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FE · 12% of the exam

Water and Wastewater practice questions

10 free Water and Wastewater questions for the FE Environmental Exam (FE) exam, each with the correct answer and an explanation for every option.

Start the questionsFree 10-question FE quiz

What this domain covers

One of the two largest domains: treating drinking water and wastewater, from demand estimates and unit processes to sludge handling.

Exam weight
12%
On a real exam
≈ 13 questions
In our bank
73 questions

Topics that come up most

  • Coagulation, sedimentation and filtration
  • Disinfection and CT values
  • Activated sludge design: SRT, F/M and oxygen demand
  • BOD kinetics and wastewater characteristics
  • Sludge treatment and water distribution

How to study it

Learn the activated sludge design relationships in the Reference Handbook and work a full example linking SRT, biomass yield and sludge wasting.

Sample Water and Wastewater questions

Try each one before opening the answer. Original questions written to the official exam objectives.

  1. Question 1

    In drinking water disinfection design, the CT value used to determine inactivation credit is defined as:

    • AApplied disinfectant dose multiplied by the contact time
    • BResidual disinfectant concentration multiplied by contact time
    • CResidual disinfectant concentration divided by the flow rate
    • DRequired log inactivation multiplied by the water temperature
    Show answer and explanation

    Answer: B) Residual disinfectant concentration multiplied by contact time

    CT is the measured residual disinfectant concentration (mg/L) times the contact time (min). The achieved CT is compared with tabulated required CT values for the target organism, temperature and pH.

    • AIncorrect. CT uses the residual concentration measured in the water, not the applied dose. Chlorine demand means the dose is higher than the residual that actually does the inactivating.
    • BCorrect. Under 40 CFR 141.2, CT is the residual disinfectant concentration C (mg/L) times the disinfectant contact time T (min), giving units of mg·min/L.
    • CIncorrect. Flow rate is used to work out the contact time (for example, pipe volume divided by peak hourly flow), but CT is a product of concentration and time, not a ratio of concentration to flow.
    • DIncorrect. Temperature and the target log inactivation set the required CT from tables, but they are not what CT is defined as.

  2. Question 2

    A clarifier discharges 2,000 m³/day over 50 m of effluent weir. The weir loading rate is:

    • A40 m³/(m·day)
    • B0.025 m³/(m·day)
    • C100,000 m³/(m·day)
    • D400 m³/(m·day)
    Show answer and explanation

    Answer: A) 40 m³/(m·day)

    • ACorrect. Q/L = 2,000/50.
    • BIncorrect. That's inverted.
    • CIncorrect. Divide, don't multiply.
    • DIncorrect. Recheck: 2,000/50.

  3. Question 3

    A town of 10,000 people uses 200 L per person per day. Average daily water demand is:

    • A200 m³/day
    • B2,000 m³/day
    • C20,000 m³/day
    • D50 m³/day
    Show answer and explanation

    Answer: B) 2,000 m³/day

    • AIncorrect. Recheck: 10,000 × 200 L.
    • BCorrect. 2,000,000 L/day.
    • CIncorrect. Recheck the conversion.
    • DIncorrect. Multiply, don't divide.

  4. Question 4

    Sludge rising in a secondary clarifier after a few hours is often caused by:

    • AExcess chlorine in the clarifier
    • BDenitrification forming nitrogen gas bubbles
    • CToo little nitrate in the mixed liquor
    • DCold influent alone
    Show answer and explanation

    Answer: B) Denitrification forming nitrogen gas bubbles

    • AIncorrect. That isn't the usual cause.
    • BCorrect. Gas lifts the sludge blanket.
    • CIncorrect. Nitrate fuels the gas formation.
    • DIncorrect. That isn't the usual cause.

  5. Question 5

    U.S. secondary treatment standards set a 30-day average limit for BOD₅ and TSS of:

    • A100 mg/L each
    • B300 mg/L each
    • C3 mg/L each
    • D30 mg/L each
    Show answer and explanation

    Answer: D) 30 mg/L each

    • AIncorrect. The limit is 30 mg/L.
    • BIncorrect. That's raw-wastewater strength.
    • CIncorrect. That's advanced treatment.
    • DCorrect. Along with 85% removal.

  6. Question 6

    Compared with a conventional activated sludge plant, the key process difference in a membrane bioreactor (MBR) is that it:

    • ASeparates biomass from effluent by membrane filtration instead of a clarifier
    • BReplaces the aeration basin with a membrane that oxidizes the organics directly
    • CRuns at a lower MLSS than conventional plants to keep the membranes from fouling
    • DDepends on improved sludge settleability to achieve a low-TSS final effluent
    Show answer and explanation

    Answer: A) Separates biomass from effluent by membrane filtration instead of a clarifier

    An MBR keeps the biological reactor but uses membranes instead of a secondary clarifier to separate solids from liquid. This allows higher MLSS, a smaller footprint and a very low-TSS effluent.

    • ACorrect. An MBR combines the activated sludge process with micro- or ultrafiltration membranes that hold back the sludge flocs. This removes the need for a secondary clarifier.
    • BIncorrect. Microorganisms in an aerated reactor still degrade the organics. The membrane only separates solids from liquid.
    • CIncorrect. MBRs usually run at much higher MLSS (about 8–12 g/L) than conventional activated sludge (about 2–4 g/L), because they don't depend on gravity settling.
    • DIncorrect. In an MBR the membrane retains the solids, so sludge settleability does not control effluent quality.

  7. Question 7

    Compared with coarse-bubble diffusers, fine-bubble diffusers generally provide:

    • ALower oxygen-transfer efficiency
    • BHigher oxygen-transfer efficiency
    • CNo oxygen transfer
    • DIdentical efficiency
    Show answer and explanation

    Answer: B) Higher oxygen-transfer efficiency

    • AIncorrect. Fine bubbles transfer more.
    • BCorrect. Smaller bubbles give more interfacial area.
    • CIncorrect. They transfer oxygen well.
    • DIncorrect. Fine bubbles transfer more.

  8. Question 8

    A clarifier must treat 4,000 m³/day at a surface overflow rate of 40 m³/(m²·day). The required surface area is:

    • A0.01 m²
    • B160,000 m²
    • C100 m²
    • D1,000 m²
    Show answer and explanation

    Answer: C) 100 m²

    Surface overflow rate is flow divided by surface area, so the required area is A = Q/SOR = 4,000/40 = 100 m².

    • AIncorrect. This is the overflow rate divided by the flow (40 ÷ 4,000), which is the inverse of the correct relationship. The units do not reduce to m².
    • BIncorrect. This multiplies flow by overflow rate (4,000 × 40). The units would be m⁶/(m²·day²), not an area.
    • CCorrect. Surface overflow rate = Q/A, so A = Q/SOR = 4,000 m³/day ÷ 40 m³/(m²·day) = 100 m².
    • DIncorrect. This is the correct division with the decimal point in the wrong place. 4,000 ÷ 40 = 100, not 1,000.

  9. Question 9

    Nitrification is the biological conversion of:

    • AAmmonia to nitrate under aerobic conditions
    • BNitrate to nitrogen gas
    • COrganic carbon to methane
    • DPhosphate to phosphorus gas
    Show answer and explanation

    Answer: A) Ammonia to nitrate under aerobic conditions

    • ACorrect. Via nitrite, by autotrophic bacteria.
    • BIncorrect. That's denitrification.
    • CIncorrect. That's methanogenesis.
    • DIncorrect. That's not nitrification.

  10. Question 10

    An anaerobic digester receives 1,500 kg/day of volatile solids (VS). A direct VS mass balance shows 900 kg/day of VS leaving in the digested sludge. What is the VS reduction, and does it meet the 40 CFR Part 503 vector attraction reduction option based on VS reduction?

    • A40%; it meets the 38% minimum reduction
    • B60%; it meets the 38% minimum reduction
    • C67%; it meets the 38% minimum reduction
    • D40%; it fails the 50% minimum reduction
    Show answer and explanation

    Answer: A) 40%; it meets the 38% minimum reduction

    VS reduction = (VS in − VS out)/VS in = (1,500 − 900)/1,500 = 40%. That exceeds the 38% minimum in 40 CFR 503.33(b)(1).

    • ACorrect. VS reduction = (1,500 − 900)/1,500 = 600/1,500 = 40%. 40 CFR 503.33(b)(1) requires the VS mass to drop by at least 38%, so this passes.
    • BIncorrect. 900/1,500 = 60% is the fraction of VS that remains, not the fraction destroyed.
    • CIncorrect. This divides the VS destroyed (600) by the outlet VS (900). The reduction is measured against the feed VS (1,500).
    • DIncorrect. The 40% figure is right, but the Part 503 VS reduction option sets a 38% minimum, not 50%. The sludge passes.

Water and Wastewater: FAQs

How much of the FE exam is Water and Wastewater?

Water and Wastewater is 12% of the FE exam by the official exam guide's weighting, which on a 110-question form works out to roughly 13 questions. The real exam's exact split varies by form.

What does the Water and Wastewater domain cover?

One of the two largest domains: treating drinking water and wastewater, from demand estimates and unit processes to sludge handling.

How should I study Water and Wastewater for FE?

Learn the activated sludge design relationships in the Reference Handbook and work a full example linking SRT, biomass yield and sludge wasting.

Where can I practise more Water and Wastewater questions?

This page has 10 free Water and Wastewater questions with answers. The ExamWize FE question bank has 73 questions in this domain. Every practice exam result breaks your score down by domain, and Review Mode re-drills the questions you missed.

Other FE topics

Also: 15 free questions across all topics, the FE cheat sheet and the study guide.

Get Water and Wastewater exam-ready

The full FE bank has 73 Water and Wastewater questions. Timed practice exams match the real exam's length and domain weights, every result shows your score per domain, and Review Mode re-drills the questions you missed.

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Not affiliated with NCEES; questions are original, written from the official exam outline, and never copied from the real exam.