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) Role of Pretreatment and Biological Treatment in Removing Colloidal and Dissolved Particulates

Colloidal (roughly 0.001–1 µm) and dissolved (truly molecular-scale, soluble) particulates are not removed by gravity settling alone: colloids are kept suspended by Brownian motion and electrostatic surface-charge repulsion (DLVO theory), while dissolved organic matter is, by definition, already in true solution with no discrete particle to settle. Pretreatment (screening, grit removal and primary sedimentation) exists to protect downstream processes by removing coarse debris and the readily-settleable fraction of solids before they can abrade equipment or hydraulically overload the biological process — but pretreatment alone does essentially nothing to the colloidal and dissolved fraction, which passes through it largely unchanged.

Biological treatment is what actually addresses that fraction. Microorganisms in an activated-sludge process (or an attached-growth process such as a trickling filter) metabolize soluble and colloidal organic matter as a substrate, converting it into new microbial biomass, carbon dioxide and water; critically, the microbial cells also excrete extracellular polymeric substances (EPS) that bioflocculate the remaining colloidal material into aggregated, settleable biological floc — without requiring a separate chemical-coagulation step. That floc, together with the biomass itself, is then removed by gravity settling in a secondary (final) clarifier, achieving a removal that pretreatment alone could never accomplish because the colloidal and dissolved material was never in a settleable form to begin with. A representative treatment train, selecting a municipal wastewater system, is:

(ii) Hardness of the Lake Water as $CaCO_3$

Given. Lake water with $Ca^{2+}=100\ \text{mg/L}$, $Mg^{2+}=80\ \text{mg/L}$, $Cu^{2+}=40\ \text{mg/L}$; atomic weights $Ca=40$, $Mg=24$, $Cu=63$, $H=1$, $C=12$, $O=16$; $Ca^{2+}$, $Mg^{2+}$ and $Cu^{2+}$ are all divalent.

Find. The total hardness of the water expressed as mg/L $CaCO_3$, and its qualitative classification (soft / moderately hard / hard).

Approach. Hardness is, by definition (Standard Methods; Sawyer, McCarty & Parkin), the sum of the concentrations of the alkaline-earth cations conventionally recognized as hardness-forming — $Ca^{2+}$ and $Mg^{2+}$ — each converted to an equivalent mass of $CaCO_3$ through the ratio of equivalent weights, $\text{mg/L as } CaCO_3 = C_{ion}\times(EW_{CaCO_3}/EW_{ion})$, where $EW=\text{(atomic or molecular weight)}/\text{valence}$; sum the two contributions and classify against the standard hardness scale.

1. Equivalent weight of $CaCO_3$. Build its molecular weight from the given atomic weights ($Ca+C+3O$), then halve for the divalent carbonate: $$\begin{aligned} MW_{CaCO_3} &= 40+12+3(16) = 100\ \text{g/mol} \\ EW_{CaCO_3} &= 100/2 = 50\ \text{g/eq} \end{aligned}$$
2. Equivalent weights of the ions. $$\begin{aligned} EW_{Ca} &= 40/2 = 20\ \text{g/eq} \\ EW_{Mg} &= 24/2 = 12\ \text{g/eq} \\ EW_{Cu} &= 63/2 = 31.5\ \text{g/eq} \end{aligned}$$
3. Convert $Ca^{2+}$ and $Mg^{2+}$ to $CaCO_3$ equivalents. $$\begin{aligned} H_{Ca} &= 100\times\dfrac{50}{20} = 250\ \text{mg/L as } CaCO_3 \\ H_{Mg} &= 80\times\dfrac{50}{12} = 333.3\ \text{mg/L as } CaCO_3 \end{aligned}$$
4. Sum for total hardness. $$\boxed{H = H_{Ca}+H_{Mg} = 250+333.3 = 583.3\ \text{mg/L as } CaCO_3.}$$ $Cu^{2+}$ is not included in the hardness sum — under the Standard Methods/Sawyer–McCarty–Parkin definition, hardness is specifically the sum of the alkaline-earth (Group 2) cations $Ca^{2+}$ and $Mg^{2+}$, not every polyvalent cation present. For reference only, if $Cu^{2+}$ were (incorrectly) folded in, it would add $40\times(50/31.5)=63.5\ \text{mg/L as } CaCO_3$, for a total of 646.8 mg/L — a common trap, since the question supplies Cu’s atomic weight alongside Ca and Mg’s.
5. Classify the water. On the standard hardness scale (soft below roughly 75–100; moderately hard roughly 75/100–150; hard above roughly 150–300 mg/L as $CaCO_3$), a total hardness of 583.3 mg/L is well above the upper boundary even of the “hard” category — on the three categories offered, this water is classified as hard (in fact very hard on finer four-tier scales that separate out anything above roughly 300 mg/L).

(iii) Engineering Methods to Reduce Thermal Impacts on Aquatic Biota

Three engineering methods to reduce the thermal impact of the nuclear plant’s condenser cooling-water discharge on biota in the receiving surface water body are: (1) cooling towers (evaporative or hybrid wet/dry) inserted between the condenser and the final discharge, which reject heat to the atmosphere rather than directly to the receiving water, substantially lowering the discharge temperature before it reaches the aquatic environment; (2) a multiport diffuser outfall, which distributes the warm discharge through many small submerged ports across the receiving water’s cross-section to achieve rapid initial dilution, meeting an in-stream temperature-rise criterion at the edge of a defined, regulator-approved mixing zone rather than concentrating the full thermal load at a single pipe; and (3) a cooling pond or reservoir with sufficient surface area and retention time to dissipate heat to the atmosphere by evaporation, convection and radiation before the water is discharged or recirculated to the condenser, which also provides operational flexibility to curtail or blend discharge during already-warm, low-flow summer conditions when thermal stress on biota is most acute.