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

Question 1: Mass Balance, Contaminant Partitioning and Disinfection (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 TP Concentration in the Lake and Outflow Stream

Given. A well-mixed lake at steady state, fed by an upstream river and losing water to both evaporation and an outflow stream, with the total phosphorus (TP) it carries subject to first-order decay:

Find. The steady-state TP concentration in the lake, and in the outflow stream.

Approach. Write an unsteady mass balance on TP for the completely-mixed lake, note that evaporation removes only water (zero TP mass flux), apply the steady-state condition, and solve for the lake concentration $C$ — which, because the lake is completely mixed, is also the outflow concentration $C_o$.

1. Write the unsteady-state TP mass balance. Evaporation is a water flux, not a TP flux, so it does not appear as a removal term for TP mass; only the outflow stream and in-lake decay remove TP, and because the lake is completely mixed its internal concentration $C$ equals the outflow concentration $C_o$: $$V\frac{dC}{dt} = Q_uC_u - Q_oC - kVC.$$
2. Apply the steady-state condition. At steady state $dC/dt=0$, so: $$0 = Q_uC_u - Q_oC - kVC \;\;\Rightarrow\;\; C(Q_o+kV) = Q_uC_u \;\;\Rightarrow\;\; C = \dfrac{Q_uC_u}{Q_o+kV}.$$
3. Substitute the given values and evaluate. Convert $C_u$ to mg/m$^3$ ($1$ mg/L $=1000$ mg/m$^3$) so all quantities share consistent units, then compute the decay term and the denominator: $$C_u = 15\ \text{mg/L} = 1.5\times10^4\ \text{mg/m}^3, \qquad kV = (0.06)(200{,}000) = 1.2\times10^4\ \text{m}^3/\text{yr}.$$ $$Q_o+kV = 3\times10^4+1.2\times10^4 = 4.2\times10^4\ \text{m}^3/\text{yr}.$$
4. Solve for the lake (and outflow) concentration. $$C = \dfrac{(10^5)(1.5\times10^4)}{4.2\times10^4} = \dfrac{1.5\times10^9}{4.2\times10^4} = 3.571\times10^4\ \text{mg/m}^3 = \boxed{35.7\ \text{mg/L}}.$$ Because the lake is completely mixed, $C_o = C = 35.7$ mg/L — the lake and the outflow stream share a single concentration.

(ii) Predicting Contaminant Partitioning Between Air, Water and Soil

When a contaminant such as benzene is released to the environment, it distributes itself across the air, water and soil/organic-carbon phases it contacts until it reaches (or approaches) thermodynamic equilibrium; that distribution is predicted from a small set of chemical-specific equilibrium partition coefficients rather than measured directly for every spill.

1. Air–water partitioning — Henry’s law. The equilibrium ratio of a chemical’s concentration in air to its concentration in water is the (dimensionless or unit) Henry’s law constant, $H = C_{air}/C_{water}$. Benzene has a moderately high $H$, so a meaningful fraction volatilizes from a surface spill or from shallow contaminated water into the air rather than staying dissolved.
2. Water–soil partitioning — the organic-carbon partition coefficient. The whole-soil partition coefficient is $K_d = K_{OC}\times f_{OC}$, where $K_{OC}$ is benzene’s (chemical-specific, roughly soil-independent) organic-carbon adsorption coefficient and $f_{OC}$ is the fraction of organic carbon in the specific soil. A soil with higher organic-carbon content sorbs proportionally more benzene for the same $K_{OC}$.
3. Source-phase behaviour — the octanol–water coefficient. Where benzene is present as a separate liquid (NAPL) at the spill source, its rate of dissolution into water and its tendency to sorb to organic matter are both predicted from its octanol–water partition coefficient $K_{OW}$ (a high $K_{OW}$ chemical is more hydrophobic and partitions more strongly to soil/NAPL than to water).
4. Calculating the equilibrium concentration in each medium. Given a known total contaminant mass $M$ released into a defined multi-media system (volumes/masses of air $V_a$, water $V_w$, soil $m_s$), the equilibrium concentrations are found by solving the mass-balance/partitioning system simultaneously: $$M = C_{water}\left(V_w + H\,V_a + K_d\,m_s\right)$$ so that $C_{water}$ is solved directly, and then $C_{air}=H\,C_{water}$ and $C_{soil}=K_d\,C_{water}$ follow from the equilibrium ratios above. This is the standard fugacity/equilibrium-partitioning approach used to predict where a spilled contaminant ends up.

(iii) Three Key Design Parameters for Drinking-Water Disinfection and Managing Distribution-System Regrowth

Effective disinfection system design turns on the applied dose, the contact time available to act on that dose, and the residual carried forward into the distribution system to prevent regrowth — all governed by Chick–Watson kinetics.

1. Disinfectant dose (net of demand). The applied chlorine (or other disinfectant) dose must first satisfy the chlorine demand — consumption by ammonia, organic matter and reduced inorganics — before any disinfecting residual concentration $C$ remains; the dose is sized as demand plus the target free/combined residual needed for the design contact time.
2. Contact time and the $Ct$ concept. Chick–Watson kinetics, $\ln(N/N_0) = -kC^nt$ ($n\approx1$ for chlorine), means inactivation is governed by the product $Ct$. Regulators specify a minimum $Ct$ for a target log-inactivation of the design pathogen at the plant’s temperature and pH; the contact basin is sized and baffled so the effective (short-circuiting-limited) $t_{10}$ still meets that $Ct$ at peak flow.
3. Target pathogen and indicator organism. Because Cryptosporidium oocysts are far more chlorine-resistant than bacteria or even Giardia cysts, the disinfectant/dose/contact-time combination must be sized for the most resistant organism realistically present (adding a UV or membrane barrier where Cryptosporidium risk is credible), while E. coli/coliforms remain the routine indicator used to verify day-to-day performance.

Bacterial regrowth in the distribution system is managed by carrying forward a persistent disinfectant residual — typically chloramines, formed by adding ammonia after free-chlorine contact, because they decay far more slowly through the pipe network than free chlorine — combined with minimizing water age (looped rather than dead-end mains, adequate turnover in storage reservoirs) and controlling biofilm-supporting organic/nutrient carryover (biological stability), since regrowth is ultimately a function of residual disinfectant, water age and available nutrients acting together.