Civil Engineering PE Water and Wastewater Treatment 5

Question 1

The F/M (food-to-microorganism) ratio in activated sludge is calculated as:

Accepted Answer

Influent BOD load (lb/day) / mass of MLVSS in aeration basin (lb)

Answer

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:

Accepted Answer

Alternating anaerobic and aerobic conditions to select for polyphosphate-accumulating organisms (PAOs)

Answer

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?

Accepted Answer

500,000 gal

Answer

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?

Accepted Answer

To add enough chlorine to oxidize all ammonia-nitrogen and achieve free chlorine residual

Answer

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?

Accepted Answer

Primary clarifiers remove settleable solids from raw sewage; secondary clarifiers separate biological floc from treated effluent

Answer

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.

Civil Engineering PE Practice Test

Civil Engineering PE Practice Test

Civil Engineering PE Water and Wastewater Treatment 5