Material Balances, Reaction Kinetics and Microbiology: 16-Civ-A3 Elementary Environmental Engineering, December 2017, Q1

Question 1: Material Balances, Reaction Kinetics 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.

Part (i) — Arsenic emission from a 1200 MW coal plant (6 marks)

Given. A pulverised-coal power plant delivering net electrical power with the following data.

Given data — Problem 1(i)
Quantity	Symbol	Value
Net electrical output	P_e	1200 MW = 1.2 × 10⁹ W
Overall (thermal) efficiency	η	0.40
Higher heating value of coal	HHV	29 × 10⁶ J/kg
Arsenic content of coal	w_As	0.1 µg/g = 0.1 × 10⁻⁶ kg/kg

Find. The mass of arsenic emitted per year (kg/yr), assuming a worst-case mass balance in which all arsenic 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 arsenic then follows the As from the fuel to the stack.

Energy balance → thermal input. The plant burns fuel to release heat at rate $\dot Q_{\text{in}}$, of which the fraction η becomes electricity: $$\dot Q_{\text{in}}=\frac{P_e}{\eta}=\frac{1200}{0.40}=3000\ \text{MW}=3.0\times10^{9}\ \text{W}.$$
Coal-burning rate. Dividing the heat rate by the heating value gives the fuel mass rate: $$\dot m_{\text{coal}}=\frac{\dot Q_{\text{in}}}{\text{HHV}}=\frac{3.0\times10^{9}\ \text{W}}{29\times10^{6}\ \text{J/kg}}=103.4\ \text{kg/s}.$$
Arsenic mass balance. With a fuel arsenic fraction $w_{\text{As}}=0.1\times10^{-6}$ and all of it partitioning to the gas stream, $$\dot m_{\text{As}}=\dot m_{\text{coal}}\,w_{\text{As}}=103.4\times0.1\times10^{-6}=1.034\times10^{-5}\ \text{kg/s}.$$
Annualise. Multiplying by the seconds in a year $(3.154\times10^{7}\ \text{s/yr})$: $$\boxed{\dot m_{\text{As}}=1.034\times10^{-5}\times3.154\times10^{7}\approx 326\ \text{kg/yr}.}$$

Check: This is the uncontrolled emission — it assumes every gram of arsenic in the coal reaches the stack. In practice most arsenic condenses onto fly ash and is captured by an electrostatic precipitator or fabric filter (typically 90–99%), so the actual stack emission is one to two orders of magnitude smaller. The 326 kg/yr figure is therefore the balance-based upper bound that sizes the control requirement.

Results — Problem 1(i)
Quantity	Value
Thermal input	3000 MW
Coal-burning rate	103.4 kg/s (≈ 3.26 × 10⁶ t/yr)
Uncontrolled arsenic emission	≈ 326 kg/yr

Part (ii) — CSTR volume for a gas-phase reaction (6 marks)

Given. The gas-phase reaction $2A+B\rightleftharpoons C$, rate law $-r_A=k_AC_A^2C_B$, carried out isothermally at $T=500\ \text{K}$ and constant $P=15\ \text{atm}$, with $k_A=10\ \text{dm}^6\,\text{mol}^{-2}$ (third order overall) and an equimolar feed (50% A, 50% B). Target conversion $X=0.95$.

Find. The CSTR (mixed-flow) reactor space-time and volume required for 95% conversion of A.

Approach. Because the total number of moles changes, work on a per-mole-of-A basis to get the expansion factor ε, express the exit concentrations of A and B in terms of conversion, evaluate the rate at the exit, and apply the CSTR design equation.

Stoichiometry and expansion factor. On a per-mole-A basis, $A+\tfrac12 B\rightarrow \tfrac12 C$, so the change in total moles per mole of A is $\delta=\tfrac12-1-\tfrac12=-1$. With inlet mole fraction of A, $y_{A0}=0.5$, $$\varepsilon=y_{A0}\,\delta=0.5\times(-1)=-0.5.$$
Inlet concentration. The total molar concentration is $C_{T0}=P/RT$, and A is half of it: $$C_{A0}=y_{A0}\frac{P}{RT}=0.5\times\frac{15}{0.0821\times500}=0.183\ \text{mol/dm}^3.$$
Exit concentrations. For a variable-density gas at constant T, P, with $\theta_B=1$ and $b/a=\tfrac12$: $$C_A=C_{A0}\frac{1-X}{1+\varepsilon X},\qquad C_B=C_{A0}\frac{1-\tfrac12 X}{1+\varepsilon X}.$$ At $X=0.95$ the denominator is $1+\varepsilon X=1-0.475=0.525$, giving $C_A=0.0174$ mol/dm³ and, since the numerator for B is also 0.525, $C_B=C_{A0}=0.183$ mol/dm³.
Rate at the exit. A CSTR operates at the exit composition, so $$-r_A=k_AC_A^2C_B=10\,(0.0174)^2(0.183)=5.53\times10^{-4}\ \text{mol}\,\text{dm}^{-3}\,\text{s}^{-1}.$$
CSTR design equation. The space-time is independent of the (unstated) feed rate: $$\boxed{\tau=\frac{C_{A0}X}{-r_A}=\frac{0.183\times0.95}{5.53\times10^{-4}}\approx 314\ \text{s}\ (\approx 5.2\ \text{min}).}$$ The volume follows once the volumetric feed $v_0$ is fixed: $V=\tau\,v_0$. For an illustrative basis $v_0=1\ \text{dm}^3/\text{s}$, $V\approx 314\ \text{dm}^3=0.31\ \text{m}^3$.

Check: The question gives no feed flow rate, so the reactor size cannot be an absolute number; the well-posed result is the space-time τ ≈ 314 s, from which $V=\tau v_0$. Per NOTE 1, the volume above adopts $v_0=1\ \text{dm}^3/\text{s}$. Also, $k_A$ is quoted only as "dm⁶/mol²" with no time unit; a third-order rate requires $\text{dm}^6\,\text{mol}^{-2}\,\text{s}^{-1}$, taken here as per second.

Results — Problem 1(ii)
Quantity	Value
Expansion factor ε	−0.5
C_A0	0.183 mol/dm³
Exit rate −r_A	5.53 × 10⁻⁴ mol dm⁻³ s⁻¹
Space-time τ	≈ 314 s (5.2 min)
Volume (v₀ = 1 dm³/s)	≈ 0.31 m³

Part (iii) — Chlorine speciation and microbial resistance (8 marks)

[Figure not reproduced: Figure A (reconstructed from the source). Concentration versus contact time for 99% kill of E. coli by three forms of chlorine; a lower line means a smaller C·t c product is needed, i.e. greater potency. See the official exam paper.]

Chlorine exists in water as three species whose disinfecting power differs by more than an order of magnitude. When chlorine gas dissolves it hydrolyses to hypochlorous acid (HOCl), which partly dissociates to the hypochlorite ion (OCl⁻) as pH rises above about 7.5; in the presence of ammonia, chlorine instead forms chloramines (NH₂Cl). Figure A plots the concentration–contact-time combinations that achieve a fixed 99% kill of E. coli: because the axes are the two factors in the C·t_c product, a line lying lower and to the left represents a smaller required dose and hence a more effective disinfectant.

The ordering in Figure A is unambiguous. HOCl is the most effective form — it is the lowest line, demanding the least concentration and contact time. HOCl is a small, electrically neutral molecule that penetrates the negatively charged bacterial cell wall readily and attacks intracellular enzymes and nucleic acids. OCl⁻ lies above it: although it is the same oxidant chemically, its negative charge is repelled by the cell surface, so it is roughly two orders of magnitude weaker. Chloramine (NH₂Cl) is the least effective, sitting highest on the plot; it is a much weaker oxidant and reacts slowly, which is precisely why it is valued as a persistent secondary (residual) disinfectant in distribution systems rather than as a primary one. The practical consequence is that free-chlorine disinfection is operated at pH below about 7.5 to keep the fraction present as HOCl high.

[Figure not reproduced: Figure B (reconstructed from the source). HOCl concentration versus contact time for 99% kill of E. coli and three enteric viruses; a higher line means the organism demands a larger C·t c and is therefore more resistant. See the official exam paper.]

Figure B fixes the disinfectant as HOCl and varies the target organism. Now a higher line means a larger C·t_c is needed for the same 99% kill, i.e. a more resistant organism. The vegetative bacterium E. coli is inactivated most easily (a low line), while the enteric viruses require substantially more exposure because their protein capsids shield the genome and they lack the metabolic machinery an oxidant most readily disrupts. Of the organisms shown, Coxsackie virus A₂ is the most resistant to HOCl disinfection — it plots highest, above poliovirus I, adenovirus and E. coli. This is the design basis for the C·t tables in drinking-water regulations: the disinfection dose is set by the most resistant pathogen of concern (in practice, protozoan cysts such as Giardia and Cryptosporidium, which are far more resistant still than any virus here), not by the readily killed indicator bacterium.

Topic. Chemical disinfection kinetics — the Chick–Watson C·t concept and chlorine speciation.

Key relations. Chick–Watson: $\ln(N/N_0)=-\Lambda\,C^{\,n}t$; for $n\approx1$ a fixed kill requires a fixed $C\!\cdot\!t$. Speciation: $\text{Cl}_2+\text{H}_2\text{O}\rightarrow \text{HOCl}+\text{H}^++\text{Cl}^-$, $\text{HOCl}\rightleftharpoons \text{H}^++\text{OCl}^-$ (pK_a ≈ 7.5).

Why this works. On log–log axes a constant C·t product plots as a straight line of slope −1; parallel lines that are shifted vertically therefore correspond to different C·t constants, so simply reading which line is lower (Figure A) or higher (Figure B) ranks potency or resistance directly. Neutral HOCl out-performs anionic OCl⁻ and weak chloramine because charge and oxidising strength govern how fast the disinfectant reaches and damages the cell.

Common pitfall. Confusing the two figures — in Figure A a low line is good (potent disinfectant), whereas in Figure B a high line is bad (resistant organism). Also, assuming more chlorine always helps: raising pH converts HOCl to the far weaker OCl⁻, so dose and pH must be considered together.

Source. Davis & Cornwell, Introduction to Environmental Engineering, disinfection chapter; GCDWQ C·t requirements for chlorine.

Question 1 of 7: Material Balances, Reaction Kinetics and Microbiology

Question 1: Material Balances, Reaction Kinetics and Microbiology (20 marks)

Part (i) — Arsenic emission from a 1200 MW coal plant (6 marks)

Part (ii) — CSTR volume for a gas-phase reaction (6 marks)

Part (iii) — Chlorine speciation and microbial resistance (8 marks)