \text{mg/L as CaCO}_3$$
7. Classify. On the three-band scale the question offers — soft below 75, moderately hard 75 to 150, hard above 150 mg/L as CaCO₃ — this water is emphatically hard. On the four-band scale in common use it exceeds the 300 mg/L threshold and is classified very hard, at roughly four times the 150 mg/L level at which softening normally becomes economic.

The result is entirely consistent with the setting the question describes. A lake receiving drainage from a rock quarry, particularly one working limestone or dolomite, dissolves carbonate minerals and picks up both calcium and magnesium; the elevated ferrous iron points to reducing conditions in the quarry sumps or in the lake sediments. At 583 mg/L this water would require lime-soda or ion-exchange softening before municipal use, and the iron would require oxidation and filtration to avoid staining and distribution-system deposits.

Final results — Problem 2(ii)

Contribution	mg/L as CaCO3
Calcium (Ca2+)	250
Magnesium (Mg2+)	333
Ferrous iron (Fe2+)	53.6
Total hardness, Ca + Mg (conventional)	583
Total hardness including Fe (strict)	637
Classification	Hard (very hard on the four-band scale)

2 (iii) — Two toxic gases from anaerobic sewers (5 marks)

Anaerobic sewer gases — impact and control

Gas	Origin	Environmental impact	Engineering control
Hydrogen sulphide, H2S	Sulphate-reducing bacteria in the slime layer of septic force mains and wet wells	Oxidised to sulphuric acid by Thiobacillus on damp crown surfaces, causing severe concrete corrosion of sewers and structures; acutely toxic and odorous, and it deadens the sense of smell above about 100 ppm so that workers lose their warning	Control at source by chemical dosing — nitrate to suppress sulphate reduction, or iron salts to precipitate sulphide — combined with forced ventilation of the wet well and treatment of the extracted air in a biofilter or chemical scrubber; reduce force-main detention time and avoid turbulent drops
Methane, CH4	Methanogenic archaea in stagnant sludge and grease deposits	A greenhouse gas roughly 28 times as potent as CO2 over a century, and explosive between 5 and 15 % by volume in air, so it presents a confined-space explosion hazard as well as a climate impact	Continuous forced ventilation with a design air-change rate, explosion-proof electrical classification and fixed lower-explosive-limit gas detection interlocked to the ventilation; where quantities justify it, collect and beneficially use the gas rather than venting it

Both controls sit within a confined-space entry programme under the applicable provincial occupational health and safety regulation: atmospheric testing before and during entry, mechanical ventilation, and attendant plus rescue provisions. The two gases are usually managed together, because the same ventilation system that dilutes methane below its lower explosive limit also carries hydrogen sulphide to the odour-control unit.

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