23-CS-3 Sustainability, Engineering and the Environment · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2016 — 11-CS-3 Sustainability, Engineering and the Environment. Closed book; approved calculator permitted. Any four questions constitute a complete paper; all questions are of equal value (25 marks each).
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.
Suspended solids: undissolved particles held in the water (contribute to turbidity). BOD: dissolved oxygen consumed by microbes decomposing organic matter—a measure of organic pollution. Embodied (virtual) water: total water used to produce a product. Hydrologic cycle: the continuous movement of water via evaporation, condensation, precipitation, infiltration, and runoff. Water table: the top of the saturated zone. Aquitard: a low-permeability layer restricting groundwater flow. Vadose zone: the unsaturated zone above the water table. Potentiometric surface (printed as "potentiometer"; the hydrogeology term intended is the potentiometric surface): the imaginary surface to which water would rise in wells tapping a confined aquifer, i.e. the hydraulic head of that aquifer.
Turbidity measures the cloudiness of water caused by suspended particles (light scattering, NTU). It relates to microbial quality because particles harbour and transport micro-organisms and, critically, shield them from disinfection, so turbid water disinfects poorly. The treatment processes that reduce turbidity are coagulation, flocculation, sedimentation, and filtration—which together aggregate and remove the suspended particles, clarifying the water so that disinfection can be effective.
With $N_0/N = 40{,}000$ over $t = 4$ days, from $N = N_0 e^{-kt}$:
Check: $e^{-2.649\times4} = e^{-10.597} = 2.5\times10^{-5} = 1/40{,}000$. (If the decay is written in base-10 form, $N = N_0\,10^{-k't}$, the equivalent constant is $k' = \log_{10}(40{,}000)/4 = 1.15\ \text{day}^{-1}$; the natural-log constant $k = 2.65\ \text{day}^{-1}$ is the conventional answer.)
Chlorination: advantage—effective, inexpensive, provides a residual; disadvantage—forms toxic disinfection by-products and residual chlorine can harm aquatic life in the receiving water (often requiring dechlorination). Ultraviolet (UV): advantage—no chemical by-products and effective against a broad range of pathogens including protozoa; disadvantage—no residual protection, requires low-turbidity effluent and reliable power. (Ozone is a third: powerful but costly and residual-free.)
Current:
Future (30 yr): population $P = 5800\,(1.005)^{30} = 5800\times1.161 \approx 6{,}736$; per-capita $q = 360(1+0.002\times30) = 360\times1.06 = 381.6$ L/person/day:
