Civil Engineering PE Water and Wastewater Treatment 5 — Questions and Answers
Question 1: The F/M (food-to-microorganism) ratio in activated sludge is calculated as:
- Effluent BOD (mg/L) / MLVSS (mg/L)
- Influent BOD load (lb/day) / mass of MLVSS in aeration basin (lb) (Correct answer)
- Influent flow (MGD) / sludge wasting rate (MGD)
- MLSS (mg/L) / HRT (days)
Correct answer: Influent BOD load (lb/day) / mass of MLVSS in aeration basin (lb)
F/M = (influent BOD, lb/day) / (MLVSS in aeration basin, lb), typically ranging from 0.05 to 0.5 d⁻¹ for conventional activated sludge.
Question 2: Phosphorus removal in a biological nutrient removal (BNR) system is achieved by:
- Nitrification followed by denitrification in anoxic zones
- Alternating anaerobic and aerobic conditions to select for polyphosphate-accumulating organisms (PAOs) (Correct answer)
- Adding alum to the anaerobic zone
- Increasing sludge age beyond 20 days
Correct answer: Alternating anaerobic and aerobic conditions to select for polyphosphate-accumulating organisms (PAOs)
PAOs release phosphorus under anaerobic conditions and take up excess phosphorus (luxury uptake) under aerobic conditions, removing it via waste sludge.
Question 3: A sedimentation basin has a detention time of 4 hours and a flow of 3 MGD. What is the volume of the basin in gallons?
- 500,000 gal (Correct answer)
- 750,000 gal
- 1,000,000 gal
- 2,000,000 gal
Correct answer: 500,000 gal
Volume = Q × t = 3 MGD × (4/24) day = 3,000,000 × 0.1667 = 500,000 gallons.
Question 4: In water treatment, what is the purpose of breakpoint chlorination?
- To form chloramines for distribution system residual
- To add enough chlorine to oxidize all ammonia-nitrogen and achieve free chlorine residual (Correct answer)
- To reduce chlorine to below 0.2 mg/L for taste control
- To lower the pH before coagulation
Correct answer: To add enough chlorine to oxidize all ammonia-nitrogen and achieve free chlorine residual
At the breakpoint, all chloramines and organic chlorine compounds are destroyed and any additional chlorine appears as free residual (HOCl/OCl⁻).
Question 5: Which of the following best describes the difference between primary and secondary clarifiers in a conventional activated sludge plant?
- Primary clarifiers remove dissolved BOD; secondary clarifiers remove suspended solids only
- Primary clarifiers remove settleable solids from raw sewage; secondary clarifiers separate biological floc from treated effluent (Correct answer)
- Primary clarifiers use chemicals; secondary clarifiers are gravity-only
- Primary clarifiers follow biological treatment; secondary clarifiers precede it
Correct answer: Primary clarifiers remove settleable solids from raw sewage; secondary clarifiers separate biological floc from treated effluent
Primary clarifiers settle raw wastewater solids (30–50% TSS, 25–40% BOD removal); secondary clarifiers follow the aeration basin to separate activated sludge biomass from treated effluent.
Question 6: What is the typical volatile suspended solids (VSS) to total suspended solids (TSS) ratio for municipal wastewater biosolids, indicating the organic fraction?
- 0.2–0.3
- 0.4–0.5
- 0.7–0.8 (Correct answer)
- 0.95–1.0
Correct answer: 0.7–0.8
Municipal biosolids typically have a VSS/TSS ratio of 0.70–0.80, indicating that 70–80% of the solids are organic (volatile) matter.
Question 7: The dissolved oxygen (DO) in an aeration basin of an activated sludge system is typically maintained at:
- 0.0–0.5 mg/L (anoxic)
- 1.0–3.0 mg/L (Correct answer)
- 5.0–8.0 mg/L
- 10–12 mg/L
Correct answer: 1.0–3.0 mg/L
Conventional activated sludge systems are designed to maintain 1.0–3.0 mg/L DO to ensure aerobic conditions without excessive energy consumption.
Civil Engineering PE Water and Wastewater Treatment 5
The F/M (food-to-microorganism) ratio in activated sludge is calculated as: