NivaarExam PrepOfficial exam papers ↗

18-Env-A1 Principles of Environmental Engineering · May 2014

Question 1 of 7: Mass and Energy Balance, Contaminant Partitioning and Microbiology

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

Notes on this paper

National Exams — May 2014 — 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 guidelines; Canadian Environmental Protection Act, 1999 (CEPA); Andrews, Canadian Professional Engineering and Geoscience (professional ethics).

Question 1: Mass and Energy Balance, Contaminant Partitioning and Microbiology (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) Steady-State TP Concentration in the Lake Outflow

Given. Lake mass balance data for total phosphorus (TP):

Given data
QuantitySymbolValue
Lake volume$V$$1\times10^{5}$ m³
Upstream river inflow$Q_u$$9\times10^{4}$ m³/yr (see the check note)
Evaporation loss$Q_e$$1\times10^{4}$ m³/yr
Lake outflow$Q_o$$8\times10^{4}$ m³/yr
Upstream TP concentration$C_u$10 mg/L
TP decay rate in lake$k$0.12 /yr

Find. The steady-state TP concentration $C$ in the lake outflow.

Check: the source states the upstream inflow as $Q_u = 9\times10^{5}$ m³/yr, but this fails the water balance $Q_u = Q_o + Q_e$ that steady-state conditions on a constant-volume lake require ($8\times10^{4} + 1\times10^{4} = 9\times10^{4}$, exactly one order of magnitude below the stated $Q_u$). The exponent is treated as a typographical slip and corrected to $Q_u = 9\times10^{4}$ m³/yr, which closes the water balance exactly; this corrected value is used throughout. $Q_e$, $Q_o$, $C_u$ and $k$ are used as given — evaporation removes pure water (no TP), so it enters the water balance but not the TP mass balance.

Approach. Model the lake as a single well-mixed (CSTR) reactor and write a steady-state mass balance on TP, with first-order decay removing mass within the lake volume.

  1. Write the unsteady TP mass balance. Accumulation equals mass in minus mass out minus decay: $$V\frac{dC}{dt} = Q_u C_u - Q_o C - kVC.$$
  2. Apply the steady-state condition. With $dC/dt = 0$, $$0 = Q_u C_u - Q_o C - kVC \quad\Rightarrow\quad C = \frac{Q_u C_u}{Q_o + kV}.$$
  3. Evaluate the decay term. $kV = (0.12)(1\times10^{5}) = 1.2\times10^{4}$ m³/yr, so the effective removal capacity is $$Q_o + kV = 8\times10^{4} + 1.2\times10^{4} = 9.2\times10^{4}\ \text{m}^3/\text{yr}.$$
  4. Substitute and solve. $$C = \frac{(9\times10^{4})(10)}{9.2\times10^{4}} = \boxed{9.78\ \text{mg/L}}.$$
QuantityValue
Effective removal capacity, $Q_o + kV$$9.2\times10^{4}$ m³/yr
Steady-state outflow TP concentration, $C$≈ 9.78 mg/L

(ii) Significance of the Octanol–Water Partition Coefficient

The octanol–water partition coefficient $K_{OW}$ is the equilibrium ratio of a chemical's concentration in n-octanol (a surrogate for lipid/fatty tissue) to its concentration in water, $K_{OW} = C_{octanol}/C_{water}$, and is normally reported as $\log K_{OW}$. It is the standard predictor of how strongly a hydrophobic organic compound will partition out of the water column and into the lipid fraction of biological tissue rather than remaining dissolved. Compounds with high $\log K_{OW}$ (PCBs, organochlorine pesticides and many other persistent organic pollutants typically have $\log K_{OW}$ in the 5–7 range) are strongly lipophilic: once a PCB molecule crosses a fish's gill or gut membrane it partitions preferentially into the fish's lipid layer rather than re-dissolving into the surrounding water, so its tissue concentration climbs far above the ambient water concentration.

This partitioning behaviour is the physical basis of bioaccumulation (uptake exceeding elimination within one organism) and biomagnification (concentration increasing at each successive trophic level, since a predator's lipid pool integrates the body burden of everything it has eaten). Engineers use $K_{OW}$ to screen contaminants of concern in site assessments and water-quality risk analyses — a compound with high $\log K_{OW}$ and environmental persistence is flagged for tissue-residue and food-chain exposure pathways even when its dissolved-phase concentration is low and would otherwise appear benign, because the receptor of concern (predatory fish, or humans consuming fish) is exposed through the lipid pathway rather than direct water contact.

(iii) Chlorine Disinfection and the C·T Concept

Chlorine (as $\text{Cl}_2$, hypochlorous acid $\text{HOCl}$, or hypochlorite $\text{OCl}^-$ depending on pH) inactivates bacteria, cysts and viruses primarily by diffusing through the cell wall/envelope and oxidatively damaging enzymes, the cell membrane and nucleic acids, disrupting the microorganism's metabolism and reproduction. $\text{HOCl}$ is the dominant and most effective species at the pH typical of drinking-water treatment (below about pH 7.5) because its small, uncharged molecule crosses the cell wall far more readily than the charged $\text{OCl}^-$ ion, so disinfection efficiency drops sharply as pH rises and the $\text{HOCl}\rightleftharpoons\text{H}^+ + \text{OCl}^-$ equilibrium shifts toward the less-effective hypochlorite form.

Because inactivation is a kinetic (time-dependent), concentration-dependent process, engineered disinfection systems are designed and regulated using the C·T concept (Chick–Watson framework): the log-inactivation achieved is a function of the product of residual disinfectant concentration $C$ and the effective contact time $T$ the water spends in the disinfection contact tank, $C\cdot T \geq (C\cdot T)_{required}$ for the target pathogen and log-removal credit. Regulators publish minimum $C\cdot T$ tables by pathogen (viruses require far lower $C\cdot T$ than the much more resistant protozoan cysts, e.g. Giardia and Cryptosporidium) and by temperature and pH. Because a shorter, higher-velocity path through a poorly baffled tank lets some water short-circuit to the outlet before the design contact time elapses, contact tanks are baffled (often serpentine, over-and-under baffling) to push the hydraulic residence-time distribution toward plug flow, and the effective $T$ used in the $C\cdot T$ calculation is the $T_{10}$ (the time within which only 10% of a tracer has broken through) rather than the nominal (volume/flow) detention time, so that the design remains conservative for the fraction of flow that moves fastest through the tank.

← Paper overview