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18-Env-A1 Principles of Environmental Engineering · May 2015

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2015 — 04-Env-A1 / Principles of Environmental Engineering. 3 hours duration; closed book with an 8.5×11 in double-sided aid sheet; Casio or Sharp approved calculator only. Any five questions constitute a complete paper (first five answers marked); all seven are solved below for completeness. Each question is worth 20 marks.

Reference texts. Davis & Cornwell, Introduction to Environmental Engineering (6th ed.); Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.); MWH’s Water Treatment: Principles and Design (3rd ed.); Guidelines for Canadian Drinking Water Quality (Health Canada); Canadian Council of Ministers of the Environment (CCME) water-quality and landfill guidelines; Canadian Environmental Protection Act, 1999 (CEPA); Impact Assessment Act, 2019 (Canada) and Alberta Environmental Protection and Enhancement Act; Andrews, Canadian Professional Engineering and Geoscience (professional ethics).

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) Combined Physical-Chemical and Biological Particle Removal (Wastewater System)

Physical-chemical and biological treatment remove different fractions of a wastewater's particulate and colloidal load, and neither alone is sufficient. Physical-chemical primary treatment (screening followed by gravity sedimentation in a primary clarifier) removes the coarse, readily-settleable fraction of solids by density difference alone — fast and low-cost, but it cannot touch the finely divided and dissolved organic matter that remains suspended or in solution. Biological secondary treatment (an aeration basin growing a mixed microbial population — activated sludge) then converts that dissolved and colloidal organic matter into new microbial cell mass and $\text{CO}_2$ through metabolism; critically, this step converts a fraction of the pollutant load that is not particulate at all into flocculant biomass that a secondary clarifier can then remove by the same gravity-settling principle as the primary stage. The two mechanisms are therefore complementary rather than redundant: physical-chemical treatment strips out what settles on its own, and biological treatment turns what does not settle on its own into something that will.

PrimaryClarifierAerationBasinSecondaryClarifierRawwastewaterSettledwastewaterMixedliquorTreatedeffluentPrimary sludge(to disposal)RAS (returnactivated sludge)Waste activatedsludge (WAS)
Figure 3.1 — Conventional activated-sludge train: physical-chemical primary clarification removes settleable solids (to disposal), the aeration basin's biological process converts dissolved/colloidal organic matter into settleable microbial floc, and a secondary clarifier removes that floc — returning most of it as return activated sludge (RAS) to sustain the biomass and wasting the surplus (WAS) to sludge handling.

(ii) Hardness of Lake Superior Water Near the Salt Mine

Given. Divalent-cation concentrations from the water analysis, with atomic weights Ca = 40, H = 1, C = 12, O = 16, Mg = 24, Fe = 56 as stated on the exam:

Given data
IonConcentrationAtomic weightValence
$Ca^{2+}$150 mg/L40 (given)2
$Mg^{2+}$800 mg/L24 (given)2
$Fe^{2+}$60 mg/L56 (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 defined as the sum of polyvalent alkaline-earth cations ($Ca^{2+}$, $Mg^{2+}$); here, unlike a trace heavy metal reported without a usable atomic weight, the question explicitly supplies $Fe$'s atomic weight (56) alongside $Ca$ and $Mg$'s and lists $Fe^{2+}$'s concentration as part of the same water-hardness dataset. This is taken as a clear instruction to include ferrous iron in the hardness total (consistent with waters near evaporite/salt deposits, where dissolved iron commonly co-occurs with $Ca^{2+}$/$Mg^{2+}$ and is conventionally folded into "total hardness" once its atomic weight is supplied for the purpose).

Approach. Convert each hardness-forming ion's mass concentration to an equivalent mass concentration as $\text{CaCO}_3$ using the ratio of equivalent weights, then sum and classify against the standard hardness scale.

  1. Equivalent weight of $\text{CaCO}_3$. Molecular weight $= 40+12+3(16) = 100\ \text{g/mol}$; with valence 2 (each $\text{CO}_3^{2-}$ neutralizes one divalent cation), $$EW_{CaCO_3} = \frac{100}{2} = 50\ \text{g/eq}.$$
  2. Convert $Ca^{2+}$ to CaCO3 equivalent. $EW_{Ca}=40/2=20\ \text{g/eq}$, so $$H_{Ca} = 150\times\frac{50}{20} = \boxed{375.0\ \text{mg/L as CaCO}_3}.$$
  3. Convert $Mg^{2+}$ to CaCO3 equivalent. $EW_{Mg}=24/2=12\ \text{g/eq}$, so $$H_{Mg} = 800\times\frac{50}{12} = \boxed{3333.3\ \text{mg/L as CaCO}_3}.$$
  4. Convert $Fe^{2+}$ to CaCO3 equivalent. $EW_{Fe}=56/2=28\ \text{g/eq}$, so $$H_{Fe} = 60\times\frac{50}{28} = \boxed{107.1\ \text{mg/L as CaCO}_3}.$$
  5. Total hardness and classification. Summing the three hardness-forming ions, $$H_{total} = H_{Ca}+H_{Mg}+H_{Fe} = 375.0+3333.3+107.1 = \boxed{3815.5\ \text{mg/L as CaCO}_3}.$$ Against the standard scale (<75 soft; 75–150 moderately hard; 150–300 hard; >300 very hard), 3815.5 mg/L is more than an order of magnitude above the 300 mg/L threshold — the water is classified as hard (at the extreme "very hard" end of that category), consistent with dissolution of evaporite salts near the salt mine dominating the water chemistry.
QuantityValue
Hardness from $Ca^{2+}$375.0 mg/L as CaCO3
Hardness from $Mg^{2+}$3333.3 mg/L as CaCO3
Hardness from $Fe^{2+}$107.1 mg/L as CaCO3
Total hardness3815.5 mg/L as CaCO3
ClassificationHard (very hard)

(iii) Downstream Mixed Temperature and Cold-Water Fishery Protection

Given. Flow rates and temperatures of the two streams mixing in the river (from the source diagram):

Given data
QuantitySymbolValue
Cooling-tower discharge flow$Q_c$200 m³/s
Cooling-tower discharge temperature$T_c$50°C
Upstream river temperature$T_s$10°C
Combined downstream flow$Q$300 m³/s
MixingZoneQs = 100 m³/sTs = 10°CQc = 200 m³/sTc = 50°CQ = 300 m³/sT = ?
Figure 3.2 — Mixing of the upstream river flow ($Q_s$, inferred by continuity as $Q-Q_c$) with the heated cooling-tower discharge ($Q_c$) to give the combined downstream flow $Q$ at unknown mixed temperature $T$.

Find. The combined downstream river temperature $T$.

Approach. The upstream river flow $Q_s$ is not stated directly but is fixed by continuity ($Q=Q_s+Q_c$); a steady-flow thermal energy balance (constant density and specific heat) on the mixing point then gives the combined temperature as a flow-weighted average.

  1. Find the upstream river flow by continuity. $$Q_s = Q-Q_c = 300-200 = 100\ \text{m}^3/\text{s}.$$
  2. Write the thermal energy balance at the mixing point. With constant $\rho$ and $c_p$, the energy balance reduces to a flow-weighted temperature average: $$Q\,T = Q_sT_s + Q_cT_c \quad\Rightarrow\quad T = \frac{Q_sT_s+Q_cT_c}{Q}.$$
  3. Substitute and solve. $$T = \frac{(100)(10)+(200)(50)}{300} = \frac{1000+10000}{300} = \boxed{36.7\,{}^{\circ}\text{C}}.$$
QuantityValue
Upstream river flow (by continuity), $Q_s$100 m³/s
Combined downstream temperature, $T$36.7 °C

A downstream temperature of 36.7°C is far above the roughly 10–19°C range most cold-water salmonid fisheries need, so the cooling-tower discharge is a serious thermal-pollution concern. Two engineering solutions to reduce it: (1) increase the cooling tower's approach/rejection performance (larger tower, added cooling cells, or a wet-and-dry hybrid tower) so that $T_c$ itself is reduced before the flow ever reaches the river, directly lowering the mixed $T$; and (2) install a multiport diffuser outfall that discharges the heated flow as many small, high-velocity jets across the river's width and depth rather than as one concentrated point source — this does not reduce the total heat load but rapidly entrains additional river water into the near-field mixing zone, lowering the peak temperature any single fish encounters and shortening the reach of river above the fishery's thermal tolerance.