Particle Characteristics, Water Chemistry and Thermal Pollution: 18-Env-A1 Principles of Environmental Engineering, May 2016, Q3

Question 3: Particle Characteristics, Water Chemistry 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) Three Particle Types and Their Removal Methods (Drinking-Water Treatment)

Choosing a drinking-water treatment facility, raw water carries particles across a wide size and settleability range, and no single unit process removes all of them efficiently:

Coarse grit/sand particles. Dense, readily-settleable material (sand, gravel-fines, debris) is removed in a grit chamber (or coarse screen ahead of it) purely by gravity settling at a controlled, low velocity — the particles are dense enough to settle on their own without any chemical conditioning.
Colloidal turbidity (fine clay/silt). These particles are too small and too electrostatically stabilized (mutually repelling negative surface charge) to settle within a practical detention time on their own. They are removed by coagulation–flocculation followed by sedimentation: a coagulant (e.g., alum) neutralizes the surface charge so particles can collide and agglomerate into larger, settleable floc, which a sedimentation basin then removes by gravity.
Very fine residual particles (post-sedimentation). The small fraction of floc and fine particulate that does not settle in sedimentation is removed by granular-media (dual-media sand/anthracite) filtration, which acts by straining, interception and adsorption within the filter bed rather than by gravity settling — a fundamentally different removal mechanism from the first two steps.

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

Given. Divalent-cation concentrations from the water analysis, with atomic weights Ca = 40, Mg = 24, Fe = 56 as stated on the exam:

Given data
Ion	Concentration	Atomic weight	Valence
$Ca^{2+}$	100 mg/L	40 (given)	2
$Mg^{2+}$	70 mg/L	24 (given)	2
$Fe^{2+}$	40 mg/L	56 (given)	2

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

Check: hardness is conventionally the sum of polyvalent alkaline-earth cations ($Ca^{2+}$, $Mg^{2+}$); here the question explicitly supplies $Fe$'s atomic weight (56) alongside $Ca$'s and $Mg$'s, and lists $Fe^{2+}$'s concentration as part of the same water-hardness dataset near a rock quarry, where dissolved ferrous iron commonly co-occurs with $Ca^{2+}$/$Mg^{2+}$. This is taken as a clear instruction to include ferrous iron in the hardness total.

Approach. Convert each cation's mass concentration to an equivalent CaCO₃ concentration using the ratio of equivalent weights, then sum the three contributions and compare the total to the standard hardness classification bands.

Compute each species' equivalent weight. Equivalent weight $= $ atomic weight $/$ valence, all three ions being divalent: $$EW_{CaCO_3} = \frac{100}{2}=50,\quad EW_{Ca}=\frac{40}{2}=20,\quad EW_{Mg}=\frac{24}{2}=12,\quad EW_{Fe}=\frac{56}{2}=28.$$
Convert each cation to mg/L as CaCO₃. Multiply each concentration by the ratio $EW_{CaCO_3}/EW_i$: $$H_{Ca} = 100\times\frac{50}{20}=250\ \text{mg/L},\qquad H_{Mg}=70\times\frac{50}{12}=291.7\ \text{mg/L},\qquad H_{Fe}=40\times\frac{50}{28}=71.4\ \text{mg/L}.$$
Sum the total hardness. $$H_{total} = 250+291.7+71.4 = \boxed{613\ \text{mg/L as CaCO}_3}.$$
Classify the water. Against the standard scale (0–75 soft, 75–150 moderately hard, 150–300 hard, >300 very hard), $H_{total}=613$ mg/L places this water firmly in the very hard category — more than double the 300 mg/L "very hard" threshold.

Quantity	Value
Hardness from $Ca^{2+}$	250.0 mg/L as CaCO₃
Hardness from $Mg^{2+}$	291.7 mg/L as CaCO₃
Hardness from $Fe^{2+}$	71.4 mg/L as CaCO₃
Total hardness	613 mg/L as CaCO₃
Classification	Very hard ($>300$ mg/L)

(iii) Downstream Mixed River Temperature and Thermal-Pollution Mitigation

Given. Cooling-tower discharge and receiving-river flows/temperatures:

Given data
Quantity	Symbol	Value
Cooling-tower discharge flow	$Q_c$	400 m³/s
Cooling-tower discharge temperature	$T_c$	60°C
Receiving-river flow	$Q_s$	600 m³/s
Receiving-river (upstream) temperature	$T_s$	5°C

Find. The fully-mixed downstream river temperature $T$.

Figure 3.1 — Cooling-tower discharge $Q_c$ at $T_c$ entering the receiving river $Q_s$ at $T_s$; the fully-mixed downstream temperature $T$ is a flow-weighted average of the two streams.

Approach. Write a steady-state thermal energy balance across the mixing point, treating the discharge and river as two streams of equal density and specific heat that fully mix.

Write the energy balance at the mixing point. With equal $\rho$ and $c_p$ for both streams, thermal energy is conserved as a flow-weighted average: $$Q_cT_c + Q_sT_s = (Q_c+Q_s)\,T.$$
Solve for $T$. $$T = \frac{Q_cT_c+Q_sT_s}{Q_c+Q_s} = \frac{(400)(60)+(600)(5)}{400+600} = \frac{24000+3000}{1000} = \boxed{27\ {}^{\circ}\text{C}}.$$

Quantity	Value
Downstream mixed temperature, $T$	27°C

A downstream temperature of 27°C is far above the 5°C ambient river temperature and would be lethal to a cold-water spawning fishery (cold-water salmonids typically require water below roughly 20°C, with spawning success further constrained below that). Two engineering solutions to reduce this impact: (1) add a second, closed-loop (wet or dry) cooling stage after the existing cooling tower so the discharge temperature $T_c$ itself is reduced before release, cutting the thermal load entering the mixing calculation directly; and (2) construct a diffuser outfall (multi-port submerged diffuser across the river cross-section, or a long mixing/cooling channel/pond before the discharge point) that spreads the heated discharge over a much larger initial dilution volume and residence time, lowering the peak near-field temperature the spawning habitat actually experiences even though the fully-mixed far-field average is unchanged.

Concept & Theory

Topic. Particle removal mechanisms in treatment trains, hardness speciation and classification, and thermal-pollution mixing at an outfall.

Key relations. Equivalent-weight hardness conversion, $H_i = C_i \times (EW_{CaCO_3}/EW_i)$; flow-weighted thermal mixing, $T=(Q_cT_c+Q_sT_s)/(Q_c+Q_s)$.

Why this works. All three sub-parts are conservation-of-mass (or energy) problems at heart: particle removal exploits different physical mechanisms (settling, coagulation, straining) to conserve mass across a barrier; hardness converts different cations onto a common CaCO₃ basis using equivalent-weight mass conservation; thermal mixing conserves enthalpy across the confluence.

Common pitfall. Omitting a cation from the hardness sum when its atomic weight is actually given (treating it as a decoy without checking); assuming filtration removes particles by "screening" alone rather than the combination of straining, interception and adsorption that actually governs filter performance; forgetting that reducing the far-field mixed temperature does not by itself protect fish in the near-field plume, which is why a diffuser/dilution solution is listed separately from reducing $T_c$ itself.

Source. Davis & Cornwell, Introduction to Environmental Engineering, ch. 5 (water chemistry/hardness) and ch. 6 (thermal pollution); MWH’s Water Treatment: Principles and Design, ch. 9–10 (coagulation, sedimentation, filtration).

Question 3 of 7: Particle Characteristics, Water Chemistry and Thermal Pollution

Question 3: Particle Characteristics, Water Chemistry and Thermal Pollution (20 marks)

(i) Three Particle Types and Their Removal Methods (Drinking-Water Treatment)

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

(iii) Downstream Mixed River Temperature and Thermal-Pollution Mitigation