Question 3 of 5: Lagoon Treatment and the BOD Test

Question 3: Lagoon Treatment and the BOD Test (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. Facultative Lagoon BOD/Nutrient Removal vs. Aerated Lagoon (10 marks)

A facultative lagoon stratifies naturally into three zones: an aerobic surface layer, where photosynthetic algae generate oxygen during daylight that supports aerobic bacteria oxidizing soluble BOD; an anaerobic bottom sludge layer, where settled solids decompose anaerobically, releasing CO₂, methane and soluble by-products; and a facultative middle zone where bacteria switch between aerobic and anaerobic metabolism depending on locally available oxygen. BOD removal is driven by a symbiotic algal-bacterial relationship: bacteria oxidize organic matter to CO₂ and mineral nutrients, algae consume that CO₂ and those nutrients through photosynthesis and release O₂, which the bacteria then use — a self-sustaining oxygen supply that needs no mechanical aeration, sustained over a long hydraulic retention time (commonly 20–180 days) that allows extensive biological stabilization. Nutrient removal is more limited: nitrogen is reduced mainly through algal biomass uptake and through ammonia volatilization at the elevated pH that algal CO₂ consumption drives, while phosphorus is removed only modestly, through algal uptake and some precipitation at high pH.

An aerated lagoon instead supplies oxygen mechanically (surface or diffused aeration) throughout the full water depth, eliminating dependence on sunlight/photosynthesis and permitting a much shorter retention time (typically 3–10 days) and greater depth (up to 4–6 m, versus 1–2.5 m for a facultative lagoon). Because mixing and oxygenation are mechanical, an aerated lagoon behaves as a complete-mix suspended-growth reactor — conceptually close to an extended-aeration activated-sludge process without solids recycle — whereas the facultative lagoon relies on quiescent gravity settling and a stratified, largely passive biology. The trade-off is that aerated lagoons need continuous electrical power and mechanical equipment (higher energy and O&M cost) but occupy a smaller footprint and deliver more consistent year-round effluent than a facultative lagoon, whose performance tracks seasonal light and temperature (ice cover and reduced algal photosynthesis noticeably degrade winter performance).

b. BOD Test — 4-Day, 5-Day and Ultimate BOD (15 marks)

Given.

Find. The 4-day BOD ($\text{BOD}_4$), 5-day BOD ($\text{BOD}_5$) and ultimate BOD ($\text{BOD}_u$) of the undiluted sewage sample.

Approach. Compute the dilution fraction, correct the observed DO depletion for the seed's own contribution, scale by the dilution fraction to get $\text{BOD}_4$ of the undiluted sample, then use the assumed first-order BOD-exertion model to back out $\text{BOD}_u$ and forward-compute $\text{BOD}_5$.

1. Dilution fraction. $$P=\frac{V_s}{V_b}=\frac{5}{300}=0.01667.$$
2. Seed-corrected oxygen depletion in the bottle. Observed depletion $=DO_i-DO_f=9.0-4.3=4.7$ mg/L; with 2% of this attributed to the seed, the sewage's own depletion is $4.7\times(1-0.02)=4.606$ mg/L.
3. 4-day BOD of the undiluted sample. Dividing by the dilution fraction scales the bottle reading up to the full-strength sample: $$\text{BOD}_4=\frac{4.606}{0.01667}=\boxed{276.4\ \text{mg/L}}.$$
4. Ultimate BOD from the assumed rate constant. With $k_1=0.10\ \text{day}^{-1}$ (base-10) and $\text{BOD}_t=\text{BOD}_u(1-10^{-k_1t})$, at $t=4$: $1-10^{-0.40}=0.6019$, so $$\text{BOD}_u=\frac{276.4}{0.6019}=\boxed{459.2\ \text{mg/L}}.$$
5. 5-day BOD. Applying the same curve at $t=5$: $1-10^{-0.50}=0.6838$, so $$\text{BOD}_5=459.2\times0.6838=\boxed{314.0\ \text{mg/L}}.$$