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Topic 9 of 10 — free theory

Water Treatment

Coagulation/flocculation, sedimentation, filtration, and disinfection with CT.

FE Civil · Environmental Engineering (6–9)PE WRE · Water Quality (5–8) + Drinking Water (6–9)

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Drinking-water treatment is a chain: each process prepares the water for the next. The exam tests the purpose, the key design parameter, and the arithmetic of each link.

The treatment train and its design parameters

Coagulation/flocculation: rapid mix then gentle stirring; design by detention time t = V/Q and the Camp number Gt

Typical: rapid mix 30–60 s; flocculation 20–40 min; Gt ≈ 104–105.

Sedimentation:   surface overflow rate = QA   weir loading = QLweir

Surface overflow rate (gpd/ft²)the controlling design parameter — particles settle if their settling velocity exceeds it
Weir loading (gpd/ft)checked separately so settled sludge is not scoured over the weirs

Filtration:   filtration rate = Q/A   (gpm/ft²);   backwash reverses the flow to clean the media

Rapid sand filters run ≈ 2–4 gpm/ft²; head loss grows as the bed clogs, triggering backwash.

Disinfection:   CT = C × T

Cdisinfectant residual concentration (mg/L)
Tcontact time (min) — use T10, the time 90% of the water exceeds, for credit
CT (mg·min/L)compared against regulatory tables for the target pathogen and disinfectant

Chick's law describes disinfection kinetics (first-order kill with contact time), but exam arithmetic is almost always the CT product.

PE depth: water-quality mass balance and distribution

Translate concentrations into loads before comparing streams. In US customary treatment calculations:

Load (lb/day) = 8.34 × Q(MGD) × C(mg/L)

For a steady completely mixed reactor with first-order decay and no internal generation:

Cout = Cin1 + kθ,   θ = V/Q

A plug-flow approximation instead gives Cout = Cine−kθ. The reactor model changes the answer even when V, Q, and k are identical.

Demand, storage, and distribution

Separate average-day, maximum-day, and peak-hour demand. Source and treatment capacity are commonly checked against maximum-day demand; distribution mains and pressure are checked against peak-hour and fire-flow conditions. Equalisation storage comes from the cumulative difference between supply and demand, with emergency and fire storage added only when the problem requires them.

PE trap: pressure at a high node depends on hydraulic grade, not simply pump discharge pressure. Subtract elevation and losses from the source HGL before converting head to psi.

Worked example Sedimentation basin loading and disinfection CT

Given:

  • Part A: rectangular basin 60 ft × 20 ft, 10 ft deep; flow Q = 1.5 MGD.
  • Part B: chlorine contactor with residual C = 1.2 mg/L and T10 = 45 min.

Solution:

  1. Part A: plan area A = 60 × 20 = 1,200 ft². Surface overflow rate = 1,500,000 gpd / 1,200 ft² = 1,250 gpd/ft² (within the usual 800–1,200+ range for conventional basins).
  2. Detention time: V = 60×20×10 = 12,000 ft³ = 89,760 gal. t = 89,760/1,500,000 days = 1.44 hr.
  3. Part B: CT = 1.2 mg/L × 45 min = 54 mg·min/L — compare with the CT table for your disinfectant and target organism.

Answer: SOR ≈ 1,250 gpd/ft², detention ≈ 1.44 hr; CT ≈ 54 mg·min/L.

Free 5-question mini-quiz

Water Treatment

Choose your answer, then check it to see the result and explanation. SI units are used unless stated otherwise.

1. A contact basin has volume 1,800 m³ and flow 0.050 m³/s. What is the theoretical detention time?

2. A sedimentation basin treats 12,000 m³/day with plan area 600 m². What is its surface overflow rate?

3. A chlorine residual is 1.5 mg/L and T10 is 30 min. What is CT?

4. A plant treats 5,000 m³/day containing 20 mg/L of a constituent. What is the incoming mass load?

5. Which observation most directly indicates that a rapid sand filter needs backwashing?

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