16-Civ-A3 Elementary Environmental Engineering · May 2018
Question 7 of 7: Particle Characteristics and Chemistry of Solutions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2018 — 16-Civ-A3 Elementary Environmental Engineering. Three hours; closed book with one candidate-prepared 8½ × 11 double-sided aid sheet; approved Casio or Sharp calculator only. Seven problems are printed, each worth 20 marks, and any five constitute a complete paper (maximum 100 marks). All seven are solved here, because the set is intended as a study resource rather than an exam script. Section marks are shown in brackets at the left margin of each question and are reproduced from the final-page Marking Scheme.
Reference texts.
Davis, M.L. & Cornwell, D.A., Introduction to Environmental Engineering, 5th ed., McGraw-Hill — material and energy balances, reactor kinetics, hardness, disinfection, water and wastewater unit processes.
Masters, G.M. & Ela, W.P., Introduction to Environmental Engineering and Science, 3rd ed., Pearson — mass/energy balances on power plants, air-emission estimation, growth–pollution linkages.
Metcalf & Eddy / AECOM, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., McGraw-Hill — wastewater unit processes and plant upgrading.
Crittenden, J.C. et al. (MWH), Water Treatment: Principles and Design, 3rd ed., Wiley — coagulation/flocculation, filtration, membranes, adsorption, disinfection.
Engineers Canada / EGBC Code of Ethics; Canadian Council of Ministers of the Environment (CCME) and the Guidelines for Canadian Drinking Water Quality (GCDWQ), Health Canada — professional duty and water-quality benchmarks.
Check: Problem 4(ii) quotes a rate constant as “20 dm6/mol2” with no time unit, and states the fundamental reaction 2A + B ⇌ C. It is solved as a forward-rate second-in-A/first-in-B rate law with $k=20\ \text{dm}^{6}\,\text{mol}^{-2}\,\text{s}^{-1}$ (the only reading that makes $-r_A=k\,C_A^{2}C_B$ dimensionally a rate); the time unit is taken as seconds per NOTE 1. Because no feed flow rate is supplied, the well-posed deliverable is the flow-independent space-time $\tau$ (with $V=\tau\,v_0$ for any stated feed basis), not an absolute volume.
Question 7: Particle Characteristics and Chemistry of Solutions (20 marks)
Part (i) — Combined role of coagulation/flocculation and filtration (10 marks)
Colloidal and fine suspended particles (clays, microorganisms, natural organic matter) carry a net negative surface charge; the resulting electrostatic repulsion keeps them stable and non-settling, and they are too small for a granular filter to capture directly. Coagulation adds a metal-salt coagulant (alum or ferric) whose hydrolysis products neutralise that charge and provide hydroxide precipitate (sweep floc). Flocculation then gently mixes the destabilised particles so they collide and aggregate into larger, settleable/filterable flocs. Sedimentation removes the bulk of the floc, and filtration through a granular (sand/anthracite) bed captures the remaining fine floc by interception, straining and attachment. The combination is essential because coagulation/flocculation converts un-filterable colloids into filterable aggregates, and filtration provides the final particle barrier — a low-turbidity filtered water is a prerequisite for effective disinfection, since particles shield pathogens from the disinfectant.
Conventional coagulation–flocculation–sedimentation–filtration train ahead of disinfection.
Part (ii) — Hardness of the copper-mine groundwater (10 marks)
Given. A groundwater analysis near a copper mine, with atomic weights supplied.
Given data — Problem 7(ii)
Cation
Concentration (mg/L)
Atomic weight
Equivalent weight (÷ valence 2)
Ca2+
20
40
20
Mg2+
30
24
12
Cu2+
45
64
32
Find. The hardness in mg/L as CaCO3, and the classification (soft / moderately hard / hard).
Approach. Convert each divalent cation to its CaCO3 equivalent through equivalent weights, using $C_{\text{as CaCO}_3}=C\times\dfrac{50}{\text{EW}}$ (EW of CaCO3 = 50). Conventional hardness is Ca + Mg only.
Calcium as CaCO3. $$\text{Ca}=20\times\frac{50}{20}=50\ \text{mg/L as CaCO}_3.$$
Magnesium as CaCO3. Magnesium’s small equivalent weight (12) makes it the dominant contributor despite the modest mg/L:
$$\text{Mg}=30\times\frac{50}{12}=125\ \text{mg/L as CaCO}_3.$$
Total (conventional) hardness. Hardness is the sum of the multivalent hardness cations Ca and Mg:
$$\boxed{\text{TH}=50+125=175\ \text{mg/L as CaCO}_3.}$$
Copper (for completeness). Cu2+ is a divalent cation and contributes to a strict all-multivalent-cation total, but it is not a conventional hardness ion:
$$\text{Cu}=45\times\frac{50}{32}=70.3\ \text{mg/L as CaCO}_3,\qquad \text{TH}+\text{Cu}=245.3\ \text{mg/L as CaCO}_3.$$
Hardness results and classification
Quantity
Value (mg/L as CaCO3)
Calcium hardness
50
Magnesium hardness
125
Total hardness (Ca + Mg)
175 — hard
Strict total including Cu2+
245.3
Classification. On the three-band scale offered by the question (soft < 75, moderately hard 75–150, hard > 150), a total hardness of 175 mg/L as CaCO3 is hard; it is likewise “hard” on the four-band GCDWQ scale (60/120/180). The reported Cu2+ of 45 mg/L is itself a contaminant of concern — far above the ≈ 1–2 mg/L aesthetic/health guideline — consistent with copper-mine drainage, and should be reported as a metals issue rather than as hardness.
Check: “Hardness” conventionally means Ca + Mg only, so the answer is 175 mg/L as CaCO3. The Cu2+ term (70.3) is shown because copper is a divalent cation that strictly contributes to total polyvalent-cation hardness, but its true significance here is toxicity, not scale-forming hardness — treat it as a copper-removal (precipitation/ion-exchange) requirement.
