Water Treatment, Decay, Conservation and BOD: 11-CS-3 Engineering Management, December 2019, Q4

Question 4: Water Treatment, Decay, Conservation and BOD (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
   ▼ [1 Screens] ── remove large debris, trash
   ▼ [2 Grit chamber] ── settle out sand and grit
   ▼ [3 Coagulation/Flocculation] ── alum neutralizes colloids; floc forms
   ▼ [4 Sedimentation] ── floc & suspended solids settle (turbidity, pathogens)
   ▼ [5 Filtration] ── sand/multimedia; fine particles, turbidity, protozoan cysts
   ▼ [6 Disinfection] ── chlorine/UV/ozone; inactivates bacteria & viruses
   ▼ To distribution (disinfectant residual maintained)

(b) Turbidity

Turbidity is the cloudiness of water caused by suspended particles (light scattering, NTU). It relates to microbial quality because particles harbour and transport micro-organisms and shield them from disinfection, so turbid water disinfects poorly—low turbidity is required for effective disinfection and as an indicator of particle/pathogen removal.

(c) Decay Coefficient

X = X_0 e^{-k_d t}\ \Rightarrow\ k_d = \frac{1}{t}\ln\!\left(\frac{X_0}{X}\right) = \frac{1}{6.0}\ln\!\left(\frac{10^{8}}{100}\right) = \frac{\ln(10^{6})}{6.0} = \frac{13.816}{6.0} \approx \boxed{2.30\ \text{day}^{-1}}

Integrating $dX/dt = -k_dX$ gives $X = X_0e^{-k_dt}$. A six-log (99.9999%) reduction in 6.0 days corresponds to $k_d \approx 2.30$ per day, so the count falls by a factor of 10 about every $\ln 10/k_d \approx 1.0$ day. (The paper prints 10⁸ cell/mL as the starting concentration; reading it as 10⁶ would give the wrong value of 1.54 per day.)

(d) Required Rate of Per-Capita Conservation

Population after 20 yr: $P_{20} = 72{,}000\,(1.012)^{20} = 72{,}000\times1.2694 \approx 91{,}400$. Required future per-capita use to hold total consumption constant:

q_{20} = q_0\,\frac{P_0}{P_{20}} = 114\times\frac{72{,}000}{91{,}400} \approx 89.8\ \text{L/person}\cdot\text{day}

Per-capita use must fall from 114 to 89.8 L (a drop of 24.2 L) linearly over 20 years. As a linear rate relative to the initial 114 L:

\text{rate} = \frac{(114-89.8)/114}{20\ \text{yr}} = \frac{0.2124}{20} \approx 0.0106\ \text{yr}^{-1} = \boxed{1.06\%\ \text{per year}}

In absolute terms the linear cut is $(114-89.8)/20 \approx 1.21$ L/person·day per year, i.e. $q(t) = 114\,(1 - 0.0106\,t)$ L/person·day. Conserva must cut per-capita water use by about 1.06% of the current value each year to offset its 1.2%/yr population growth and keep total consumption flat.

(e) Biochemical Oxygen Demand

BOD (biochemical oxygen demand) is the amount of dissolved oxygen consumed by micro-organisms as they biologically decompose the organic matter in water over a specified time. It measures the biodegradable organic material in the wastewater—its organic "strength." BOD is reduced in treatment in two main ways: primary sedimentation settles out settleable organic solids (removing ~30% of BOD), and secondary (biological) treatment—the activated-sludge process or trickling filters—uses micro-organisms to consume the dissolved and colloidal organic matter (removing most of the remaining BOD), after which the biomass is settled in a secondary clarifier.

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