FE Environmental Exam Prep
By the Castleport Test Prep Editorial Team · Last verified October 7, 2026
Start FE Environmental exam prep with the problem below, then work through the 24 original questions or use the topic map and 8-week plan to choose your next session.
24 worked FE Environmental practice problems
Original Castleport practice problems, not official NCEES questions. Work each one before opening the solution.
1. Daily mass load
Water and Wastewater · FEENV-01
A stream flows continuously at 2,400 m³/day and carries a dissolved constituent at a constant 35 mg/L. What mass of the constituent passes the sampling point each day?
- A. 0.084 kg/day
- B. 84 kg/day
- C. 84,000 kg/day
- D. 84,000,000 kg/day
Show answer and explanation
Answer: B — 84 kg/day.
Load = concentration × flow, with matching volume units: 35 mg/L × 2,400 m³/day × 1,000 L/m³ × (1 kg / 1,000,000 mg) = 84 kg/day. Quick check: 35 mg/L is 0.035 kg/m³, and 0.035 × 2,400 = 84.
Why the other answers miss:
- A: Skips the m³ → L conversion, so it's 1,000 times too small.
- C: That's the load in grams, labeled as kilograms.
- D: That's the load in milligrams, labeled as kilograms.
Source: USGS SIR 2006-5188, Section 8: loads = streamflow × concentration, with unit conversion
2. Theoretical detention time (numerical response)
Water and Wastewater · FEENV-02
A basin holds 480 m³ at its operating level. Water enters and leaves at a constant 0.040 m³/s, and the volume stays constant. What is the theoretical detention time, in minutes?
Your answer: ______ minutes
Show answer and explanation
Answer: 200 minutes.
τ = V/Q = 480 m³ ÷ 0.040 m³/s = 12,000 s = 200 minutes.
Common wrong answers:
- 12,000: That's seconds. The flow is per second, so the quotient is in seconds until you convert.
- 3.33: That's the answer in hours.
- Q/V: Flow divided by volume has units of 1/time, so it can't be a detention time.
This is theoretical detention time only. It isn't a tracer residence time or an effective disinfection contact time. The real exam also uses fill-in-the-blank questions like this one.
Source: EPA Disinfection Profiling and Benchmarking Technical Guidance Manual (EPA 815-R-20-003, June 2020), §4.4.2, Eq. 4-2, p. 26
3. Clarifier overflow rate
Water and Wastewater · FEENV-03
A circular primary clarifier 30 m in diameter treats 10,000 m³/d. What is its surface overflow rate?
- A. 3.5 m³/(m²·d)
- B. 11.1 m³/(m²·d)
- C. 14.1 m³/(m²·d)
- D. 44.4 m³/(m²·d)
Show answer and explanation
Answer: C — 14.1 m³/(m²·d).
Surface area A = π(15 m)² = 706.9 m². Overflow rate = Q/A = 10,000 ÷ 706.9 = 14.1 m³/(m²·d). Depth doesn't enter the overflow rate.
Why the other answers miss:
- A: Uses πD² (2,827 m²) instead of πr².
- B: Divides by D² (900 m²) with no π.
- D: Uses r² (225 m²) and forgets π.
Source: EPA Operator Webinar Series, Primary Clarifiers (August 24, 2023), PDF p. 10, “Surface Overflow Rate (SOR),” Q/A
4. Ultimate BOD
Water and Wastewater · FEENV-04
A wastewater's 5-day BOD is 200 mg/L. Its first-order BOD rate constant is k = 0.23 d⁻¹ (base e). What is the ultimate BOD?
- A. 137 mg/L
- B. 200 mg/L
- C. 293 mg/L
- D. 870 mg/L
Show answer and explanation
Answer: C — 293 mg/L.
BOD₅ = L₀(1 − e^(−kt)), so L₀ = 200 ÷ (1 − e^(−0.23×5)) = 200 ÷ (1 − e^(−1.15)) = 200 ÷ 0.683 = 293 mg/L. Ultimate BOD always exceeds BOD₅.
Why the other answers miss:
- A: Multiplies by 0.683 instead of dividing.
- B: Treats 5-day BOD as if it were the ultimate BOD.
- D: Divides by k, which isn't part of the relationship.
Source: Rowan University, Environmental Engineering I study guide: Oxygen Demand, PDF p. 2, BODₜ = L₀(1 − e^(−kt)); L₀ = ultimate BOD
5. Pipe contraction (US units)
Fluid Mechanics and Hydraulics · FEENV-05
Water flows steadily through a full 8.0-inch pipe at a mean velocity of 1.50 ft/s. The pipe narrows to 4.0 inches. Assume constant density, no leaks or branches, and no storage between the sections. What is the mean velocity in the smaller pipe?
- A. 0.375 ft/s
- B. 1.50 ft/s
- C. 3.00 ft/s
- D. 6.00 ft/s
Show answer and explanation
Answer: D — 6.00 ft/s.
Continuity: A₁v₁ = A₂v₂. Area scales with diameter squared, so v₂ = 1.50 × (8.0/4.0)² = 6.00 ft/s. Inches cancel in the ratio, so you don't need to convert them.
Why the other answers miss:
- A: Flips the area ratio. A smaller pipe must speed the flow up, not slow it down.
- B: Keeps the velocity unchanged, which would mean less flow gets through.
- C: Uses the diameter ratio without squaring it.
Source: NASA Glenn Research Center, Conservation of Mass, steady-flow mass-flow equality; constant density
6. Manning's equation (SI)
Fluid Mechanics and Hydraulics · FEENV-06
A concrete channel (n = 0.013) has a flow area of 3.0 m², a wetted perimeter of 5.0 m, and a bed slope of 0.001 m/m. Assume steady, uniform flow, so the energy slope equals the bed slope. What is the discharge by Manning's equation in SI units?
- A. 0.16 m³/s
- B. 4.38 m³/s
- C. 5.19 m³/s
- D. 7.74 m³/s
Show answer and explanation
Answer: C — 5.19 m³/s.
R = A/P = 3.0/5.0 = 0.6 m. Q = (1/n)·A·R^(2/3)·S^(1/2) = (1/0.013)(3.0)(0.6^(2/3))(0.001^(1/2)) = 76.92 × 3.0 × 0.7114 × 0.03162 = 5.19 m³/s.
Why the other answers miss:
- A: Uses S instead of √S.
- B: Uses R to the first power instead of R^(2/3).
- D: Puts the US-units 1.49 constant into an SI calculation.
Source: FHWA, Design of Roadside Drainage Channels, §§3.2–3.2-1, printed p. 12 / PDF p. 24, Eqs. 2–5; USACE HEC-RAS, 1D Saint-Venant Equations, opening clear-water Manning equation in SI units and friction-slope definition
7. Pump brake power
Fluid Mechanics and Hydraulics · FEENV-07
A pump delivers 0.10 m³/s of water against a total dynamic head of 30 m at 75% efficiency. What shaft (brake) power does it need? Use ρ = 1,000 kg/m³ and g = 9.81 m/s².
- A. 22.1 kW
- B. 29.4 kW
- C. 39.2 kW
- D. 392 kW
Show answer and explanation
Answer: C — 39.2 kW.
Water power = ρgQH = 1,000 × 9.81 × 0.10 × 30 = 29,430 W = 29.4 kW. Brake power = water power ÷ efficiency = 29.4 ÷ 0.75 = 39.2 kW.
Why the other answers miss:
- A: Multiplies by efficiency. A pump always needs more input power than it delivers to the water.
- B: That's the water (hydraulic) power, not the shaft power.
- D: A decimal-place error.
Source: Penn State ME 325 pump laboratory introduction, Eqs. 3–4: water horsepower = ρgHV̇; efficiency = water horsepower ÷ brake horsepower
8. Rational method peak runoff
Surface Water Resources and Hydrology · FEENV-08
A 10-acre commercial site has a runoff coefficient C = 0.5. For the selected design frequency and a duration equal to the site's time of concentration, the rainfall intensity is 2 in/hr. Assume rainfall is uniform across the site and the rational method applies. What is the approximate peak runoff?
- A. 1 ft³/s
- B. 5 ft³/s
- C. 10 ft³/s
- D. 20 ft³/s
Show answer and explanation
Answer: C — 10 ft³/s.
Using the customary approximation, Q ≈ CIA = 0.5 × 2 × 10 = 10 ft³/s. In US units, in/hr times acres gives ft³/s directly (to within about 1%). The exact unit-conversion factor is 43,560/(12 × 3,600) = 1.0083, giving 10.08 ft³/s. With I in mm/hr and A in hectares, use Q = CIA/360 for Q in m³/s.
Why the other answers miss:
- A: Drops the area (C × I only).
- B: Drops the intensity (C × A only).
- D: Drops the runoff coefficient (I × A only).
Source: FHWA Urban Drainage Design, HEC-22 Fourth Edition (February 2024), §4.2.2, Eq. 4.1, printed pp. 23–24 / PDF pp. 55–56: units and rational-method assumptions
9. Hydrograph volume by the trapezoidal rule
Mathematics · FEENV-09
An outfall's measured flows at hourly intervals are 0, 4, 6, and 2 m³/s (at t = 0, 1, 2, and 3 h). Using the trapezoidal rule, what total volume was discharged over the 3 hours?
- A. 11 m³
- B. 36,000 m³
- C. 39,600 m³
- D. 43,200 m³
Show answer and explanation
Answer: C — 39,600 m³.
T = Δt[(f₀ + f₃)/2 + f₁ + f₂] = 1 h × [(0 + 2)/2 + 4 + 6] = 11 (m³/s)·h. Convert hours to seconds: 11 × 3,600 = 39,600 m³. The area under a hydrograph is a volume; the hydrograph's highest point is the peak flow, which is a different answer.
Why the other answers miss:
- A: Forgets to convert hours to seconds.
- B: Drops the end-point term (0 + 2)/2.
- D: Adds every ordinate at full weight (12 × 3,600) instead of halving the end points.
Source: OpenStax Calculus Volume 2, §3.6 Numerical Integration, Theorem 3.4, trapezoidal rule
10. Seepage velocity vs. Darcy flux
Groundwater, Soils, and Sediments · FEENV-10
An aquifer has a hydraulic conductivity K = 10 m/d, a hydraulic gradient of 0.005, and an effective porosity of 0.25. What is the average linear (seepage) velocity of the groundwater?
- A. 0.0125 m/d
- B. 0.05 m/d
- C. 0.2 m/d
- D. 2.0 m/d
Show answer and explanation
Answer: C — 0.2 m/d.
Darcy flux q = Ki = 10 × 0.005 = 0.05 m/d. Water moves only through the connected pores, so the seepage velocity is v = q/nₑ = 0.05 ÷ 0.25 = 0.2 m/d. If a question asks for total discharge instead, multiply the flux by the cross-sectional area: Q = KiA.
Why the other answers miss:
- A: Multiplies by porosity instead of dividing.
- B: That's the Darcy flux (specific discharge), not the average linear groundwater velocity.
- D: A factor-of-ten slip.
Source: The Groundwater Project, Darcy's Law as a basis for measuring groundwater velocity, v = Ki/nₑ = q/nₑ
11. ppm to mg/m³
Air Quality and Control · FEENV-11
Convert 9 ppm (by volume) of carbon monoxide to mg/m³ at 25 °C and 1 atm. The molecular weight of CO is 28.01.
- A. 0.32 mg/m³
- B. 10.3 mg/m³
- C. 11.3 mg/m³
- D. 252 mg/m³
Show answer and explanation
Answer: B — 10.3 mg/m³.
At 25 °C and 1 atm, the ideal-gas molar volume is approximately 24.45 L/mol. mg/m³ = ppm × MW ÷ 24.45 = 9 × 28.01 ÷ 24.45 = 10.3 mg/m³.
Why the other answers miss:
- A: Divides ppm by the molecular weight.
- C: Uses 22.4 L/mol, the molar volume at 0 °C, for a 25 °C problem.
- D: Leaves out the molar volume entirely.
Source: CDC/NIOSH Occupational Exposure Banding e-Tool, Conversion Calculator, “mg/m³ to ppm” or “ppm to mg/m³”: equations at 25 °C and 1 atmosphere
12. Electrostatic precipitator efficiency
Air Quality and Control · FEENV-12
An electrostatic precipitator has 5,000 m² of collection area, treats 250 m³/s of gas, and has an effective drift (migration) velocity of 0.10 m/s. What is its collection efficiency by the Deutsch–Anderson equation?
- A. 13.5%
- B. 63.2%
- C. 86.5%
- D. 98.2%
Show answer and explanation
Answer: C — 86.5%.
η = 1 − e^(−wA/Q) = 1 − e^(−0.10 × 5,000 / 250) = 1 − e^(−2) = 86.5%.
Why the other answers miss:
- A: That's the penetration (e^(−2)), the fraction that gets through, not the efficiency.
- B: Uses an exponent of 1.
- D: Uses an exponent of 4, as if the collection area were doubled.
Source: EPA Air Pollution Control Cost Manual, Section 6, Chapter 3 (Electrostatic Precipitators), §3.1.4.3, Eqs. 3.17–3.18
13. Total hardness as CaCO₃
Environmental Chemistry · FEENV-13
A water contains 80 mg/L Ca²⁺ and 24 mg/L Mg²⁺. What is its total hardness, expressed as mg/L CaCO₃?
- A. 104 mg/L
- B. 200 mg/L
- C. 299 mg/L
- D. 597 mg/L
Show answer and explanation
Answer: C — 299 mg/L.
Convert each ion with its equivalent-weight ratio. Calcium: 80 × (50.04/20.04) = 80 × 2.497 = 199.8. Magnesium: 24 × (50.04/12.15) = 24 × 4.118 = 98.8. Total = 299 mg/L as CaCO₃.
Why the other answers miss:
- A: Adds the ion concentrations without converting either one to CaCO₃.
- B: Converts calcium only and leaves out magnesium.
- D: Mixes CaCO₃'s molar mass (100.09) with the ions' equivalent weights, which doubles the answer.
Source: EPA 2015 Multi-Sector General Permit, Appendix J (Calculating Hardness), p. J-2: mg/L CaCO₃ = 2.497(Ca) + 4.118(Mg)
14. First-order decay
Environmental Chemistry · FEENV-14
A contaminant starts at 100 mg/L in a closed, well-mixed batch vessel of constant volume and decays by first-order kinetics with k = 0.2 d⁻¹. What concentration remains after 5 days?
- A. 0 mg/L
- B. 36.8 mg/L
- C. 50 mg/L
- D. 63.2 mg/L
Show answer and explanation
Answer: B — 36.8 mg/L.
C = C₀e^(−kt) = 100 × e^(−0.2 × 5) = 100 × e^(−1) = 36.8 mg/L. Check that kt has no units: d⁻¹ × d.
Why the other answers miss:
- A: Treats the decay as linear: 100 × (1 − 0.2 × 5) = 0.
- C: Assumes 5 days is the half-life. The half-life is actually ln 2 ÷ 0.2 = 3.47 days.
- D: That's the amount removed, not the amount remaining.
Source: UCLA, Stenstrom & Rosso, Fundamentals of Chemical Reactor Theory, p. 3, Eqs. 10–12: first-order decay in a constant-volume batch reactor
15. Henry's law and air stripping
Environmental Chemistry · FEENV-15
Four hypothetical dissolved organics have these dimensionless Henry's constants (H = C_air/C_water): W = 0.002, X = 0.05, Y = 0.3, Z = 1.2. All else equal, which is removed most effectively by air stripping?
- A. W
- B. X
- C. Y
- D. Z
Show answer and explanation
Answer: D — Z.
A higher Henry's constant means the compound partitions more strongly into air at equilibrium. At the same air-to-water ratio, Z strips most readily.
Why the other answers miss:
- A: W has the lowest Henry's constant, so it partitions least strongly into air at equilibrium.
- B: X's Henry's constant is lower than Z's.
- C: Z's Henry's constant is four times higher than Y's.
The compounds and values are invented for practice.
Source: EPA CEAM Learn2Model, Henry's Law Constants, “Unit Choices for the Henry’s Law Constant”; FRTR Remediation Technologies Screening Matrix, Air Stripping, “Applicability” and “Implementability Considerations”
16. Geometric population projection
Fundamental Principles · FEENV-16
A town of 40,000 grows geometrically at 2% per year. What is its projected population in 20 years?
- A. 40,800
- B. 56,000
- C. 59,400
- D. 88,000
Show answer and explanation
Answer: C — 59,400.
Geometric growth compounds: P = P₀(1 + r)ᵗ = 40,000 × 1.02²⁰ = 59,438, or about 59,400.
Why the other answers miss:
- A: That's only one year of growth.
- B: Uses arithmetic (straight-line) growth: 40,000 + 20 × 800.
- D: Compounds at 4% instead of 2% (40,000 × 1.04²⁰ ≈ 87,600).
Source: OpenStax Precalculus 2e, §4.1 Exponential Functions, Example 3: population growth P(t) = P₀(1 + r)ᵗ
17. CSTR residence time
Fundamental Principles · FEENV-17
A first-order reaction has k = 0.5 h⁻¹. What hydraulic residence time does a single ideal CSTR (completely mixed reactor) at steady state need to remove 90%?
- A. 1.8 h
- B. 4.6 h
- C. 9.0 h
- D. 18 h
Show answer and explanation
Answer: D — 18 h.
For a CSTR, C/C₀ = 1/(1 + kτ). With C/C₀ = 0.10: 1 + kτ = 10, so τ = 9 ÷ 0.5 = 18 h.
Why the other answers miss:
- A: Divides 0.9 by k, which isn't the CSTR relationship.
- B: That's the plug-flow answer, ln(10) ÷ 0.5 = 4.6 h. Here the CSTR needs about four times as long.
- C: Forgets to divide by k.
Source: UCLA, Stenstrom & Rosso, Fundamentals of Chemical Reactor Theory, p. 4, Eq. 16 (CSTR); p. 5, Eq. 22 (plug flow / first-order decay)
18. Annual landfill volume
Solid and Hazardous Waste · FEENV-18
A community of 50,000 people generates 2.0 kg of solid waste per person per day. The waste is compacted in a landfill to 600 kg/m³. How much landfill volume does the waste consume each year? Use a 365-day year. Ignore cover soil.
- A. 167 m³
- B. 6,100 m³
- C. 60,800 m³
- D. 36,500,000 m³
Show answer and explanation
Answer: C — 60,800 m³.
Mass per year = 50,000 people × 2.0 kg/(person·day) × 365 days = 36,500,000 kg. Volume = 36,500,000 ÷ 600 kg/m³ = 60,800 m³/yr. Actual landfill capacity also accounts for applicable cover requirements, so watch for that in the stem.
Why the other answers miss:
- A: That's one day's volume.
- B: Off by a factor of ten.
- D: That's the yearly mass in kilograms, not a volume.
Source: Derived calculation using V = m/ρ and the stated inputs. OpenStax Chemistry 2e, §1.4 Measurements, “Density”; 40 CFR 258.21, cover requirements and approved alternatives
19. Which law covers hazardous waste cradle to grave?
Ethics and Professional Practice · FEENV-19
Which federal law gives EPA authority to control hazardous waste from generation through transportation, treatment, storage, and disposal?
- A. CERCLA
- B. RCRA
- C. SDWA
- D. CAA
Show answer and explanation
Answer: B — RCRA.
The Resource Conservation and Recovery Act (RCRA) gives EPA its "cradle-to-grave" authority over hazardous waste.
Why the other answers miss:
- A: CERCLA (Superfund) deals with cleanup of releases and contaminated sites.
- C: The Safe Drinking Water Act protects public drinking water.
- D: The Clean Air Act regulates air emissions.
Source: EPA, Summary of the Resource Conservation and Recovery Act, “cradle to grave” hazardous waste authority. Related EPA summaries: CERCLA, SDWA, and CAA
20. Hazard quotient
Health Hazards and Risk Assessment · FEENV-20
For a hypothetical noncarcinogen, a resident's chronic oral intake is 0.003 mg/(kg·d). The matching oral reference dose (RfD) is 0.002 mg/(kg·d). What is the hazard quotient, and what does it indicate?
- A. 0.67, so no further evaluation is needed
- B. 1.5, so it exceeds the health guideline and needs further evaluation
- C. 0.000006, so the risk is negligible
- D. 1.5, which is a cancer risk of 1.5 in a million
Show answer and explanation
Answer: B — 1.5, so it exceeds the health guideline and needs further evaluation.
HQ = intake ÷ RfD = 0.003 ÷ 0.002 = 1.5. An HQ above 1 exceeds the health guideline and calls for a closer toxicological look. It is a ratio, not a probability of illness.
Why the other answers miss:
- A: Inverts the ratio (RfD ÷ intake).
- C: Multiplies the two values.
- D: Confuses a noncancer hazard quotient with cancer risk, which uses a slope factor instead.
Source: ATSDR Public Health Assessment Guidance, Calculating Hazard Quotients and Cancer Risk Estimates, “Estimating Hazard Quotients”: HQ = exposure dose ÷ RfD; HQ > 1 calls for in-depth toxicological effects analysis
21. Present worth of annual savings
Engineering Economics · FEENV-21
A plant upgrade saves $50,000 at the end of each year for 20 years. At an effective annual interest rate of 6%, what is the present worth of those savings?
- A. $311,800
- B. $573,500
- C. $1,000,000
- D. $1,839,300
Show answer and explanation
Answer: B — $573,500.
P/A = [(1.06)²⁰ − 1] ÷ [0.06 × (1.06)²⁰] = 11.47. P = $50,000 × 11.47 = $573,500.
Why the other answers miss:
- A: Discounts the 20-year total as a single lump sum at year 20.
- C: Ignores the time value of money.
- D: That's the future worth (F/A = 36.79), not the present worth.
Source: NIST, Discount Factor Tables for Life-Cycle Cost Analyses (NISTIR 89-4203), printed pp. 1–2 / PDF pp. 9–10, Eq. 4 and definitions of the periodic rate and end-of-period payments
22. Sample standard deviation
Probability and Statistics · FEENV-22
Five turbidity readings are 10, 12, 14, 16, and 18 NTU. What is the sample standard deviation?
- A. 2.83 NTU
- B. 3.16 NTU
- C. 8.0 NTU
- D. 10.0 NTU
Show answer and explanation
Answer: B — 3.16 NTU.
The mean is 14. The squared deviations (16, 4, 0, 4, 16) sum to 40. For a sample, divide by n − 1: s = √(40 ÷ 4) = 3.16 NTU.
Why the other answers miss:
- A: Divides by n, which gives the population standard deviation.
- C: That's the range (18 − 10).
- D: That's the variance (s² = 10 NTU²), not the standard deviation.
Source: OpenStax Introductory Statistics 2e, §2.7 Measures of the Spread of the Data, sample variance divides by n − 1; population by N
23. Carnot efficiency
Thermodynamics · FEENV-23
A heat engine operates between reservoirs at 600 K and 300 K. What is its maximum possible (Carnot) thermal efficiency?
- A. 33%
- B. 50%
- C. 67%
- D. 100%
Show answer and explanation
Answer: B — 50%.
η = 1 − T_c/T_h = 1 − 300/600 = 50%. Always use absolute temperatures (kelvin or rankine).
Why the other answers miss:
- A: Divides the temperature difference by the sum of the temperatures: 300 ÷ 900.
- C: Uses T_c ÷ (T_h + T_c) in place of T_c ÷ T_h: 1 − 300/900.
- D: Divides the temperature difference by T_c. A 100%-efficient heat engine would violate the second law of thermodynamics.
Source: OpenStax University Physics Volume 2, §4.5 The Carnot Cycle, Eq. 4.5, e = 1 − T_c/T_h (kelvin)
24. CO₂ from burning methane
Energy and Environment · FEENV-24
Complete combustion of methane follows CH₄ + 2O₂ → CO₂ + 2H₂O. How much CO₂ does 1.0 kg of methane produce? Use molar masses CH₄ = 16.04 and CO₂ = 44.01 g/mol.
- A. 0.36 kg
- B. 1.0 kg
- C. 2.0 kg
- D. 2.74 kg
Show answer and explanation
Answer: D — 2.74 kg.
One mole of CH₄ makes one mole of CO₂, so the mass ratio is 44.01 ÷ 16.04 = 2.74 kg CO₂ per kg CH₄. The extra mass comes from oxygen taken out of the air.
Why the other answers miss:
- A: Inverts the ratio.
- B: Assumes the CO₂ weighs the same as the fuel that burned.
- C: Uses a coefficient of 2 (from O₂ or H₂O) instead of the molar-mass ratio.
Source: OpenStax Chemistry 2e, §4.1 Writing and Balancing Chemical Equations, Figure 4.2, methane combustion; NIST Chemistry WebBook, carbon dioxide, molecular weight 44.0095; NIST Chemistry WebBook, methane, molecular weight 16.0425
Where to focus your FE Environmental study
Start FE Environmental exam prep with the two biggest areas. NCEES puts 12–18 of the 110 questions on Fluid Mechanics and Hydraulics and another 12–18 on Water and Wastewater. Use the map below to see where the rest of the questions sit.
| Priority | NCEES knowledge areas (published question range) |
|---|---|
| 1. Largest | Fluid Mechanics and Hydraulics (12–18) · Water and Wastewater (12–18) |
| 2. Large | Surface Water Resources and Hydrology (9–14) · Groundwater, Soils, and Sediments (8–12) · Air Quality and Control (8–12) |
| 3. Core | Fundamental Principles (7–11) · Environmental Chemistry (7–11) · Solid and Hazardous Waste (7–11) |
| 4. Smaller ranges | Mathematics (5–8) · Ethics and Professional Practice (5–8) · Engineering Economics (5–8) · Probability and Statistics (4–6) · Health Hazards and Risk Assessment (4–6) · Energy and Environment (4–6) · Thermodynamics (3–5) |
These ranges come from the NCEES FE Environmental exam specifications, effective since the July 2020 exams. The priority groups are our way of ordering them. Each exam also includes some unscored pretest questions you can't identify, so treat the ranges as ranges, not exact counts (NCEES Examinee Guide, p. 11).
What your score on these 24 tells you
Count your misses by knowledge area, not just overall. Use the published question ranges and recurring errors to choose your next study block; the small, unequal samples here do not support precise comparisons of readiness between areas. Treat your score as feedback on these 24 problems. It isn't an NCEES scaled score, and it can't predict whether you'll pass. Getting one problem right also doesn't mean you've mastered that area.
Your 8-week FE Environmental study plan
The plan assumes about 8 hours a week for 8 weeks (64 hours). That's our editorial plan, not an official or validated number of hours. If you can't explain a method without looking at the solution, give that block more time.
| Week | Focus | Tasks (about 8 hours) | By the end of the week |
|---|---|---|---|
| 1 | Water and Wastewater; handbook setup | Download the current FE Reference Handbook in MyNCEES and skim its table of contents (1 h). Review loading, mass balance, clarifiers, disinfection, activated sludge, and sludge handling (2 h). Work problems (3 h). Fix your errors and retry them (2 h). | Problems 1–4 done; units written on every factor of a load calculation |
| 2 | Fluid Mechanics and Hydraulics | Review continuity, Bernoulli, pipe friction (Darcy-Weisbach, Hazen-Williams), Manning, pumps and blowers in series and parallel, and weirs and orifices (2 h). Work problems in both SI and US units (3 h). Fix errors (2 h). Practice handbook searches (1 h). | Problems 5–7 done; a note on any head/pressure mix-ups |
| 3 | Surface Water Resources and Hydrology; Groundwater, Soils, and Sediments | Review runoff, detention sizing, routing, Streeter-Phelps, and water budgets. Then review Darcy flux vs. seepage velocity and Thiem, Theis, Jacob, and Dupuit drawdown (2 h). Work problems (3 h). Fix errors (2 h). Handbook searches (1 h). | Problems 8–10 done; volume, peak flow, and flux kept separate |
| 4 | Air Quality and Control | Review ppm ↔ mg/m³, emission rates, stability classes and dispersion, scrubbers and adsorbers, cyclones, baghouses, ESPs, and indoor-air box models (2 h). Work problems (3 h). Fix errors (2 h). Handbook searches (1 h). | Problems 11–12 done; a table matching each pollutant phase to a control device |
| 5 | Environmental Chemistry; Fundamental Principles | Review stoichiometry, equilibrium and pH, kinetics, organic nomenclature, Henry's law and Kow, population projections, and reactors (2 h). Work problems (3 h). Fix errors (2 h). Handbook searches (1 h). | Problems 13–17 done; CSTR vs. plug-flow explained in your own words |
| 6 | Solid and Hazardous Waste; Health Hazards and Risk Assessment; Energy and Environment | Review landfill volume and leachate, waste compatibility, site characterization, hazard quotient and cancer risk, exposure pathways, and greenhouse gas and energy balances (2 h). Work problems (3 h). Fix errors (2 h). Handbook searches (1 h). | Problems 18, 20, and 24 done; a hazard quotient interpreted correctly |
| 7 | Mathematics; Probability and Statistics; Engineering Economics; Ethics and Professional Practice; Thermodynamics | Review these areas and practice selecting the right model before working on speed. Cover cash-flow timelines, sample vs. population statistics, CWA/CAA/RCRA/CERCLA/SDWA/NEPA/OSHA, and first- and second-law problems (8 h total, about half on fresh practice problems). | Problems 9, 19, 21, 22, and 23 done |
| 8 | Mixed timed practice and logistics | Do mixed sets across all 15 areas under time pressure (4 h). Retry every problem in your error log (2 h). Confirm your calculator model, ID, and appointment details (1 h). Review the governing equations behind your remaining errors (1 h). | Every error-log entry retried; logistics checked |
Week 7 lists problem 9 again on purpose. It's a math problem in a hydrology setting, so redo it as math review.
One study hour, start to finish
- 10 minutes: Recall the governing idea and find it in the handbook.
- 30 minutes: Work a few problems without peeking at solutions.
- 15 minutes: For each miss, write down what went wrong, including model and unit errors.
- 5 minutes: Choose one fix and schedule a retry for a later day.
Adjust the plan
- Only 4 hours a week: Spread each week over two calendar weeks. That's 16 weeks for the same 64 hours.
- Recent coursework and strong basics: Compress to 4 weeks at about 16 hours per week by pairing weeks 1–2, 3–4, 5–6, and 7–8. Don't skip an area just because one practice problem felt easy.
- Out of school for a while: Add one or two weeks of math, units, and chemistry basics before week 1. Speed won't fix a wrong equation.
- Retaking after a fail: Start from your NCEES diagnostic report. Put first the areas where you scored well below passing examinees, especially areas with more questions. The report's 0–15 scale isn't a percentage correct, and it doesn't tell you how far you were from passing (NCEES Examinee Guide, pp. 20–21).
Keep an error log
| What to record | Example (problem 5) |
|---|---|
| Problem and date | Problem 5, week 2 |
| NCEES area | Fluid Mechanics and Hydraulics |
| Error type (concept, model choice, algebra, units, calculator entry, reading, pacing) | Model setup |
| What exactly went wrong | Used 8/4 instead of (8/4)² |
| Fix | Sketch both circles. Area goes with diameter squared. |
| Handbook search word that found it | continuity |
| Retry date and result | Week 4: new diameter ratio, solved correctly |
Write down only what the retry actually showed: that this specific error is fixed. One correct retry doesn't mean the whole area is mastered.
Using the handbook and the clock
On exam day you'll have the current FE Reference Handbook as a searchable PDF beside each question. Search with the box on the left side of the reference. Ctrl+F doesn't work. You can't bring notes or formula sheets. Pearson gives you two reusable booklets and three markers for scratch work (NCEES Examinee Guide, pp. 8–10). You can download the same handbook free from MyNCEES. It's for your personal use, and NCEES doesn't allow it to be reposted or redistributed without written permission (NCEES Help Center).
For each practice problem:
- Name what's being asked and write its units before you search.
- Pick the model. Is it loading, detention, continuity, decay, Darcy flux, or something else?
- Search the concept and read the variable definitions and assumptions next to the equation.
- Set it up with units and convert only what you need to.
- Sanity-check the direction, size, and units of your answer.
Keep a running list of the search words that found each equation. By week 8 you'll know where things are, and on exam day search speed is time.
Pace yourself. The exam has 320 minutes for 110 questions, about 2 minutes 55 seconds each on average. That's a budget for the whole exam, not a limit per question. The exam has two sections, which are not individually timed. After about half the questions, you review and submit them, and then you can't go back to them (Examinee Guide, p. 11). Wrong answers cost nothing, so answer every question before you submit a section (NCEES exam scoring). Some questions ask for more than one answer, a typed number, a click on a graphic, or a drag-and-drop, and none gets partial credit (NCEES computer-based testing). Read every instruction line.
FE Environmental exam facts
| Detail | Current rule |
|---|---|
| Questions | 110, including unscored pretest questions |
| Testing time | 5 hours 20 minutes, plus an optional 25-minute scheduled break |
| Appointment length | The NCEES FE page describes a 6-hour appointment. The May 2026 Examinee Guide lists 5 hours 55 minutes. Follow your appointment confirmation. |
| Question types | Mostly single-answer multiple choice, plus multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank. No partial credit. |
| Reference | Current FE Reference Handbook, onscreen and searchable |
| Scoring | Scaled score. No deduction for wrong answers. NCEES doesn't publish the passing score. |
| Results | Usually within 7–10 days. Pass/fail, with a diagnostic report if you don't pass. |
| NCEES fee | $225. Your licensing board may charge its own application fee. |
| Attempts | Once per testing window (Jan–Mar, Apr–Jun, Jul–Sep, Oct–Dec), up to 3 times in any 12 months. Some boards are stricter. |
| Calculators (2026 list) | HP 33s, HP 35s, Casio fx-115 and fx-991 models, TI-30X and TI-36X models. One calculator in the room; an onscreen TI-30XS is also available. |
| Reschedule or cancel an appointment | At least 48 hours ahead; $50 fee paid to Pearson |
| Eligibility and licensure | Your licensing board decides who can sit for the exam. Passing the FE doesn't by itself license you to practice engineering. |
Sources: NCEES FE exam page, NCEES Examinee Guide (May 2026), pp. 2–3, 5–8, 11, 14, 16, and NCEES 2026 approved calculator list (member board memo, Oct. 20, 2025).
Ready to book? Check your licensing board's requirements on the NCEES FE exam page, then follow our NCEES exam registration steps. If you need testing accommodations, you must request them during registration, so read how to request NCEES exam accommodations first.
Quick answers
How long should I study for FE Environmental? The plan above uses about 64 hours over 8 weeks for someone with recent coursework. Stretch it if you've been out of school, and use your missed problems to decide which topics need more work.
What's the passing score? NCEES doesn't publish one. Your number of correct answers becomes a scaled score that adjusts for small differences between exam forms, and that scaled score is compared to the passing standard.
Is the handbook enough? It gives you the equations, not the judgment. You still need to recognize which model fits, handle the units, and find the equation fast with the left-side search box.
Is FE Environmental the right discipline for me, or should I take FE Civil? Pick the discipline that matches your coursework and the PE exam you're aiming for. FE Environmental leans heavily on water, wastewater, air, waste, and chemistry. If you're still deciding, compare the FE disciplines before you register.
Sources and verification
Last verified: October 7, 2026. We checked the NCEES FE exam page, the FE Environmental exam specifications, the May 2026 NCEES Examinee Guide, NCEES's scoring and computer-based-testing pages, NCEES's 2026 calculator list, and NCEES's July 2026 notice of upcoming exam changes, which lists no FE changes. The 24 practice problems are original. We recalculated the numerical answers, checked the methods against the linked technical sources, and reviewed the conceptual answers against their cited sources.
AI tools assisted drafting, source checks, and calculation checks; see our editorial standards.
Official NCEES sources
- NCEES FE exam page
- FE Environmental CBT Exam Specifications, effective July 2020 (PDF)
- NCEES Examinee Guide, May 2026 (PDF)
- NCEES exam scoring
- NCEES computer-based testing and alternative item types
- NCEES reference handbook access
- NCEES member board memo, Oct. 20, 2025 (2026 calculators)
- NCEES member board memo, July 20, 2026 (future exam changes)
The technical source for each practice problem is linked in its explanation.
Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify the subjects we cover; trademarks belong to their respective owners. Our practice problems are original and are not official exam questions. This page doesn't guarantee an exam result or decide your eligibility or licensure.