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

Question 7: 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) Coagulation and Flocculation in Water/Wastewater Treatment

Coagulation is the chemical destabilization of colloidal particles: a coagulant (e.g., alum, ferric chloride, or a cationic polymer) is dosed and rapidly, intensely mixed into the water so it neutralizes the negative surface charge that normally keeps colloidal particles apart (charge neutralization) and/or compresses the electrical double layer surrounding each particle, removing the electrostatic repulsion that otherwise prevents particles from approaching one another. Flocculation is the subsequent physical process of gentle, prolonged mixing (low velocity gradient $G$, longer detention time than the rapid-mix step) that promotes particle–particle collisions among the now-destabilized colloids, allowing them to aggregate into larger, denser flocs. Together, coagulation removes the repulsive barrier and flocculation supplies the collision opportunity for aggregation, converting turbidity-causing colloidal particles — which are individually far too small and too slow-settling to remove by gravity alone — into flocs large and dense enough to be removed efficiently by downstream sedimentation and/or filtration. This coagulation–flocculation–sedimentation–filtration sequence is the core physicochemical pretreatment train in conventional water treatment, and an analogous chemically-enhanced coagulation step is also used in wastewater treatment to enhance phosphorus and solids removal.

(ii) Hardness of Lake Huron Water as CaCO₃

Given. Average ion concentrations near the salt quarry, with atomic weights Ca = 40, Mg = 24, Cu = 64, C = 12, O = 16 (so CaCO₃ molar mass = 100):

Find. Total hardness in mg/L as CaCO₃, and the water's hardness classification.

Approach. Total hardness is conventionally defined (Standard Methods) as the sum of the Ca²⁺ and Mg²⁺ equivalents expressed as CaCO₃ — Cu²⁺ is not a hardness-forming cation under this definition (it is not one of the alkaline-earth divalent cations that hardness classically refers to) and is a decoy value in this dataset, not part of the hardness sum. Convert each hardness-causing ion's concentration to a CaCO₃-equivalent using the ratio of equivalent weights, then sum and compare to standard hardness classification bands.

1. Convert Ca²⁺ to CaCO₃ equivalent. $$H_{Ca} = 100\ \text{mg/L} \times \frac{50}{20} = 250\ \text{mg/L as CaCO}_3.$$
2. Convert Mg²⁺ to CaCO₃ equivalent. $$H_{Mg} = 80\ \text{mg/L} \times \frac{50}{12} = 333.3\ \text{mg/L as CaCO}_3.$$
3. Sum for total hardness (Cu²⁺ excluded — not a hardness ion). $$\boxed{H_{total} = H_{Ca} + H_{Mg} = 250 + 333.3 = 583.3\ \text{mg/L as CaCO}_3.}$$
4. Classify against standard hardness bands. Soft < 75, moderately hard 75–150, hard 150–300, very hard > 300 mg/L as CaCO₃. At 583.3 mg/L, the water is classified as very hard.

(iii) Downstream River Temperature After Cooling-Tower Discharge

Given. A cooling-tower discharge mixes with the upstream river before the downstream measurement point.

Find. The downstream river temperature $T$ immediately after complete mixing.

Approach. The upstream river flow is the balance of the total downstream flow not accounted for by the cooling-tower discharge ($Q_s = Q - Q_c$); apply a steady-state thermal energy balance (equivalent to a flow-weighted average temperature, since density and specific heat are effectively constant over this range) across the mixing point.

1. Find the upstream river flow rate by continuity. $$Q_s = Q - Q_c = 500 - 400 = 100\ \text{m}^3/\text{s}.$$
2. Write the steady-state thermal energy balance at the mixing point. With constant $\rho$ and $c_p$, the energy balance reduces to a flow-weighted temperature average: $$QT = Q_cT_c + Q_sT_s.$$
3. Solve for the downstream temperature. $$T = \frac{Q_cT_c + Q_sT_s}{Q} = \frac{(400)(70) + (100)(20)}{500} = \frac{28000+2000}{500} = \boxed{60\ ^\circ\text{C}}.$$

A downstream temperature of 60 °C is far above any cold-water fishery tolerance (typically well under 20 °C for salmonids), so mitigation is required. Two engineering solutions: (1) reduce the discharge temperature at the source by adding a supplementary cooling stage (a wet or dry cooling tower cell, or an engineered cooling pond) so $T_c$ is lowered before the cooling water ever reaches the river, which is the most direct fix since it reduces the driving temperature difference itself; and (2) diffuse the discharge to increase effective initial dilution using a multi-port diffuser across the river cross-section instead of a single discharge point, which increases the effective mixing flow at the point of discharge and lowers the peak near-field temperature the fishery is exposed to, even though the far-field mixed temperature is governed by the same overall energy balance.