Question 3 of 5: Environmental Fate, Risk Assessment and Ecological Impact

Question 3: Environmental Fate, Risk Assessment and Ecological Impact (25 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.

Given.

Find. (a) Equilibrium chemical concentration in water, sediment, and biota at the discharge point. (b) Whether the drinking-water outflow meets a 10 ppt (by mass) standard. (c) The biota concentration relative to a 15 ppm LC50. (d) A discussion of the model's simplifications.

Approach. Treat the discharge point as a well-mixed, steady-state three-compartment equilibrium partitioning problem: assume the dissolved concentration C_w sets the sediment and biota concentrations via the given partition coefficients (C_sed=K_dC_w, C_biota=BCF·C_w), then close a total-mass balance across the river's daily mass flows of water, suspended sediment, and biota to solve for C_w. The river velocity is not needed for this equilibrium snapshot (it only sets travel time, relevant to part (d)); it is given context, not a required input.

1. Daily mass flow discharged to the river. After 80% WWTP removal: $$\dot m_{river}=4000\times(1-0.80)=\boxed{800\ \text{kg/day}}$$
2. Daily mass "flow" of each compartment carried by the river. Water mass flow (ρ=1000 kg/m³): $\dot M_w=800{,}000\ \text{m}^3/\text{day}\times1000\ \text{kg/m}^3=8.00\times10^8\ \text{kg/day}$. Suspended-sediment mass flow, from the 30 mg solids/kg-water loading (already on a per-kg-water basis): $$\dot M_{sed}=30\times10^{-6}\ \text{kg/kg}\times8.00\times10^8\ \text{kg/day}=\boxed{24{,}000\ \text{kg/day}}$$ Biota mass flow, converting 120 g/100 m³ to a per-kg-water basis (divide by the water density 1000 kg/m³ to move from "per m³" to "per kg," then by 1000 to move g→kg): $$\dot M_{biota}=\frac{120\ \text{g}}{100\ \text{m}^3}\times\frac{1}{1000\ \text{kg/m}^3}\times\frac{1\ \text{kg}}{1000\ \text{g}}\times8.00\times10^8\ \text{kg/day}=\boxed{960\ \text{kg/day}}$$
3. (a) Equilibrium partitioning — solve for C_w. The total discharged mass equals the sum held in each compartment (all expressed per kg of that compartment, tied together by K_d and BCF): $$800\times10^6\ \text{mg/day}=C_w\big(\dot M_w+K_d\dot M_{sed}+BCF\cdot\dot M_{biota}\big) =C_w\big(8.00\times10^8+110(24{,}000)+40(960)\big)$$ $$C_w=\frac{800\times10^6}{8.02678\times10^8}=\boxed{0.997\ \text{mg/kg water}\ (\approx1.00\ \text{mg/L})}$$ $$C_{sed}=110\times0.997=\boxed{109.6\ \text{mg/kg sediment}}\qquad C_{biota}=40\times0.997=\boxed{39.9\ \text{mg/kg biota}}$$ Despite the sizeable partition coefficients, sediment and biota together carry only about 0.33% and 0.005% of the total discharged mass respectively — their loadings (mass per unit river flow) are simply too small relative to the water mass flow for the partitioning to shift much mass out of solution; essentially all (≈99.67%) of the chemical remains dissolved.
4. (b) Drinking-water outflow vs. the 10 ppt standard. The DW plant removes 95% of the 800 kg/day arriving (assuming negligible loss/degradation over the 50 km reach, addressed further in part (d)), leaving 40 kg/day in the finished water, diluted in the same river-water mass flow: $$C_{DW,out}=\frac{40\times10^6\ \text{mg/day}}{8.00\times10^8\ \text{kg/day}}=0.0500\ \text{mg/kg} =\boxed{50{,}000\ \text{ppt}}$$ This is roughly 5000 times the 10 ppt regulated maximum. Performance modifications are clearly needed — meeting 10 ppt requires an overall (WWTP×DW combined) removal efficiency of 99.9998%, versus the 99.0% currently achieved (1−0.20×0.05). The fraction passing through both plants must fall from 1% to 0.0002% — a 5000-fold cut, between three and four orders of magnitude — so a major upgrade in combined treatment performance (upgrading one or both facilities, e.g. advanced oxidation, activated-carbon polishing, or membrane treatment at the DW plant) is required; neither facility can be assumed adequate as currently operated.
5. (c) Ecological impact vs. the 15 ppm LC50. The equilibrium biota concentration computed in step 3, 39.9 mg/kg (ppm), is compared directly against the stated LC50: $$\frac{C_{biota}}{LC_{50}}=\frac{39.9}{15}=\boxed{2.66}$$ Biota at or near the discharge point are predicted to bio-concentrate to roughly 2.7 times the LC50 — i.e., if local organisms are exposed long enough to approach this equilibrium partitioning, mortality could plausibly exceed 50% of the exposed population near the outfall. This is a serious acute-toxicity red flag even though it is a "worst case" (equilibrium, zero degradation) estimate; true risk likely decreases with distance downstream as dilution, degradation and further dispersion reduce C_w, but the near-field ecological risk indicated here warrants immediate further site-specific investigation (caged-organism bioassays, sediment/tissue sampling) rather than being dismissed on the basis of this screening-level number alone.

(d) Gross simplifications in this analysis

This screening calculation makes several simplifications that a comprehensive assessment would need to relax. It assumes instantaneous, complete equilibrium partitioning between water, sediment and biota with no kinetic limitation, when real sorption/uptake and depuration are rate-limited processes that may not reach equilibrium within the 50 km/≈23-hour travel time implied by the given 0.6 m/s river velocity (50 km ÷ 0.6 m/s ≈ 23.1 h one-way to the intake). It ignores chemical degradation entirely (biodegradation, hydrolysis, photolysis), any of which would reduce C_w before it reaches the drinking-water intake — a first-order decay term k combined with the travel time would need to be incorporated, and requires a fate half-life the problem does not supply. It treats the river as a single well-mixed compartment with no additional dilution from tributaries or groundwater exchange, no sediment resuspension/deposition dynamics (net burial vs. resuspension of contaminated sediment can be a long-term secondary source), and a single lumped biota compartment ignoring species-specific bioaccumulation, trophic-level biomagnification (predator fish typically carry higher body burdens than prey), and lipid-content variability in BCF. It uses a single acute LC50 endpoint without addressing chronic/sublethal toxicity (reproductive, developmental) at concentrations well below the LC50, and does not address human exposure pathways beyond direct drinking-water ingestion (fish consumption, recreational contact). A more comprehensive approach would couple a river fate-and-transport model (advection-dispersion with a first-order decay/sorption-desorption kinetic term, calibrated against field water/sediment/tissue sampling) to a full ecological and human-health risk assessment using species- and endpoint-specific toxicity data (e.g. NOEC/chronic values, not just acute LC50) and probabilistic (rather than single-point) exposure estimates.