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18-Env-A4 Water and Wastewater Engineering · Undated paper

Question 5 of 5: BOD with Seed Correction; Anaerobic Digestion Stages

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

Notes on this paper

National Exams — May 2019 — 18-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved Casio/Sharp calculator permitted. Question 1 is compulsory; the paper instructs candidates to attempt any three of the remaining four (100 marks total); all five are solved below for completeness.

Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — BOD kinetics, activated-sludge/clarifier design, anaerobic digestion; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — alkalinity chemistry, disinfection, coagulation/flocculation; MWH’s Water Treatment: Principles and Design (3rd ed.) — ozonation, turbidity; Standard Methods for the Examination of Water and Wastewater — alkalinity titration (2320B).

Question 5: BOD with Seed Correction; Anaerobic Digestion Stages (25 marks: a 15, b 10)

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. Seed-corrected BOD5 and ultimate BOD (15 marks)

Given. A raw sewage sample (5 mL) is diluted to 300 mL in a BOD bottle and incubated at the standard 20 °C; DO falls from 8.0 mg/L to 5.0 mg/L over the incubation, of which 5% of the observed depletion is attributed to seed already present in the sample.

QuantityValue
Sample volume $V_s$5 mL
Bottle (dilution) volume $V_b$300 mL
Initial DO8.0 mg/L
Final DO5.0 mg/L
Seed share of depletion5%
Check: the question states the DO was read "after 3 days of incubation" but then attributes the 5% seed share to "this 4-day period". Since only one DO reading is given (at the stated 3-day mark), this solution treats the incubation/measurement period as 3 days throughout — consistent with the only depletion actually measured — and reads "this 4-day period" as a printing slip for "this 3-day period." No 4-day DO reading exists anywhere in the question to use instead.

Find. The standard BOD5 (5-day, 20 °C) and the ultimate BOD ($BOD_u$) of the undiluted sample.

Approach. Scale the seed-corrected DO depletion by the dilution factor to get the 3-day BOD; since the exam supplies no deoxygenation rate constant, assume the typical raw-domestic-sewage value and use first-order BOD kinetics to convert the 3-day reading to $BOD_u$, then re-apply the same kinetics at 5 days for the standard BOD5 (no temperature correction is needed here, since the test already ran at the standard 20 °C).

  1. Dilution fraction. $$P=\frac{V_s}{V_b}=\frac{5}{300}=0.01667.$$
  2. Seed-corrected 3-day depletion. Raw DO drop $\Delta DO=8.0-5.0=3.0\ \text{mg/L}$; with 5% attributed to the seed, the sewage's own share is $$\Delta DO_{net}=3.0\times(1-0.05)=2.85\ \text{mg/L}.$$
  3. 3-day BOD at 20 °C. $$BOD_3=\frac{\Delta DO_{net}}{P}=\frac{2.85}{0.01667}=\boxed{171.0\ \text{mg/L}}.$$
  4. Rate constant (assumed). No deoxygenation rate constant is supplied, so a typical value for raw domestic sewage is assumed (Check assumption): $$k_1(20^\circ C)=0.10\ \text{d}^{-1}\ (\text{base-10}).$$
  5. Ultimate BOD. From the first-order model $BOD_t=BOD_u\left(1-10^{-k_1t}\right)$ at $t=3$ days: $$BOD_u=\frac{BOD_3}{1-10^{-k_1(3)}}=\frac{171.0}{1-10^{-0.30}}=\frac{171.0}{0.4988}=\boxed{342.8\ \text{mg/L}}.$$
  6. Standard BOD5. Re-applying the same first-order model at the standard $t=5$ days: $$BOD_5=BOD_u\left(1-10^{-k_1(5)}\right)=342.8\times\left(1-10^{-0.50}\right)=342.8\times0.6838=\boxed{234.4\ \text{mg/L}}.$$
QuantityResult
Dilution fraction $P$0.01667
3-day BOD171.0 mg/L
Assumed rate constant $k_1$0.10 d-1 (base-10)
Ultimate BOD, $BOD_u$342.8 mg/L
Standard BOD5 (20 °C)234.4 mg/L

b. Four key steps of anaerobic sludge digestion (10 marks)

Anaerobic digestion of sludge proceeds through four sequential microbial stages, each stage's products serving as the substrate for the next: (1) Hydrolysis — extracellular enzymes secreted by fermentative bacteria break large, insoluble polymers (proteins, carbohydrates, lipids) in the sludge solids into soluble monomers and oligomers (amino acids, simple sugars, long-chain fatty acids) that can cross a microbial cell membrane; this is often the rate-limiting step for a sludge with a high proportion of particulate/complex solids. (2) Acidogenesis (fermentation) — a diverse population of fermentative bacteria takes up the soluble monomers and ferments them into short-chain volatile fatty acids (acetic, propionic, butyric acid), along with alcohols, hydrogen and carbon dioxide. (3) Acetogenesis — acetogenic bacteria further convert the longer-chain volatile fatty acids and alcohols from stage 2 into acetate, hydrogen and $CO_2$, the only substrates the final-stage organisms can use; this stage depends on a low partial pressure of hydrogen (maintained by the hydrogen-consuming methanogens working alongside it) to remain thermodynamically favourable. (4) Methanogenesis — strictly anaerobic methanogenic archaea, the most sensitive and slowest-growing group in the consortium, convert acetate (via aceticlastic methanogens, roughly two-thirds of the biogas methane) and hydrogen/$CO_2$ (via hydrogenotrophic methanogens, the remaining third) into methane and carbon dioxide biogas, the process's usable end product. Because methanogens are the slowest-growing and most sensitive stage (to pH swings, temperature shocks, oxygen intrusion and toxic/inhibitory compounds), digester upset almost always appears first as methanogen inhibition and a resulting volatile-fatty-acid build-up, which is why VFA/alkalinity monitoring is the standard early-warning indicator of a failing digester.

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