Water Treatment, Decay, Conservation and IPAT: 11-CS-3 Engineering Management, May 2019, Q4

Question 4: Water Treatment, Decay, Conservation and IPAT (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.

(a) Drinking-Water Treatment Flow Diagram

 Raw surface water (river/lake intake)
   ▼ [1 Screens] ── remove large debris: logs, leaves, fish, trash
   ▼ [2 Grit chamber] ── settle out sand, silt and grit that would abrade pumps
   ▼ [3 Coagulation/Flocculation] ── alum/ferric salt neutralizes charged colloids
   │                                 (clay, colour, organics); gentle mixing grows floc
   ▼ [4 Sedimentation] ── floc and suspended solids settle (turbidity, attached microbes)
   ▼ [5 Filtration] ── sand/anthracite media: fine particles, residual turbidity,
   │                   protozoan cysts (Giardia, Cryptosporidium)
   ▼ [6 Disinfection] ── chlorine/UV/ozone inactivates bacteria and viruses
   ▼ To distribution (chlorine residual maintained)

The order moves from the largest contaminants to the smallest: physical removal of coarse material, chemical conditioning so that colloids can settle, removal of the floc, polishing by filtration, and finally disinfection, which works best once the particles that shield micro-organisms are gone.

(b) Bacterial Decay Time

X = X_0e^{-k_dt}\ \Rightarrow\ t = \frac{1}{k_d}\ln\!\left(\frac{X_0}{X}\right) = \frac{1}{2.9}\ln\!\left(\frac{10^{6}}{10}\right) = \frac{\ln(10^{5})}{2.9} = \frac{11.513}{2.9} \approx \boxed{3.97\ \text{days}}

Integrating $dX/dt = -k_dX$ gives $X = X_0e^{-k_dt}$. A five-log (99.999%) reduction at $k_d = 2.9$ per day takes about 4 days; the count falls tenfold roughly every $\ln 10/2.9 \approx 0.79$ day.

(c) Turbidity

Turbidity is the cloudiness of water caused by suspended particles (clay, silt, organic matter, microbes) that scatter light; it is measured in NTU. It relates to microbial quality because particles harbour and transport micro-organisms and shield them from disinfection, so turbid water disinfects poorly; low filtered-water turbidity is therefore required for effective disinfection and serves as an indicator of pathogen (e.g. Giardia, Cryptosporidium) removal.

(d) Required Rate of Per-Capita Conservation

To hold total consumption constant, future population × future per-capita use must equal today's. Population after 20 yr:

P_{20} = 7500\,(1.015)^{20} = 7500\times1.3469 \approx 10{,}101\ \text{people}

Required future per-capita use:

q_{20} = q_0\,\frac{P_0}{P_{20}} = \frac{204}{1.3469} \approx 151.5\ \text{L/person}\cdot\text{day}

Per-capita use must fall linearly from 204 to 151.5 L/person·day (a drop of 52.5 L) over 20 years, so $q(t) = 204 - rt$ with

r = \frac{204-151.5}{20\ \text{yr}} \approx \boxed{2.63\ \text{L/person}\cdot\text{day per year}} \;=\; \frac{0.2575}{20} \approx \boxed{1.29\%\ \text{of today's use per year}}

Conserville must cut per-capita water use by about 2.6 L/person·day every year (≈1.3% of the current value each year) to offset its 1.5%/yr population growth and keep total consumption at today's level at year 20.

(e) The IPAT Equation

P = Population (number of people); A = Affluence (consumption per person, e.g. GDP or goods per capita); T = Technology (environmental impact per unit of consumption). The function is multiplicative:

I = P \times A \times T

Impact therefore grows with population and affluence and can be reduced only if the technology term (impact per unit of consumption) falls faster than the product P×A rises.

(f) Definition (any one)

Coliform: a group of rod-shaped bacteria (including E. coli) found in the intestines of warm-blooded animals, used as indicator organisms: their presence in drinking water signals fecal contamination and the likely presence of pathogens. (Alternatives: virtual water—the total volume of water used to produce a product or service; aquifer—a saturated, permeable geological formation that stores and transmits usable quantities of groundwater; groundwater—water held in the saturated zone beneath the water table, in the pores and fractures of soil and rock.)

Question 4 of 5: Water Treatment, Decay, Conservation and IPAT