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23-CS-3 Sustainability, Engineering and the Environment · December 2017

Question 4 of 5: Water Treatment, Bacterial Decay, Water Balance and IPAT

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

Notes on this paper

National Exams — December 2017 — 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, Water Balance 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) Surface-Water Treatment Flow Diagram

 Raw surface water
      │
      ▼
[1 Intake & Screening] ── removes large debris, trash
      │
      ▼
[2 Coagulation]  ── add alum; neutralizes colloid charge
      │
      ▼
[3 Flocculation] ── gentle mixing; colloids form settleable floc
      │
      ▼
[4 Sedimentation]── floc & suspended solids settle (turbidity, some pathogens)
      │
      ▼
[5 Filtration]   ── sand/multimedia; fine particles, turbidity, protozoan cysts
      │
      ▼
[6 Disinfection] ── chlorine/UV/ozone; inactivates bacteria & viruses
      │
      ▼
[7 Storage & Distribution] ── disinfectant residual maintained
      │
      ▼
   To consumers

(b) Bacterial Decay Calculation

$$t = \frac{1}{k}\ln\!\left(\frac{N_0}{N}\right) = \frac{1}{2.4}\ln\!\left(\frac{10^{6}}{1}\right) = \frac{13.816}{2.4} \approx \boxed{5.76\ \text{days}}$$

(c) Sustainable Groundwater Extraction (Water Balance)

The paper prints "[1 ha = 10,000 m³]"; this is a typographical slip for 1 ha = 10,000 m², since a hectare is an area. First, the annual rainfall volume on the farm (100 cm = 1 m depth):

$$V_{\text{rain}} = 800\ \text{ha}\times10{,}000\ \tfrac{\text{m}^2}{\text{ha}}\times1\ \text{m} = 8.0\times10^{6}\ \text{m}^3/\text{yr}$$

Half percolates to the aquifer, giving natural recharge $R = 4.0\times10^{6}\ \text{m}^3/\text{yr}$. Let the extraction be $W$. Of the extracted water, 75% is lost to evapotranspiration and 25% returns to the aquifer. For no change in groundwater level, aquifer inflow must equal outflow:

$$\underbrace{R + 0.25\,W}_{\text{inflow}} = \underbrace{W}_{\text{outflow}} \;\Rightarrow\; R = 0.75\,W \;\Rightarrow\; W = \frac{4.0\times10^{6}}{0.75}$$
$$W \approx \boxed{5.33\times10^{6}\ \text{m}^3/\text{yr}}$$

The farmer can sustainably extract about 5.3 million m³ per year: the net loss (0.75 W, the evapotranspired portion) then just equals the 4×10⁶ m³ natural recharge, leaving the water table stable.

(d) The IPAT Equation

The three terms are Population (P), Affluence (A) (consumption per person), and Technology (T) (environmental impact per unit of consumption). They combine by multiplication: $I = P \times A \times T$. Impact grows with more people and higher per-capita consumption but can be reduced by cleaner, more efficient technology.