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18-Env-A4 Water and Wastewater Engineering · May 2015

Question 3 of 5: pH in Water Treatment; BOD Test Calculation

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

Notes on this paper

National Exams / EGBC — May 2015 — 04-ENV-A4 Water and Wastewater Engineering. Three-hour exam; Question 1 is compulsory (25 marks) and any three of the remaining four questions are required (25 marks each); all five are solved below for completeness. Closed book, one double-sided aid sheet permitted, approved calculator permitted.

Reference texts: Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — nitrogen and solids characterization, BOD test theory, nitrification/alkalinity, disinfection chemistry, activated-sludge and sludge-processing design; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — the Streeter–Phelps oxygen sag, pH and coagulation–flocculation chemistry, jar testing; MWH's Water Treatment: Principles and Design (3rd ed.) — granular filtration (headloss, backwash) and chemical phosphorus removal.

Question 3: pH in Water Treatment; BOD Test Calculation (25 marks: a 10, b 15)

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. pH: definition and significance for disinfection and coagulation–flocculation (10 marks)

pH is the negative base-10 logarithm of the hydrogen-ion activity, $\text{pH}=-\log_{10}[H^+]$, a dimensionless measure on a 0–14 scale of how acidic (pH<7) or basic (pH>7) a water is at 25 °C. For disinfection, pH controls the speciation of aqueous chlorine between hypochlorous acid ($HOCl$, $pK_a\approx7.5$ at 20 °C) and hypochlorite ion ($OCl^-$); $HOCl$ is roughly 80–100 times more effective a biocide than $OCl^-$ because its neutral charge lets it penetrate microbial cell walls far more readily, so disinfection is markedly less efficient at high pH (where $OCl^-$ dominates) and a higher CT (concentration × time) is needed to compensate. For coagulation–flocculation, each coagulant has a narrow optimum pH range where its metal-hydroxide floc has minimum solubility and forms an effective, rapidly settling precipitate through charge neutralization and sweep-floc capture (roughly pH 5.5–7.5 for alum, pH 5–8.5 for ferric salts); outside that band the metal stays in soluble form or forms a fragile floc, and coagulant is wasted without achieving turbidity/colour removal. Operators therefore control (and often pre-adjust with lime, soda ash, or acid) pH ahead of both disinfection and coagulation to keep each process in its effective window.

b. BOD test — 3-day, 5-day and ultimate BOD (15 marks)

Given. A raw sewage sample is diluted in a standard BOD bottle and incubated at 20 °C; 5% of the observed 3-day oxygen depletion is attributed to the seed organisms already present in the sample (not the sewage itself).

QuantityValue
Sample volume $V_s$5 mL
Total bottle (diluted) volume $V_t$300 mL
Initial DO, $DO_i$7.5 mg/L
DO after 3 days, $DO_f$4.3 mg/L
Incubation time, temperature3 days at 20 °C
Seed contribution to depletion5%

Find. The 3-day BOD ($\text{BOD}_3$), 5-day BOD ($\text{BOD}_5$) and ultimate BOD ($\text{BOD}_u$) of the undiluted sewage sample.

Check: the exam gives only a single 3-day dilution-test reading, with no deoxygenation rate constant $k_1$. To extrapolate from $\text{BOD}_3$ to $\text{BOD}_5$ and $\text{BOD}_u$, a typical base-10 rate constant for raw domestic sewage at 20 °C, $k_1=0.10\ \text{day}^{-1}$ (Metcalf & Eddy / Davis & Cornwell typical range 0.05–0.30 day⁻¹), is assumed and stated explicitly here.

Approach. Compute the dilution fraction, correct the observed DO depletion for the seed's own contribution, scale by the dilution to get $\text{BOD}_3$ of the undiluted sample, then use the assumed first-order BOD-exertion model $\text{BOD}_t=\text{BOD}_u(1-10^{-k_1t})$ to back out $\text{BOD}_u$ and forward-compute $\text{BOD}_5$.

  1. Dilution fraction. $P=\dfrac{V_s}{V_t}=\dfrac{5}{300}=0.01667.$
  2. Seed-corrected oxygen depletion in the bottle. Observed depletion $=DO_i-DO_f=7.5-4.3=3.2$ mg/L; with 5% of this attributed to the seed, the sewage's own depletion is $3.2\times(1-0.05)=3.04$ mg/L.
  3. 3-day BOD of the undiluted sample. Dividing by the dilution fraction scales the bottle reading up to the full-strength sample: $$\text{BOD}_3=\frac{3.04}{0.01667}=\boxed{182.4\ \text{mg/L}}.$$
  4. Ultimate BOD from the assumed rate constant. With $k_1=0.10\ \text{day}^{-1}$ (base-10) and the standard exertion curve $\text{BOD}_t=\text{BOD}_u(1-10^{-k_1t})$, at $t=3$: $1-10^{-0.30}=0.4988$, so $$\text{BOD}_u=\frac{182.4}{0.4988}=\boxed{365.7\ \text{mg/L}}.$$
  5. 5-day BOD. Applying the same curve at $t=5$: $1-10^{-0.50}=0.6838$, so $$\text{BOD}_5=365.7\times0.6838=\boxed{250.0\ \text{mg/L}}.$$
QuantityResult
Dilution fraction $P$0.01667
3-day BOD, $\text{BOD}_3$182.4 mg/L
5-day BOD, $\text{BOD}_5$250.0 mg/L
Ultimate BOD, $\text{BOD}_u$365.7 mg/L