Question 4 of 7: Particle Characteristics, Chemistry of Solutions and Thermal Pollution

Question 4: Particle Characteristics, Chemistry of Solutions and Thermal Pollution (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) Physical-Chemical Processes for Two Classes of Wastewater Particulates

Municipal wastewater carries particulates ranging from coarse, dense settleable solids to fine colloidal solids stabilized by surface charge; each class needs a different physical-chemical process to remove it effectively:

1. Screening and grit removal for coarse, settleable (grit/debris) particulates. Bar screens physically strain out rags, plastics and large debris, and a grit chamber (using controlled velocity so dense inorganic grit settles while lighter organics stay suspended and continue downstream) removes sand and grit. This is purely mechanical/physical and protects downstream pumps and process equipment from abrasion and clogging, addressing the coarse, dense particulate fraction before it can interfere with biological or chemical treatment.
2. Coagulation-flocculation ahead of sedimentation for fine/colloidal particulates. Fine colloidal solids (organic fines, some fraction of the BOD-bearing material) are too small and too electrostatically stabilized to settle under gravity alone. Adding a coagulant (e.g., alum or a ferric salt) neutralizes the particles’ surface charge (the chemical step), allowing gentle mixing (flocculation) to grow them into larger, settleable floc that is then removed by sedimentation (the physical step) — a combined physical-chemical process specifically targeting the particulate size range that screening and plain sedimentation cannot reach.

(ii) Hardness of the Lake Erie Water Near the Rock Quarry

Given. Divalent-cation concentrations from the water analysis, with atomic weights $Ca=40$, $Mg=24$, $Cu=64$, $H=1$, $C=12$, $O=16$ as stated on the exam:

Find. The total hardness of the water expressed as mg/L CaCO3, and its qualitative classification (soft, moderately hard or hard).

Approach. Hardness is, by the Standard Methods/Sawyer–McCarty definition, the sum of the alkaline-earth cations $Ca^{2+}$ and $Mg^{2+}$ only, each converted to an equivalent mass of CaCO3 through the ratio of equivalent weights; $Cu^{2+}$ is a transition-metal cation supplied alongside Ca and Mg as a deliberate decoy and is not a hardness-forming ion under this definition.

1. Equivalent weight of CaCO3. Build its molecular weight from the given atomic weights ($Ca+C+3O$), then halve for the divalent carbonate: $$MW_{CaCO_3} = 40+12+3(16) = 100\ \text{g/mol}, \qquad EW_{CaCO_3} = 100/2 = 50\ \text{g/eq}.$$
2. Equivalent weights of the cations. Equivalent weight $=$ atomic weight $\div$ valence, all three ions being divalent: $$EW_{Ca} = 40/2 = 20\ \text{g/eq}, \qquad EW_{Mg} = 24/2 = 12\ \text{g/eq}, \qquad EW_{Cu} = 64/2 = 32\ \text{g/eq}.$$
3. Convert $Ca^{2+}$ and $Mg^{2+}$ to CaCO3 equivalents. $$H_{Ca} = 30\times\dfrac{50}{20} = 75.0\ \text{mg/L as CaCO}_3, \qquad H_{Mg} = 40\times\dfrac{50}{12} = 166.7\ \text{mg/L as CaCO}_3.$$
4. Sum for total hardness. $$\boxed{H = H_{Ca}+H_{Mg} = 75.0+166.7 = 241.7\ \text{mg/L as CaCO}_3.}$$ $Cu^{2+}$ is not included in this sum: under the Standard Methods/Sawyer–McCarty definition, hardness is specifically the Group 2 alkaline-earth pair $Ca^{2+}+Mg^{2+}$, not every polyvalent cation a water analysis happens to report. For reference only, $Cu^{2+}$ would contribute $20\times(50/32) = 31.25\ \text{mg/L as CaCO}_3$ if (incorrectly) included, giving 272.9 mg/L — a common trap, since the question deliberately supplies copper’s atomic weight alongside calcium’s and magnesium’s.
5. Classify the water. Against the standard hardness scale (0–75 soft, 75–150 moderately hard, 150–300 hard, >300 very hard), $H=241.7$ mg/L places this water in the hard category.

(iii) Thermal Impacts on the Downstream Cold-Water Fishery and Engineering Reduction Methods

Two potential thermal impacts of the cooling-tower discharge on the downstream cold-water fishery:

1. Reduced dissolved-oxygen availability compounded with elevated fish metabolic demand. Warmer water holds less dissolved oxygen at saturation, at the same time as the fish’s own metabolic (and therefore oxygen) demand rises with temperature — a double stress that can push a cold-water species (which already has a narrow thermal tolerance) into physiological stress or lethal conditions even without any other pollutant present.
2. Disruption of spawning cues and life-cycle timing. Many cold-water species (e.g., salmonids) rely on a specific seasonal temperature signal to trigger spawning, egg incubation and migration; an artificially warmed discharge can shift or suppress these cues, and can also allow warm-water competitor or predator species to encroach on habitat that would otherwise remain thermally unsuitable for them.

Two engineering methods to reduce the thermal impact:

1. Additional (closed-loop) cooling capacity or a cooling pond ahead of discharge. Expanding the cooling tower’s capacity, or routing the blowdown through a lined cooling pond with sufficient surface area and residence time, actively removes heat from the water before it ever reaches the river, meeting the temperature target at the point of discharge itself rather than relying on the river to dilute the heat away.
2. A submerged multi-port diffuser outfall for rapid initial dilution. Distributing the warm discharge through many small ports across the river’s cross-section achieves fast initial mixing, so the temperature rise is confined to a small, regulator-approved mixing zone rather than creating a long, continuous warm plume through the fishery’s habitat — a lower-cost option than added cooling capacity, though it depends on the river having adequate flow to provide sufficient dilution, particularly during low-flow summer conditions when the fishery is already most heat-stressed.