Question 1 of 7: Mass Balance, Disinfection, and Contaminant Partitioning

Problem 1: Mass Balance, Disinfection, and Contaminant Partitioning (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 NH₃ concentration in the mine's industrial stream

Given.

Find. The NH₃ concentration C_u in the industrial (mine) inflow that sustains C₀ = 50 mg/L at steady state.

Approach. Write a steady-state mass balance on NH₃ around the lake as a continuously-stirred reactor (CSTR) with a first-order decay sink, then solve for the unknown inflow concentration.

1. Set up the steady-state CSTR balance. Accumulation = In − Out − Decay = 0: $$V\frac{dC}{dt} = Q_u C_u - Q_0 C_0 - kVC_0 = 0$$ Because the lake is modelled as well-mixed, the internal (and outflow) concentration is C₀ everywhere in the reactor.
2. Solve for C_u. Rearranging, $$C_u = \frac{Q_0 C_0 + kVC_0}{Q_u} = \frac{C_0\left(Q_0 + kV\right)}{Q_u}$$
3. Substitute the given values. $kV = (0.2\ \text{yr}^{-1})(1\times10^{5}\ \text{m}^3) = 2\times10^{4}\ \text{m}^3/\text{yr}$, so $$C_u = \frac{50\ \text{mg/L}\,\times\,\left(1\times10^{5} + 2\times10^{4}\right)\ \text{m}^3/\text{yr}}{1\times10^{5}\ \text{m}^3/\text{yr}} = 50\ \text{mg/L}\times 1.2 = \boxed{60\ \text{mg/L}}$$

(ii) Primary vs. secondary disinfectants

A primary disinfectant is applied at the treatment plant to achieve the inactivation credit (the "CT" dose) required to destroy pathogenic bacteria, viruses and protozoa (e.g., Giardia, Cryptosporidium) in the raw or partially treated surface water before it leaves the plant. Ozone (O₃), ultraviolet (UV) light, and free chlorine are the common primary disinfectants used in Canadian surface-water treatment plants; ozone and UV are favoured where the raw water carries high natural organic matter (NOM), because reacting free chlorine with NOM generates regulated disinfection by-products (DBPs) such as trihalomethanes (THMs) and haloacetic acids (HAAs). Ozone and UV, however, leave no lasting residual once the water enters the pipe network.

A secondary disinfectant is instead chosen for its ability to persist as a measurable, stable residual as the water travels through the distribution system, protecting against recontamination, biofilm regrowth and cross-connections between the treatment plant and the consumer's tap. Free chlorine or, more commonly on longer/looped Canadian distribution systems, chloramines (formed by adding ammonia after chlorine) serve this role because chloramines decay much more slowly than free chlorine and produce far fewer DBPs over long residence times, even though they are weaker, slower oxidants at the point of application. Canadian utilities following the Guidelines for Canadian Drinking Water Quality therefore commonly practice a two-stage strategy: a strong primary disinfectant (ozone/UV/free chlorine) achieves the CT inactivation credit at the plant, and a milder, persistent secondary disinfectant (chloramine or a small free-chlorine residual) is maintained throughout distribution, monitored at representative sampling points to ensure a detectable residual reaches every consumer.

(iii) Aqueous/solid-phase partitioning of 1,4-dichlorobenzene

Given.

Find. The fraction of contaminant mass residing in the aqueous phase versus sorbed to the solid (aquifer matrix) phase.

Approach. Use the supplied Kow–Koc correlation to get the organic-carbon partition coefficient, scale it by f_oc to get the bulk soil–water distribution coefficient K_p, then combine K_p with the aquifer's porosity and bulk density to split total contaminant mass between the pore water and the solid matrix.

1. Carbon sorption coefficient. $$\log(K_{oc}) = 0.69\log(K_{ow}) + 0.22 = 0.69(3.6) + 0.22 = 2.704$$ $$K_{oc} = 10^{2.704} \approx 506\ \text{L/kg}$$
2. Soil–water distribution coefficient. $$K_p = K_{oc}\times f_{oc} = 506\ \text{L/kg} \times 0.002 = \boxed{1.01\ \text{L/kg}\ (\approx 1.01\ \text{mL/g})}$$ A K_p just above 1 mL/g indicates a moderately hydrophobic compound (consistent with log K_ow = 3.6) that partitions preferentially, but not overwhelmingly, onto the solid phase.
3. Convert K_p to a mass-fraction split. For a unit bulk volume of saturated aquifer, the dissolved mass per volume is $nC_{aq}$ and the sorbed mass per volume is $\rho_b K_p C_{aq}$, where $\rho_b$ is the dry bulk density. Taking a typical mineral particle density $\rho_s = 2.65\ \text{g/cm}^3$ (not printed on the exam — flagged below) with the given porosity, $\rho_b = \rho_s(1-n) = 2.65(0.5) = 1.325\ \text{g/cm}^3$. The fractions are $$F_{water} = \frac{n}{n + \rho_b K_p},\qquad F_{solid} = \frac{\rho_b K_p}{n + \rho_b K_p}$$ $$F_{water} = \frac{0.5}{0.5 + (1.325)(1.01)} = \frac{0.5}{1.840} = \boxed{0.272\ (27.2\%)}$$ $$F_{solid} = 1 - F_{water} = \boxed{0.728\ (72.8\%)}$$