23-Chem-B10 Life Cycle Assessment (LCA) · December 2013
Question 3 of 5: Environmental Fate, Risk Assessment and Ecological Impact
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-Chem-B10, Life Cycle Assessment (LCA) — December 2013. 3 hours, Closed-Book
Exam (Casio/Sharp approved calculator and one double-sided aid sheet permitted). Question 1 is
mandatory; any three (3) of the remaining four (Questions 2–5) constitute a complete 100-mark
paper, and only the first four questions as they appear in the answer book are marked. All five
questions (and, in Question 5, all five sub-parts) are solved below for completeness.
Reference texts: Baumann & Tillman, The Hitch Hiker's Guide to LCA;
Graedel & Allenby, Industrial Ecology and Sustainable Engineering; Kemp, Pinch
Analysis and Process Integration, 2nd ed.; Mackay, Multimedia Environmental Models: The
Fugacity Approach, 2nd ed.; Davis & Cornwell, Introduction to Environmental
Engineering.
Drinking-water (DW) plant removal of remaining chemical
90%
Distance to DW intake
50 km downriver
River flow rate
900,000 m³/day
River velocity
0.5 m/s
Suspended-sediment loading
25 mg solids / kg water
Biota loading
125 g biota / 100 m³ water
Soil/sediment–water partition coeff. Kd
100 kg/kg
Bio-concentration factor (BCF)
50 kg/kg
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 Cw sets the sediment
and biota concentrations via the given partition coefficients (Csed=KdCw,
Cbiota=BCF·Cw), then close a total-mass balance across the river's daily
mass flows of water, suspended sediment, and biota to solve for Cw. 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.
Daily mass flow discharged to the river. After 85% WWTP removal:
$$\dot m_{river}=3500\times(1-0.85)=\boxed{525\ \text{kg/day}}$$
Daily mass "flow" of each compartment carried by the river. Water mass flow (ρ=1000
kg/m³): $\dot M_w=900{,}000\ \text{m}^3/\text{day}\times1000\ \text{kg/m}^3=9.00\times10^8\ \text{kg/day}$.
Suspended-sediment mass flow, from the 25 mg solids/kg-water loading:
$$\dot M_{sed}=25\times10^{-6}\ \text{kg/kg}\times9.00\times10^8\ \text{kg/day}=\boxed{22{,}500\ \text{kg/day}}$$
Biota mass flow, converting 125 g/100 m³ to a per-kg-water basis (1 m³ water ≡ 1000 kg):
$$\dot M_{biota}=\frac{125\ \text{g}}{100\ \text{m}^3}\times\frac{1\ \text{m}^3}{1000\ \text{kg}}\times9.00\times10^8\ \text{kg/day}=\boxed{1125\ \text{kg/day}}$$
(a) Equilibrium partitioning — solve for Cw. The total discharged
mass equals the sum held in each compartment (all expressed per kg of that compartment, tied together
by Kd and BCF):
$$525\times10^6\ \text{mg/day}=C_w\big(\dot M_w+K_d\dot M_{sed}+BCF\cdot\dot M_{biota}\big)
=C_w\big(9.00\times10^8+100(22{,}500)+50(1125)\big)$$
$$C_w=\frac{525\times10^6}{9.023\times10^8}=\boxed{0.582\ \text{mg/kg water}\ (\approx0.58\ \text{mg/L})}$$
$$C_{sed}=100\times0.582=\boxed{58.2\ \text{mg/kg sediment}}\qquad C_{biota}=50\times0.582=\boxed{29.1\ \text{mg/kg biota}}$$
Despite the sizeable partition coefficients, sediment and biota carry only 0.25% and 0.006% of the
total chemical 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 (≈99.7%) of the chemical remains dissolved.
(b) Drinking-water outflow vs. the 10 ppt standard. The DW plant removes 90% of the
525 kg/day arriving (assuming negligible loss/degradation over the 50 km reach, addressed further in
part (d)), leaving 52.5 kg/day in the finished water, diluted in the same river-water mass flow:
$$C_{DW,out}=\frac{52.5\times10^6\ \text{mg/day}}{9.00\times10^8\ \text{kg/day}}=0.0583\ \text{mg/kg}\ (5.83\times10^{-8}\ \text{kg/kg})
=\boxed{58{,}333\ \text{ppt}}$$
This is roughly 5800 times the 10 ppt regulated maximum. Performance
modifications are clearly needed — meeting 10 ppt requires an overall (WWTP×DW combined)
removal efficiency of 99.9997%, versus the 98.5% currently achieved (1−0.15×0.10). The fraction of
the chemical passing both plants must fall from 1.5% to about 0.00026% — a roughly 5800-fold (nearly
four-orders-of-magnitude) improvement 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.
(c) Ecological impact vs. the 15 ppm LC50. The equilibrium biota concentration
computed in step 3, 29.1 mg/kg (ppm), is compared directly against the stated LC50:
$$\frac{C_{biota}}{LC_{50}}=\frac{29.1}{15}=\boxed{1.94}$$
Biota at or near the discharge point are predicted to bio-concentrate to roughly twice 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; the true risk likely decreases with distance downstream as dilution, degradation and further
dispersion reduce Cw, 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.
Quantity
Value
Mass discharged to river (post-WWTP)
525 kg/day
Cwater at discharge point
0.582 mg/kg (≈0.58 mg/L)
Csediment at discharge point
58.2 mg/kg
Cbiota at discharge point
29.1 mg/kg (ppm)
Drinking-water outflow concentration
58,333 ppt (≈5800× the 10 ppt limit)
Overall removal needed vs. currently achieved
99.9997% needed vs. 98.5% achieved
Cbiota / LC50
1.94 (≈2× LC50 — acute risk near outfall)
(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/≈28-hour travel time implied by the given 0.5 m/s
river velocity (50,000 m ÷ 0.5 m/s = 100,000 s ≈ 28 h to the intake). It ignores chemical
degradation entirely (biodegradation, hydrolysis, photolysis), any of which would reduce Cw
before it reaches the drinking-water intake — a first-order decay term k 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.