Question 2 of 5: BOD Statistics, Oxygen-Demanding Wastes, and Air Pollutant Classification

Problem 2: BOD Statistics, Oxygen-Demanding Wastes, and Air Pollutant Classification (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)(i) Meaning and Measurement of BOD

Biochemical Oxygen Demand (BOD) is the mass of dissolved oxygen consumed by micro-organisms as they biologically oxidize the biodegradable organic matter in a water sample over a specified incubation period — conventionally 5 days at 20 °C (BOD₅) — and it is used as a surrogate measure of a water or wastewater’s biodegradable organic pollution strength.

It is measured by the dilution/seed method: a diluted portion of the sample (mixed with aerated dilution water, and seeded with acclimated micro-organisms and nutrients where needed) is placed in a sealed, air-free BOD bottle and incubated in the dark at 20 °C. Dissolved oxygen is measured immediately (DO_i) with a DO probe or Winkler titration, and again after 5 days (DO_f); BOD = (DO_i − DO_f) × dilution factor, corrected for the oxygen consumed by the blank/seed alone.

(a)(ii) Best Estimate of the BOD Concentration

Given. n = 9 replicate BOD measurements (table above): 9.8, 10.2, 8.6, 9.6, 10.6, 10.4, 8.9, 9.9, 10.2 mg/L; negligible bias assumed.

Find. The best (point) estimate of the true BOD concentration.

Approach. With negligible bias, the sample mean is the best unbiased point estimate of the population mean.

1. Sum and average the nine replicates. $$\bar{x} = \dfrac{9.8+10.2+8.6+9.6+10.6+10.4+8.9+9.9+10.2}{9} = \dfrac{88.2}{9} = \boxed{9.80\ \text{mg/L}}$$

(a)(iii) 95% Confidence Interval for the True Mean

Given. The same n = 9 values, sample mean ̄x = 9.80 mg/L.

Find. The 95% confidence interval for the true (population) mean BOD.

Approach. With n < 30 and the population standard deviation unknown, the confidence interval uses Student’s t-distribution with n−1 = 8 degrees of freedom, not the normal (z) distribution.

1. Sample standard deviation. Summing squared deviations from ̄x = 9.80 gives Σ(x−̄x)² = 3.62, so $$s = \sqrt{\dfrac{\Sigma(x-\bar{x})^{2}}{n-1}} = \sqrt{\dfrac{3.62}{8}} = 0.673\ \text{mg/L}$$
2. Margin of error. With t_0.025,8 = 2.306 (two-tailed, 95%, df = 8): $$E = t_{0.025,8}\,\dfrac{s}{\sqrt{n}} = (2.306)\dfrac{0.673}{\sqrt{9}} = 0.517\ \text{mg/L}$$
3. Confidence interval. $$\bar{x} \pm E = 9.80 \pm 0.52 \ \Rightarrow\ \boxed{[9.28,\ 10.32]\ \text{mg/L}}$$

(b) Impact of Oxygen-Demanding Wastes on Rivers

Discharging oxygen-demanding (high-BOD) waste into a river sets up the classic dissolved-oxygen (DO) sag: downstream of the outfall, bacterial oxidation of the organic load consumes dissolved oxygen (deoxygenation) faster than atmospheric reaeration can replace it, so DO falls to a minimum (the critical deficit point) some distance downstream before recovering as the organic load is exhausted and reaeration catches up — the Streeter–Phelps DO-sag behaviour. The ecological consequences track that DO profile: coldwater/sensitive fish species become stressed below roughly 4–5 mg/L DO and are killed outright as DO approaches 1–2 mg/L or less; the healthy, diverse benthic community is replaced by pollution-tolerant, low-oxygen species (e.g. sludge worms, midge larvae); if DO is driven to zero, the reach turns anaerobic/septic, producing odours (H₂S), black sludge deposits, and loss of essentially all higher aquatic life. Because a river’s self-purification capacity is finite, repeated or cumulative oxygen-demanding discharges along a watercourse can prevent recovery between outfalls entirely.

(c) Primary vs. Secondary Air Pollutants

Primary pollutants are emitted directly from an identifiable source into the atmosphere in the same chemical form in which they cause harm — for example, sulphur dioxide (SO₂) emitted directly from fossil-fuel combustion, or carbon monoxide (CO) from incomplete combustion.

Secondary pollutants are not emitted at all; they are formed in the atmosphere itself through chemical (often photochemical) reactions between primary pollutants and/or normal atmospheric constituents. The textbook example is ground-level ozone (O₃), formed from nitrogen oxides (NO_x) and volatile organic compounds (VOCs) reacting in sunlight to produce photochemical smog.

So the SO₂/O₃ pair is the clean example pair: SO₂ is primary (comes straight out of the stack), O₃ is secondary (only exists because of an atmospheric reaction downwind).