16-Civ-A3 Elementary Environmental Engineering · May 2018
Question 4 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, May 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.
Metcalf & Eddy / AECOM, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., McGraw-Hill — wastewater unit processes and plant upgrading.
Crittenden, J.C. et al. (MWH), Water Treatment: Principles and Design, 3rd ed., Wiley — coagulation/flocculation, filtration, membranes, adsorption, 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.
Check: Problem 4(ii) quotes a rate constant as “20 dm6/mol2” with no time unit, and states the fundamental reaction 2A + B ⇌ C. It is solved as a forward-rate second-in-A/first-in-B rate law with $k=20\ \text{dm}^{6}\,\text{mol}^{-2}\,\text{s}^{-1}$ (the only reading that makes $-r_A=k\,C_A^{2}C_B$ dimensionally a rate); the time unit is taken as seconds per NOTE 1. Because no feed flow rate is supplied, the well-posed deliverable is the flow-independent space-time $\tau$ (with $V=\tau\,v_0$ for any stated feed basis), not an absolute volume.
Question 4: Material Balances, Reaction Kinetics and Microbiology (20 marks)
Part (i) — Lead emission from a 2000 MW coal plant (6 marks)
Given. A coal-fired power plant is defined by the following data.
Given data — Problem 4(i)
Quantity
Symbol
Value
Net electrical output
Pe
2000 MW = 2.0 × 109 W
Overall (thermal) efficiency
η
0.50
Higher heating value of coal
HHV
30 × 106 J/kg
Lead content of coal
wPb
0.1 µg/g = 0.1 × 10−6 kg/kg
Find. The mass of lead emitted per year (kg/yr), on a worst-case mass balance in which all lead in the fuel leaves with the flue gas (no capture in a precipitator or scrubber).
Approach. An energy balance converts electrical output to a coal-burning rate; a species mass balance on lead then follows the Pb from fuel to stack, and the rate is annualised.
Energy balance → thermal input. Fuel heat is released at rate $\dot Q_{\text{in}}$, of which the fraction $\eta$ becomes electricity:
$$\dot Q_{\text{in}}=\frac{P_e}{\eta}=\frac{2000}{0.50}=4000\ \text{MW}=4.0\times10^{9}\ \text{W}.$$
Coal-burning rate. Dividing the heat rate by the heating value gives the fuel mass rate:
$$\dot m_{\text{coal}}=\frac{\dot Q_{\text{in}}}{\text{HHV}}=\frac{4.0\times10^{9}\ \text{W}}{30\times10^{6}\ \text{J/kg}}=133.3\ \text{kg/s}.$$
Lead mass balance. With a fuel lead fraction $w_{\text{Pb}}=0.1\times10^{-6}$ and all of it partitioning to the gas stream,
$$\dot m_{\text{Pb}}=\dot m_{\text{coal}}\,w_{\text{Pb}}=133.3\times0.1\times10^{-6}=1.333\times10^{-5}\ \text{kg/s}.$$
Annualise. Multiplying by the seconds in a year $(3.154\times10^{7}\ \text{s/yr})$:
$$\boxed{\dot m_{\text{Pb}}=1.333\times10^{-5}\times3.154\times10^{7}\approx 4.2\times10^{2}\ \text{kg/yr}\;(\approx 420\ \text{kg/yr}).}$$
Check: This is the uncontrolled emission — it assumes every gram of fuel lead reaches the stack. In practice most lead condenses onto fly ash and is captured by an electrostatic precipitator or fabric filter (typically 90–99%), so the actual stack emission is roughly 4–40 kg/yr. The mass balance sizes the challenge to the control device rather than the release to atmosphere.
Part (ii) — CSTR for the gas-phase reaction 2A + B ⇌ C (6 marks)
Given. A constant-T, constant-P gas-phase CSTR with a variable number of moles.
Given data — Problem 4(ii)
Quantity
Symbol
Value
Reaction / rate law
—
2A + B ⇌ C, $-r_A=k\,C_A^{2}C_B$
Temperature / pressure
T, P
500 K, 15 atm
Rate constant
k
20 dm6 mol−2 s−1
Feed composition
yA0, yB0
0.60 A, 0.40 B
Target conversion
X
0.99
Find. The CSTR volume (as the flow-independent space-time $\tau$, with $V=\tau v_0$) for 99% conversion of A.
Approach. Because total moles change, express $C_A$ and $C_B$ in terms of conversion with a mole-change factor $\varepsilon$, evaluate the rate at the exit composition (a CSTR operates at exit conditions), then apply the CSTR design equation.
Mole-change factor. On a per-mole-A basis the reaction is $A+\tfrac12 B\rightarrow \tfrac12 C$, so $\delta=\tfrac12-1-\tfrac12=-1$ and
$$\varepsilon=y_{A0}\,\delta=0.60\times(-1)=-0.60.$$
Inlet concentrations. From the ideal-gas law with $R=0.082057\ \text{dm}^3\text{atm}\,\text{mol}^{-1}\text{K}^{-1}$:
$$C_{T0}=\frac{P}{RT}=\frac{15}{0.082057\times500}=0.3656\ \tfrac{\text{mol}}{\text{dm}^3},\qquad C_{A0}=y_{A0}C_{T0}=0.2193\ \tfrac{\text{mol}}{\text{dm}^3}.$$
The feed ratio is $\theta_B=y_{B0}/y_{A0}=0.6667$.
Exit concentrations at X = 0.99. With $C_A=C_{A0}\dfrac{1-X}{1+\varepsilon X}$ and $C_B=C_{A0}\dfrac{\theta_B-\tfrac12 X}{1+\varepsilon X}$, and denominator $1+\varepsilon X = 1-0.594 = 0.406$:
$$C_A=0.2193\times\frac{0.01}{0.406}=5.40\times10^{-3}\ \tfrac{\text{mol}}{\text{dm}^3},\qquad C_B=0.2193\times\frac{0.6667-0.495}{0.406}=9.28\times10^{-2}\ \tfrac{\text{mol}}{\text{dm}^3}.$$
Rate at the exit. $$-r_A=k\,C_A^{2}C_B=20\times(5.40\times10^{-3})^{2}\times9.28\times10^{-2}=5.41\times10^{-5}\ \tfrac{\text{mol}}{\text{dm}^3\text{s}}.$$
CSTR design equation (space-time). With $\tau=V/v_0$ and $C_{A0}X=(-r_A)\tau$,
$$\boxed{\tau=\frac{C_{A0}\,X}{-r_A}=\frac{0.2193\times0.99}{5.41\times10^{-5}}\approx 4.0\times10^{3}\ \text{s}\;(\approx 67\ \text{min}),\qquad V=\tau\,v_0\approx 4.0\ \text{m}^3\ \text{per (dm}^3\text{/s) of feed.}}$$
Check: No feed flow rate is given, so an absolute volume cannot be stated; the space-time $\tau\approx4.0\times10^{3}$ s is the flow-independent result, and $V=\tau v_0$ (e.g. $\approx4.0\ \text{m}^3$ for a 1 dm3/s feed). The large $\tau$ is expected: a CSTR runs entirely at the low exit concentration where a second-order-in-A rate is slowest, so it needs far more volume than a plug-flow reactor for the same 99% conversion.
Part (iii) — UV disinfection: resistance order and 99.9% doses (8 marks)
A UV dose–response (fluence–response) curve plots log inactivation against UV dose; the steeper the curve, the less resistant the organism, because a smaller dose achieves the same log kill. Reading the four series in the figure at the 99.9% (3-log) level gives the ranking and doses below. The figure is reconstructed as an inline plot from the described data.
UV dose–response curves. The dashed line marks 3-log (99.9%) inactivation; where each curve crosses it gives the required dose.
Order of resistance (highest to lowest):B. subtilis spores > Rotavirus > Total coliform (wastewater) > E. coli. Spores and viruses are far more UV-resistant than vegetative bacteria; the bacterial spore is the most resistant of the four. The reads at the 99.9% (3-log) level are collected below.
Approximate UV dose for 99.9% (3-log) inactivation
Microorganism
Relative resistance
UV dose for 3-log (mJ/cm²)
B. subtilis spores
Highest
≈ 70–75
Rotavirus
High
≈ 30
Total coliform (wastewater)
Moderate (tails)
≈ 13
E. coli
Lowest
≈ 9
Check: Doses are read graphically from the described curve and are approximate. The total-coliform curve tails (approaches a plateau near 4.5–5 log), so pushing beyond 3-log for that population needs disproportionately more dose — a practical caution when a wastewater UV system must meet a strict effluent limit.
