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

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) — Second-order removal in a CSTR

Given. A completely stirred tank reactor removing ferrous iron by second-order oxidation, with the data below.

Find. The mean residence time $t=\tau$ (in minutes) and the CSTR volume $V$.

Approach. Write a steady-state mass balance on Fe²⁺ around the well-mixed tank; because the tank is completely mixed, the reaction proceeds everywhere at the exit concentration.

1. Steady-state CSTR balance. Accumulation is zero, so inflow minus outflow equals what reacts, evaluated at the exit concentration $C$: $$Q\,(C_0-C)=(-r)\,V=k\,C^{2}\,V.$$
2. Solve for the residence time. With $\tau=V/Q$, $$\tau=\frac{C_0-C}{k\,C^{2}}=\frac{(1.0\times10^{-5}-5.0\times10^{-6})}{5000\,(5.0\times10^{-6})^{2}}=\frac{5.0\times10^{-6}}{1.25\times10^{-7}}.$$ $$\boxed{\tau=40\ \text{s}=0.667\ \text{min}}$$
3. Size the reactor. $V=Q\,\tau=(1\ \text{m}^3/\text{s})(40\ \text{s})=\boxed{40\ \text{m}^3}.$
4. Contrast with the batch/PFR hint. The supplied integral $\tfrac{1}{C_0}-\tfrac{1}{C}=-kt$ is the batch (equivalently plug-flow) result, in which the reaction begins at the high inlet concentration: $t=\dfrac{1}{k}\left(\dfrac{1}{C}-\dfrac{1}{C_0}\right)=\dfrac{2.0\times10^{5}-1.0\times10^{5}}{5000}=20\ \text{s}$, giving $V=20\ \text{m}^3$. The CSTR needs exactly twice the plug-flow volume because a second-order rate is slowest at the low exit concentration, at which the entire mixed tank must operate.

Part (ii) — First-order conversion in a CSTR

Given. A pseudo-first-order decomposition of CHP in a CSTR: $k=4$ hr−¹, $Q=30$ m³/hr, target conversion $X=0.90$.

Find. The CSTR volume $V$ for 90% conversion at steady state.

Approach. Apply the steady-state CSTR balance for a first-order reaction, again evaluated at the exit concentration $C=C_0(1-X)$.

1. First-order CSTR balance. $Q\,C_0\,X=k\,C\,V=k\,C_0(1-X)\,V$, so the required residence time is $$\tau=\frac{X}{k\,(1-X)}=\frac{0.90}{4\,(1-0.90)}=\frac{0.90}{0.40}=2.25\ \text{hr}.$$
2. Size the reactor. $V=Q\,\tau=(30\ \text{m}^3/\text{hr})(2.25\ \text{hr})=\boxed{67.5\ \text{m}^3}.$
3. Plug-flow contrast. A PFR achieving the same 90% removal would need only $\tau_{PFR}=-\dfrac{\ln(1-X)}{k}=\dfrac{-\ln 0.10}{4}=0.576\ \text{hr}$, i.e. $V_{PFR}=17.3\ \text{m}^3$ — roughly a quarter of the CSTR, the penalty back-mixing imposes at high conversion.

Part (iii) — How a disinfectant inactivates pathogens

Disinfection removes the infectivity of waterborne pathogens (bacteria such as E. coli, protozoan cysts, viruses) rather than physically removing the organisms. Taking chlorine as the selected agent: when chlorine gas or hypochlorite is added to water it hydrolyses to hypochlorous acid, $\text{Cl}_2+\text{H}_2\text{O}\rightleftharpoons \text{HOCl}+\text{H}^{+}+\text{Cl}^{-}$, and HOCl dissociates to the hypochlorite ion, $\text{HOCl}\rightleftharpoons \text{H}^{+}+\text{OCl}^{-}$. The small, neutral HOCl molecule penetrates the cell wall and oxidises enzymes, membrane proteins and nucleic acids, destroying the metabolic machinery the organism needs to reproduce. Because HOCl is a far stronger disinfectant than OCl⁻, effectiveness falls sharply as pH rises above about 7.5.

The controlling design variable is the product of concentration and contact time, the CT concept: the log-inactivation achieved is governed by $CT=C\cdot t$, where $C$ is the residual disinfectant concentration (the "dose" $D$ available at the pathogen) and $t$ is the contact time in the disinfection basin. Chick–Watson kinetics express this as $\ln(N/N_0)=-\Lambda\,C^{\,n}\,t$, so a higher residual, a longer contact time, or both, increase the kill. In practice regulators publish required $CT$ values for each target log-removal, pathogen, temperature and pH (for example, Health Canada’s GCDWQ and the U.S. EPA Surface Water Treatment Rule tables). A plant meets its treatment goal by holding a defined residual through a baffled contact chamber sized for a known $t_{10}$ contact time; raising the dose lets the basin be smaller, while a larger basin lets the dose be lower — the two are traded against one another at a fixed $CT$. Chlorine additionally provides a lasting residual into the distribution system, which UV cannot, at the cost of taste, odour and disinfection by-product (trihalomethane) formation.