16-Civ-A3 Elementary Environmental Engineering · December 2018
Question 1 of 7: Material Balances, Reaction Kinetics and Microbiology
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 16-Civ-A3 Elementary Environmental Engineering. Three hours; closed book with one candidate-prepared 8½ × 11 double-sided aid sheet; approved Casio or Sharp calculator only. Seven problems are printed, each worth 20 marks, and any five constitute a complete paper (maximum 100 marks). All seven are solved here, because the set is intended as a study resource rather than an exam script. Section marks are shown in brackets at the left margin of each question and are reproduced from the final-page Marking Scheme.
Reference texts.
Davis, M.L. & Cornwell, D.A., Introduction to Environmental Engineering, 5th ed., McGraw-Hill — material and energy balances, reactor kinetics, hardness, disinfection, water and wastewater unit processes.
Masters, G.M. & Ela, W.P., Introduction to Environmental Engineering and Science, 3rd ed., Pearson — mass/energy balances on power plants, air-emission estimation, growth–pollution linkages.
Crittenden, J.C. et al. (MWH), Water Treatment: Principles and Design, 3rd ed., Wiley — coagulation/flocculation, filtration, membranes, disinfection.
Engineers Canada / EGBC Code of Ethics; Canadian Council of Ministers of the Environment (CCME) and the Guidelines for Canadian Drinking Water Quality (GCDWQ), Health Canada — professional duty and water-quality benchmarks.
Question 1: Material Balances, Reaction Kinetics and Microbiology (20 marks)
Part (i) — Lead emission from a coal-fired power plant
Given. A coal-fired plant delivering electrical power with the fuel and lead data tabulated below.
Given data — Part (i)
Net electrical output, Pe
2000 MW = 2000 × 106 W
Overall (thermal-to-electric) efficiency, η
0.50
Higher heating value of coal, HHV
30 × 106 J/kg
Lead content of coal, wPb
5 µg/g = 5 × 10−6 kg Pb / kg coal
Find. The annual lead mass emitted to the atmosphere, in kg/yr, on an uncontrolled (all fuel lead reaches the stack) basis.
Approach. An energy balance converts the electrical demand to a fuel firing rate; a mass balance on the trace metal then carries the fuel lead through to the stack, and the result is annualised.
Fuel energy (firing) rate from the electrical output. The efficiency links electrical output to the thermal energy that must be released by the fuel:
$$\dot{Q}_{in}=\frac{P_e}{\eta}=\frac{2000\times10^{6}\ \text{W}}{0.50}=4000\times10^{6}\ \text{W}$$
Coal mass rate from the heating value. Dividing the firing rate by the energy released per kilogram of coal:
$$\dot{m}_{coal}=\frac{\dot{Q}_{in}}{\text{HHV}}=\frac{4000\times10^{6}\ \text{J/s}}{30\times10^{6}\ \text{J/kg}}=133.3\ \text{kg/s}$$
Lead mass rate. The lead is a conserved trace species carried in proportion to the coal burned:
$$\dot{m}_{Pb}=\dot{m}_{coal}\,w_{Pb}=133.3\times\left(5\times10^{-6}\right)=6.67\times10^{-4}\ \text{kg/s}$$
Annualise. With 3.1536 × 107 s in a year (365 d):
$$\boxed{\ \dot{m}_{Pb}=6.67\times10^{-4}\ \tfrac{\text{kg}}{\text{s}}\times3.1536\times10^{7}\ \tfrac{\text{s}}{\text{yr}}\approx2.10\times10^{4}\ \text{kg/yr}\ }$$
The plant would release on the order of 21,000 kg (21 tonnes) of lead per year if every atom of fuel lead escaped up the stack.
Check: this is the uncontrolled worst-case emission — it assumes 100% of the fuel lead partitions to the flue gas and passes the stack. In practice an electrostatic precipitator or fabric filter (often followed by a wet scrubber) captures roughly 90–99% of particulate-bound metals, so the emitted figure is one to two orders of magnitude lower; the balance above sizes the load that the control train must handle, not the amount that actually leaves the plant.
Part (ii) — Variable-mole gas-phase CSTR
Given. The reversible gas reaction 2A + B ⇌ C run in a CSTR at constant temperature and pressure, treated as effectively irreversible to the stated conversion.
Given data — Part (ii)
Rate law
−rA = kACA2CB
Rate constant, kA
100 dm6/mol2 (time unit assumed s−1)
Temperature / pressure
400 K / 10 atm
Feed composition
yA0 = 0.40, yB0 = 0.60
Target conversion of A, X
0.90
Find. The CSTR volume (equivalently the space-time τ) required for 90% conversion of A.
Steady-state CSTR: the well-mixed control volume runs at the exit composition, so the reaction rate is evaluated at CA, CB leaving the tank.
Approach. Because the total mole count changes as A and B combine into C at constant T and P, the volumetric flow expands; exit concentrations are written with the expansion factor ε, the rate is evaluated at the exit, and the algebraic CSTR mole balance gives the space-time directly.
Feed concentration of A. The total molar concentration follows from the ideal-gas law, and CA0 is its mole fraction:
$$C_{tot}=\frac{P}{RT}=\frac{10}{(0.08206)(400)}=0.3047\ \tfrac{\text{mol}}{\text{dm}^3},\qquad C_{A0}=y_{A0}C_{tot}=0.1219\ \tfrac{\text{mol}}{\text{dm}^3}$$
Expansion factor. On a per-mole-A basis the stoichiometry is A + ½B → ½C, so the change in total moles per mole of A reacted is δ = ½ − 1 − ½ = −1:
$$\varepsilon=y_{A0}\,\delta=0.40\,(-1)=-0.40$$
Rate at the exit condition. The well-mixed tank reacts everywhere at the outlet concentrations:
$$-r_A=k_A C_A^2 C_B=100\,(0.01904)^2(0.1999)=7.25\times10^{-3}\ \tfrac{\text{mol}}{\text{dm}^3\text{s}}$$
CSTR mole balance (space-time). For a CSTR, V = FA0X/(−rA), and dividing by the feed flow gives the space-time:
$$\boxed{\ \tau=\frac{C_{A0}X}{-r_A}=\frac{(0.1219)(0.90)}{7.25\times10^{-3}}=15.1\ \text{s}\ }$$
The required reactor volume scales with the throughput: V = τ v0. For a convenient basis of v0 = 1 dm3/s of feed, V = 15.1 dm3 (0.0151 m3); doubling the feed doubles the tank.
Check: two source ambiguities are handled per NOTE 1. (1) The rate constant is printed as “100 dm6/mol2” with no time unit; a third-order rate must carry s−1, so kA = 100 dm6 mol−2 s−1 is adopted. (2) No feed volumetric flow is given, so the well-posed answer is the space-time (residence time) τ = 15.1 s; the absolute volume follows for any specified throughput. As a check, an ideal PFR to the same conversion would need less volume than the CSTR because it is not forced to run entirely at the slow, low outlet concentration.
Part (iii) — Membrane removal of bacteriophage
Given. Raw water C0 = 106 mL−1; membrane-treated water C = 10 mL−1. Find. Percent reduction, log-reduction value (LRV), and a comparison of membrane vs chlorine disinfection.
A 5-log removal (99.999%) is a strong result — comparable to the credits regulators grant well-operated membrane barriers for protozoa and viruses. Advantages of the membrane: it is an absolute physical barrier that removes (not merely inactivates) protozoan cysts, bacteria and many viruses regardless of their chlorine resistance; it produces no chlorinated disinfection by-products (THMs, HAAs); it needs no chemical dosing, storage or handling; and it has a compact footprint with consistent, monitorable performance (integrity testing gives a direct log-removal check). Disadvantages: a membrane leaves no disinfectant residual, so a separate residual (chlorine or chloramine) is still required to protect the distribution system against regrowth and recontamination; it is subject to fouling and requires cleaning and higher energy and capital cost; a breach or pinhole passes water un-disinfected, so continuous integrity monitoring is essential; and it removes organisms by size exclusion rather than killing them, so the reject stream is a concentrated waste. Chlorine, by contrast, is cheap and provides a lasting residual but forms by-products and is weak against Cryptosporidium. In practice the two are complementary: membrane filtration for the physical/protozoan barrier, followed by a small chlorine (or UV) dose for residual protection.
