Water and Wastewater Treatment Flashcards
7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Water and Wastewater Treatment flashcards as text
The F/M (food-to-microorganism) ratio in activated sludge is calculated as:
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.
Phosphorus removal in a biological nutrient removal (BNR) system is achieved by:
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.
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?
Answer: 500,000 gal
Volume = Q × t = 3 MGD × (4/24) day = 3,000,000 × 0.1667 = 500,000 gallons.
In water treatment, what is the purpose of breakpoint chlorination?
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⁻).
Which of the following best describes the difference between primary and secondary clarifiers in a conventional activated sludge plant?
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.
What is the typical volatile suspended solids (VSS) to total suspended solids (TSS) ratio for municipal wastewater biosolids, indicating the organic fraction?
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.
The dissolved oxygen (DO) in an aeration basin of an activated sludge system is typically maintained at:
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.