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

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 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 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 TN Concentration in the Lake and Outflow

Given. Lake mass-balance data for total nitrogen (TN), as printed on the exam:

Given data
QuantitySymbolValue
Lake volume$V$$1\times10^{5}$ m³
River inflow$Q_u$$1\times10^{6}$ m³/yr
Evaporation loss$Q_e$$5\times10^{4}$ m³/yr
Lake outflow$Q_o$$3\times10^{4}$ m³/yr
Inflow TN concentration$C_u$15 mg/L
TN decay rate in lake$k$0.1 /yr

Find. The steady-state TN concentration $C$ in the (well-mixed) lake and in its outflow stream.

Check: the printed inflow $Q_u = 1\times10^{6}$ m³/yr is far larger than $Q_o+Q_e = 8\times10^{4}$ m³/yr, so these four flows would not close a water balance on a lake of literally constant volume. Note 1 on the exam cover page invites the candidate to state any assumption made where doubt exists; the reading used here is the least speculative one available: $Q_u$, $Q_o$, $C_u$ and $k$ are taken exactly as given and applied in the standard well-mixed (CSTR) TN mass balance, since evaporation carries no TN mass and does not enter that balance regardless of how the total water budget resolves. No value is silently substituted for another.

Approach. Model the lake as a single well-mixed (CSTR) reactor and write a steady-state mass balance on TN, with first-order decay removing mass within the lake volume; the outflow leaves at the lake's own (uniform) concentration.

  1. Write the unsteady TN mass balance. Accumulation equals mass in minus mass out minus decay: $$V\frac{dC}{dt} = Q_uC_u - Q_oC - kVC.$$
  2. Apply the steady-state condition. With $dC/dt=0$, $$0 = Q_uC_u - Q_oC - kVC \quad\Rightarrow\quad C = \frac{Q_uC_u}{Q_o+kV}.$$
  3. Evaluate the decay term. $kV = (0.1)(1\times10^{5}) = 1\times10^{4}$ m³/yr, so the effective removal capacity is $$Q_o+kV = 3\times10^{4} + 1\times10^{4} = 4\times10^{4}\ \text{m}^3/\text{yr}.$$
  4. Substitute and solve. $$C = \frac{(1\times10^{6})(15)}{4\times10^{4}} = \boxed{375\ \text{mg/L}}.$$ Because the lake is modelled as a single well-mixed reactor, the outflow leaves at the same concentration as the lake itself.
QuantityValue
Effective removal capacity, $Q_o+kV$$4\times10^{4}$ m³/yr
Steady-state lake TN concentration, $C$375 mg/L
Steady-state outflow TN concentration375 mg/L (same as the lake, well-mixed)

(ii) Partition Coefficient ($K_d$) and Contaminant Fate on Suspended Solids

The solid–liquid partition coefficient $K_d = C_s/C_w$ describes the equilibrium ratio between a contaminant's concentration sorbed onto suspended solids, $C_s$ (mass contaminant per mass solids), and its concentration remaining dissolved in the water column, $C_w$ (mass per volume). A high $K_d$ means the contaminant strongly favours the solid phase — it sorbs onto the surfaces of clay, organic detritus and other suspended particulate matter rather than staying dissolved — while a low $K_d$ means the contaminant stays largely in solution.

During surface flow, $K_d$ therefore governs a contaminant's transport pathway: a high-$K_d$ (strongly sorbing) contaminant travels with the suspended-solids load, so its fate tracks the fate of the particles — it settles out wherever the particles settle (in pools, behind obstructions, in a downstream reservoir or estuary), can be resuspended and re-transported during high-flow events, and ultimately accumulates in bed sediment, where it becomes available to benthic organisms rather than to the water-column community. A low-$K_d$ contaminant instead remains dissolved, travels essentially at the velocity of the water itself, and is far more available for direct uptake by aquatic organisms through the water they respire. Engineers use $K_d$ (together with the organic-carbon-normalized $K_{oc}$ for hydrophobic organics) to predict whether a spill or chronic discharge will behave as a sediment-transport problem requiring dredging/capping remediation, or as a water-column problem requiring source control and dilution/treatment of the flow itself.

(iii) UV and Chlorine Disinfection: Dose, Contact Time and Pathogen Inactivation

Chlorine (as $\text{Cl}_2$, hypochlorous acid $\text{HOCl}$, or hypochlorite $\text{OCl}^-$ depending on pH) inactivates bacteria, cysts and viruses chemically: it diffuses through the cell wall or viral capsid and oxidatively damages enzymes, the cell membrane and nucleic acids, disrupting the organism's metabolism and ability to reproduce. Ultraviolet light inactivates the same organisms physically rather than chemically: UV photons (typically around 254 nm) are absorbed directly by nucleic acids (DNA/RNA), forming pyrimidine dimers that block the organism's ability to replicate and repair itself, without adding any chemical residual to the water.

Because both mechanisms are dose- and time-dependent, engineered systems size the disinfection step from a dose (D)–time (T) relationship, though the two technologies define "dose" differently. For chlorine, the relevant product is $C\cdot T$ — residual disinfectant concentration $C$ multiplied by the contact time $T$ the water spends in a (usually baffled, serpentine) contact tank — and regulators publish minimum $C\cdot T$ tables by pathogen, temperature and pH; the effective $T$ used is the hydraulically conservative $T_{10}$, not the nominal volume/flow detention time, so that short-circuiting flow paths are still adequately treated. For UV, the analogous quantity is the fluence or UV dose $D = I\cdot t$ — irradiance $I$ (intensity, mW/cm²) multiplied by exposure time $t$ — delivered as water passes through the reactor past the lamps; UV reactors are validated (biodosimetrically) to guarantee a minimum delivered dose across the full range of flow and UV-transmittance conditions, since (unlike chlorine) UV leaves no residual to protect the water further downstream in the distribution system.

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