Water Treatment, Decay, Mass Balance and Demand: 11-CS-3 Engineering Management, December 2018, Q4

Question 4: Water Treatment, Decay, Mass Balance and Demand (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) Potable-Water Treatment Flow Diagram

 Raw surface water
      │
      ▼
[1 Screens] ── remove large debris, leaves, trash
      │
      ▼
[2 Grit chamber] ── settle out sand and grit
      │
      ▼
[3 Coagulation/Flocculation] ── alum neutralizes colloid charge; floc forms
      │
      ▼
[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
      │
      ▼
   To distribution (disinfectant residual maintained)

Screens remove gross debris; the grit chamber removes abrasive sand; coagulation-flocculation-sedimentation remove colloidal turbidity and attached pathogens; filtration polishes out fine particles and chlorine-resistant cysts; disinfection inactivates remaining bacteria and viruses.

(b) Decay Coefficient

k = \frac{1}{t}\ln\!\left(\frac{N_0}{N}\right) = \frac{1}{2.0}\ln\!\left(\frac{10^{5}}{10}\right) = \frac{\ln(10^{4})}{2.0} = \frac{9.210}{2.0} \approx \boxed{4.61\ \text{day}^{-1}}

(c) Selenium Mass Balance

 Upstream                  Bypass reach                     Downstream
 Q_u = 5.0 m3/s  ──►(withdrawal)── Q_b = 4.0 m3/s ──►(mixing)──► Q_d = 4.5 m3/s
 C_u = 0.0014 mg/L        │                              ▲        C_d = ?
                   Q_w = 1.0 m3/s                  Q_r = 0.5 m3/s
                          ▼                        C_r = 1.0 mg/L
                  [ Irrigated field ] ─────────────────┘
                          │
                   0.5 m3/s lost to ground and plants

Water balance: Q_b = Q_u − Q_w = 5.0 − 1.0 = 4.0 m³/s and Q_d = Q_b + Q_r = 4.0 + 0.5 = 4.5 m³/s. Selenium balance at the mixing point (steady state, conservative pollutant, so mass in = mass out; mg/s = m³/s × mg/L × 1000):

$$Q_b C_u + Q_r C_r = Q_d C_d$$

\dot m = (4.0)(0.0014)(1000) + (0.5)(1.0)(1000) = 5.6 + 500 = 505.6\ \text{mg/s}

C = \frac{505.6}{4.5\times1000} \approx 0.112\ \text{mg/L} \approx \boxed{0.11\ \text{ppm}}

The steady-state selenium concentration is about 0.11 ppm (1 mg/L = 1 ppm in dilute water), dominated by the concentrated irrigation run-off. The question also says the field is the "only source" of selenium; if the small upstream background is treated as zero, C_d = 500/4,500 = 0.111 mg/L — the same 0.11 ppm to two figures, about 80 times the upstream level.

(d) Future Water Demand

P = 5500\,(1.014)^{20} = 5500\times1.321 \approx 7{,}263\ \text{people}

q = 400\,(1 + 0.004\times20) = 400\times1.08 = 432\ \text{L/person/day}

Q = 7{,}263\times432 \approx 3{,}138{,}000\ \text{L/day} \approx \boxed{3.14\ \text{ML/day}}

(e) Annual Toilet-Flushing Water for a Household of Four

Per person toilet use = 25% × 204 = 51 L/day; household of four = 4 × 51 = 204 L/day:

V = 204\ \tfrac{\text{L}}{\text{day}} \times 365\ \tfrac{\text{day}}{\text{yr}} = 74{,}460\ \text{L/yr} \approx \boxed{74.5\ \text{m}^3/\text{yr}}

Question 4 of 5: Water Treatment, Decay, Mass Balance and Demand