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11-CS-3 Engineering Management · December 2013

Question 4 of 5: Water Treatment, Bacterial Decay and Demand Forecasting

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2013 — 11-CS-3 Sustainability, Engineering and the Environment. Open book; non-communicating calculator permitted. Any four questions constitute a complete paper; all questions are of equal value (25 marks each).

Question 4: Water Treatment, Bacterial Decay and Demand Forecasting (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) Surface-Water Treatment Flow Diagram

 Raw surface water
      │
      ▼
[1 Intake & Screening] ── removes large debris, leaves, fish, trash
      │
      ▼
[2 Coagulation]  ── add alum/coagulant; neutralizes charge on colloids
      │
      ▼
[3 Flocculation] ── gentle mixing; colloids aggregate into settleable floc
      │
      ▼
[4 Sedimentation]── floc & suspended solids settle out (turbidity, some pathogens)
      │
      ▼
[5 Filtration]   ── sand/multimedia; removes fine particles, remaining turbidity,
      │              protozoan cysts (Giardia, Cryptosporidium)
      ▼
[6 Disinfection] ── chlorine / UV / ozone; inactivates bacteria & viruses
      │              (+ pH adjustment, fluoridation as required)
      ▼
[7 Clear well / Storage & Distribution] ── disinfectant residual maintained
      │
      ▼
   To consumers

Screening removes gross debris; coagulation–flocculation–sedimentation remove colloidal particles and turbidity (and attached pathogens); filtration removes fine particles and disinfection-resistant cysts; disinfection inactivates remaining bacteria and viruses, with a residual carried into distribution to guard against re-contamination.

(b) Bacterial Decay Calculation

$$t = \frac{1}{k}\ln\!\left(\frac{N_0}{N}\right) = \frac{1}{2.5}\ln\!\left(\frac{10^{7}}{10}\right) = \frac{\ln(10^{6})}{2.5} = \frac{13.816}{2.5} \approx \boxed{5.53\ \text{days}}$$

About 5.5 days are required for the viable bacteria to fall from 10⁷ to 10 cell/mL.

(c) Turbidity

Turbidity measures the cloudiness of water caused by suspended particles—it quantifies how much the particles scatter light (reported in nephelometric turbidity units, NTU). It is related to microbial quality in two ways: the particles can shelter and transport micro-organisms, so high turbidity often signals higher microbial contamination; and, critically, suspended particles shield micro-organisms from disinfection, so turbid water disinfects poorly. Low turbidity is therefore required both as an indicator of particle (and pathogen) removal and to ensure disinfection can reach the pathogens.

(d) Future Water-Demand Forecast

Population (exponential growth, 1.5%/yr for 20 yr):

$$P = P_0(1+r)^n = 5000\,(1.015)^{20} = 5000 \times 1.347 \approx 6{,}735\ \text{people}$$

Per-capita demand (linear growth, 0.5%/yr of the initial 400 L for 20 yr):

$$q = 400\,\bigl(1 + 0.005 \times 20\bigr) = 400 \times 1.10 = 440\ \text{L/person/day}$$

Total future demand:

$$Q = P \times q = 6{,}735 \times 440 \approx 2{,}963{,}000\ \text{L/day} \approx \boxed{2.96\ \text{ML/day}}$$

The town should plan for roughly 3 ML/day at the end of the 20-year design period.