Free PE Civil Water Resources and Environmental Practice Test
These 40 original practice questions cover all 12 NCEES knowledge areas of the PE Civil: Water Resources and Environmental (WRE) exam, and every question has a worked solution. Pick your answer, then open the solution; the questions are unofficial and written by Castleport, not taken from any NCEES exam.
Practice questions
Question 1: Pump electrical input
Hydraulics–Closed Conduit · Multiple choice · SI
A pump moves 0.040 m³/s of water from one large open reservoir to another whose water surface is 18 m higher. Total hydraulic losses are 7 m at this flow. Neglect velocity head at the reservoir surfaces. The combined pump-and-motor efficiency is 70%. Use ρ = 1,000 kg/m³ and g = 9.81 m/s².
What electrical input power is required?
Show answer and worked solution
Answer: C. 14.0 kW
The pump has to supply the elevation rise plus the losses: H = 18 + 7 = 25 m.
Power delivered to the water: ρgQH = 1,000 × 9.81 × 0.040 × 25 = 9,810 W.
Electrical input: 9,810 ÷ 0.70 = 14,014 W ≈ 14.0 kW.
Why the other choices are wrong
- A. Uses only the 18 m static rise and skips the efficiency.
- B. Stops at hydraulic (water) power. The motor has to draw more than it delivers.
- D. Divides by 0.70 twice. The 70% is already the combined pump-and-motor figure.
Basis: DOE/Hydraulic Institute, Pump Life Cycle Costs (2001), input-power equation with pump and motor efficiency, p. 6.
Question 2: Critical path duration
Project Planning · Multiple choice
Activity durations are in workdays. All links are finish-to-start with zero lag, on one calendar, with enough resources for parallel work.
| Activity | Duration | Predecessors |
|---|---|---|
| A | 2 | — |
| B | 5 | A |
| C | 3 | A |
| D | 4 | B and C |
| E | 4 | C |
| F | 1 | D and E |
What is the minimum project duration?
Show answer and worked solution
Answer: C. 12 workdays
Forward pass (early finish): A = 2; B = 2 + 5 = 7; C = 2 + 3 = 5; D starts after both B and C, so 7 + 4 = 11; E = 5 + 4 = 9; F starts after both D and E, so 11 + 1 = 12 workdays.
The critical (longest) path is A–B–D–F.
Why the other choices are wrong
- A. Starts D when C finishes (day 5) and ignores that B runs to day 7.
- B. Drops activity F at the end.
- D. Adds every duration as if nothing ran in parallel.
Basis: FHWA TA 5080.15, Contract Time Determination, Section 9, Critical Path Method: longest path and early start.
Question 3: Lower present cost
Project Planning · Multiple choice · USD
Two systems give equal service for 10 years. System A costs $90,000 now plus $18,000 at the end of each year. System B costs $125,000 now plus $12,000 at the end of each year. Use a 5% annual discount rate, no escalation, and no salvage value.
Which system has the lower present cost?
Show answer and worked solution
Answer: B. System B, by about $11,300
Uniform series present worth factor: (P/A, 5%, 10) = [(1.05)¹⁰ − 1] / [0.05 × (1.05)¹⁰] = 7.7217.
- A: 90,000 + 18,000 × 7.7217 = $228,991
- B: 125,000 + 12,000 × 7.7217 = $217,661
B is lower by about $11,330.
Why the other choices are wrong
- A. First cost alone ignores ten years of operating cost.
- C. Compares undiscounted totals ($270,000 vs $245,000). Money later is worth less now.
- D. There's no rounding that closes an $11,000 gap.
Basis: NIST Handbook 135 (2025), Life-Cycle Costing Manual, §3.2.2.1 Annually Recurring Uniform Amounts, p. 28; §17.2.2.1 UPV factor, p. 252.
Question 4: Effective overburden stress
Soil Mechanics · Multiple choice · SI
A level site has 2 m of soil above the water table (moist unit weight 18 kN/m³) over saturated soil (20 kN/m³). Pore pressure is hydrostatic, with no capillary suction and no surcharge. Use γw = 9.81 kN/m³.
What is the vertical effective stress 5 m below the ground surface?
Show answer and worked solution
Answer: B. 66.6 kPa
Total stress: 2(18) + 3(20) = 96 kPa.
Pore pressure (3 m below the water table): 3 × 9.81 = 29.43 kPa.
Effective stress: 96 − 29.43 = 66.6 kPa.
Why the other choices are wrong
- A. Subtracts water pressure over the full 5 m, but only 3 m is below the water table.
- C. Total stress. It ignores pore pressure.
- D. Adds the pore pressure instead of subtracting it.
Basis: FHWA LRFD Steel Bridge Design Example, Design Step P, Design Step P.1: wet unit weight above the water table, effective unit weight below.
Question 5: Rankine active thrust
Soil Mechanics · Multiple choice · USCS
A 12-ft-high vertical wall retains level, dry, cohesionless backfill with unit weight 120 lb/ft³ and friction angle φ = 30°. The wall can move enough to develop active pressure. Neglect wall friction, surcharge, and water.
Using Rankine theory, what is the resultant lateral thrust per foot of wall?
Show answer and worked solution
Answer: A. 2.88 kips/ft
Ka = (1 − sin 30°) / (1 + sin 30°) = 0.5 / 1.5 = 1/3.
Pressure is triangular, so the resultant is Pa = ½ Ka γ H² = ½ × (1/3) × 120 × 12² = 2,880 lb/ft = 2.88 kips/ft, acting H/3 = 4 ft above the base.
Why the other choices are wrong
- B. Uses the at-rest coefficient K0 = 1 − sin φ = 0.5. That fits a wall that can't move, not this one.
- C. Uses tan(45° − φ/2) = 0.577 without squaring it.
- D. Forgets the ½ for the triangular pressure diagram.
Basis: USACE EM 1110-2-2502, Retaining and Flood Walls (1989), §3-12b: active force (Eq. 3-11) and Coulomb reducing to Rankine for a vertical wall, level backfill, no wall friction (Eq. 3-15), pp. 3-18 to 3-19.
Question 6: Relative compaction
Materials · Numeric entry · USCS
A field test on compacted fill measures a moist unit weight of 125 lb/ft³ at a water content of 12%. The laboratory maximum dry unit weight for this material (same test basis) is 117 lb/ft³.
What is the relative compaction, in percent? Round to one decimal place.
Enter a number in %.
Show answer and worked solution
Answer: 95.4%
Dry unit weight: γd = γ / (1 + w) = 125 / 1.12 = 111.6 lb/ft³.
Relative compaction: 111.6 / 117 × 100 = 95.4%.
Accepted: 95.3 to 95.5.
Common mistake: Dividing the moist unit weight by the lab dry maximum gives 106.8%. Always compare dry to dry.
Basis: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 5, §5.3.2, Eq. 5.1 (dry unit weight) and Eq. 5.4 (relative compaction).
Question 7: Void ratio of a saturated clay
Materials · Multiple choice
A fully saturated clay sample has a water content of 32% and a specific gravity of solids of 2.70.
What is its void ratio?
Show answer and worked solution
Answer: D. 0.864
Use Se = wGs. Saturated means S = 1, so e = wGs = 0.32 × 2.70 = 0.864.
Why the other choices are wrong
- A. Divides w by Gs instead of multiplying.
- B. Uses the water content itself as the void ratio.
- C. That's the porosity, n = e / (1 + e) = 0.864 / 1.864.
Basis: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 5, Table 5-8 soil identities and Eq. 5.2: Gs·w = S·e.
Question 8: Conservative mixing
Analysis and Design · Multiple choice · SI
Two steady inflows meet at a junction: 30 L/s at 10 mg/L and 20 L/s at 40 mg/L of the same conservative dissolved constituent. Assume complete mixing, no reaction, and no other inflow.
What is the downstream concentration?
Show answer and worked solution
Answer: A. 22 mg/L
Mass in equals mass out: C = (30 × 10 + 20 × 40) / (30 + 20) = 1,100 / 50 = 22 mg/L.
Why the other choices are wrong
- B. Simple average. It ignores that the cleaner stream carries more flow.
- C. Swaps the flows, weighting the 40 mg/L stream by 30 L/s.
- D. Adds the concentrations. Concentrations don't add; masses do.
Basis: EPA EPANET 2.2 User Manual, EPA/600/R-20/133, §12.2 water quality mixing at junctions (flow-weighted).
Question 9: Suspended solids removed
Analysis and Design · Multiple choice · USCS
A treatment step receives 1.8 MGD. Influent TSS is 180 mg/L and effluent TSS is 30 mg/L. Assume the same flow in and out. Use lb/day = MGD × mg/L × 8.34.
How many pounds of TSS are removed per day?
Show answer and worked solution
Answer: C. 2,252 lb/day
Removed = 1.8 × (180 − 30) × 8.34 = 2,252 lb/day of dry solids.
That's a dry-solids mass, not the weight or volume of wet sludge.
Why the other choices are wrong
- A. Leaves out the 8.34 conversion.
- B. That's the effluent load (1.8 × 30 × 8.34).
- D. That's the influent load, not the amount removed.
Basis: Florida DEP, Wastewater Treatment Formulas (April 2024), "Pounds" formula: flow (MGD) × concentration (mg/L) × 8.34.
Question 10: Contracted rectangular weir
Analysis and Design · Multiple choice · USCS
A sharp-crested rectangular weir has a 3.0-ft crest and end contractions on both sides. The measured head is 0.80 ft. The installation meets the Francis-equation conditions; neglect approach velocity. Use the Francis equation, Q = 3.33 (L − 0.1nh) h^1.5, where n is the number of end contractions (L and h in ft, Q in ft³/s).
What is the flow?
Show answer and worked solution
Answer: B. 6.77 ft³/s
Two contractions, so n = 2 and the effective length is L − 0.2h = 3.0 − 0.16 = 2.84 ft.
Q = 3.33 × 2.84 × 0.80^1.5 = 3.33 × 2.84 × 0.7155 = 6.77 ft³/s.
Check: h = 0.80 ft is less than L/3 = 1.0 ft, inside the equation's range.
Why the other choices are wrong
- A. Uses h¹ instead of h^1.5.
- C. Treats the weir as suppressed (n = 0).
- D. Counts only one end contraction (n = 1).
Basis: USBR Water Measurement Manual, Ch. 7, Sec. 9, Eq. 7-4, fully contracted rectangular weir, and its h ≤ L/3 limit.
Question 11: Major and minor pipe losses
Hydraulics–Closed Conduit · Multiple choice · SI
Water flows full at 0.050 m³/s through 300 m of 0.200-m-diameter pipe. The Darcy friction factor is 0.024. Fittings and valves have a combined loss coefficient ΣK = 3.0, based on the pipe velocity. Use g = 9.81 m/s².
What is the total head loss?
Show answer and worked solution
Answer: C. 5.04 m
Area = π(0.2)²/4 = 0.03142 m². Velocity = 0.050 / 0.03142 = 1.592 m/s. Velocity head = 1.592² / (2 × 9.81) = 0.1291 m.
Total loss = (fL/D + ΣK) × V²/2g = (0.024 × 300 / 0.2 + 3.0) × 0.1291 = 39 × 0.1291 = 5.04 m.
Why the other choices are wrong
- A. Uses V instead of V² in the velocity head.
- B. Friction only; it drops the minor losses.
- D. Multiplies the friction factor by 4, a Fanning-to-Darcy conversion that doesn't apply because f is already a Darcy factor.
Basis: EPA EPANET 2.2 User Manual, EPA/600/R-20/133, Table 3.1 head-loss formulas, p. 18; §3.1 minor losses, p. 19.
Question 12: Force main friction loss
Hydraulics–Closed Conduit · Multiple choice · USCS
A 10-in.-diameter force main, 2,000 ft long, carries 1.2 ft³/s. Hazen-Williams C = 120. Use h = 4.727 L Q^1.852 / (C^1.852 D^4.871), with h, L, and D in feet and Q in ft³/s.
What is the friction head loss?
Show answer and worked solution
Answer: D. 4.54 ft
D = 10/12 = 0.8333 ft.
h = 4.727 × 2,000 × 1.2^1.852 / (120^1.852 × 0.8333^4.871) = 4.54 ft.
Velocity check: V = 1.2 / 0.5454 = 2.2 ft/s, a normal force-main velocity.
Why the other choices are wrong
- A. Divides by 2.31. That converts feet of water to psi (1.97 psi), but the question asks for head in feet.
- B. Uses C = 100 instead of the given 120.
- C. Multiplies by 2.31, converting in the wrong direction.
Basis: EPA EPANET 2.2 User Manual, EPA/600/R-20/133, Table 3.1, Hazen-Williams resistance coefficient A = 4.727 C^−1.852 d^−4.871 L (ft, cfs), p. 18.
Question 13: Pump operating point
Hydraulics–Closed Conduit · Numeric entry · SI
The pump curve is Hp = 45 − 800Q² and the system curve is Hs = 9 + 2,800Q², with heads in meters and Q in m³/s. Both curves are valid over the flow range of interest.
What is the operating flow, in m³/s? Round to three decimal places.
Enter a number in m³/s.
Show answer and worked solution
Answer: 0.100 m³/s
The pump operates where its curve meets the system curve: 45 − 800Q² = 9 + 2,800Q², so 36 = 3,600Q² and Q = 0.100 m³/s (positive root).
Check: both curves give 37 m at 0.100 m³/s.
Accepted: 0.099 to 0.101.
Common mistake: Using the shutoff head (45 m) or the static head (9 m) alone. Neither tells you the flow; only the intersection does.
Basis: DOE/Hydraulic Institute, Pump Life Cycle Costs (2001), duty point at the intersection of the pump and system curves, p. 10, Figure 1.
Question 14: Manning discharge in a trapezoid
Hydraulics–Open Channel · Multiple choice · SI
A trapezoidal channel has a 2.00-m bottom width, side slopes of 1.5H:1V, flow depth 0.80 m, Manning n = 0.030, and slope 0.0016. Flow is steady and uniform.
What is the discharge?
Show answer and worked solution
Answer: A. 2.22 m³/s
Area: A = (b + zy)y = (2 + 1.5 × 0.8) × 0.8 = 2.56 m².
Wetted perimeter: P = b + 2y√(1 + z²) = 2 + 2 × 0.8 × 1.803 = 4.884 m.
Hydraulic radius: R = 2.56 / 4.884 = 0.5241 m.
Manning (SI): Q = (1/n) A R^(2/3) S^(1/2) = (1/0.030) × 2.56 × 0.6503 × 0.04 = 2.22 m³/s.
Why the other choices are wrong
- B. Uses R instead of R^(2/3).
- C. Takes the wetted perimeter as if the sides were vertical (2 + 2 × 0.8), which inflates R.
- D. Applies the 1.49 U.S. customary coefficient to SI inputs.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), §4.3.1 Manning's equation, pp. 4-3 to 4-4.
Question 15: Flow regime
Hydraulics–Open Channel · Multiple choice · SI
A rectangular channel 3.0 m wide carries 6.0 m³/s at a depth of 0.50 m. Use g = 9.81 m/s².
What is the flow regime?
Show answer and worked solution
Answer: C. Supercritical
V = Q / (by) = 6.0 / (3.0 × 0.50) = 4.0 m/s.
Fr = V / √(gy) = 4.0 / √(9.81 × 0.50) = 1.81 > 1, supercritical.
Cross-check: critical depth yc = (q²/g)^(1/3) = (2²/9.81)^(1/3) = 0.74 m. Actual depth 0.50 m is below it.
Why the other choices are wrong
- A. Subcritical needs Fr < 1 (depth above 0.74 m here).
- B. Critical would be Fr = 1, depth 0.74 m.
- D. Depth and velocity are enough. Slope tells you the normal depth, not the regime of this flow.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), Froude number definition, Glossary p. xiv; regime criteria, p. 3-1.
Question 16: What a hydraulic jump changes
Hydraulics–Open Channel · Select all that apply
A stationary hydraulic jump forms in a horizontal rectangular channel of constant width carrying steady flow, with no inflow or leakage along the jump.
Select all true statements.
Show answer and worked solution
Answer: A, B, C
Credit requires exactly these three; any extra or missing choice scores zero.
- A: continuity. Nothing enters or leaves, so Q is unchanged.
- B: a jump is the transition from supercritical to subcritical flow.
- C: the turbulent roller converts mechanical energy to heat. That's why jumps are used in stilling basins.
Why the other statements are false
- D. Reversed. Depth increases through the jump.
- E. Momentum is conserved across the jump; specific energy is not. The drop in specific energy is the head lost in the jump.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), §4.6.4 hydraulic jump, p. 4-32.
Question 17: Culvert control
Hydraulics–Open Channel · Multiple choice · USCS
At the design flow, a culvert analysis gives a headwater of 6.2 ft assuming inlet control and 7.4 ft assuming outlet control.
Which control governs, and what headwater should be used for design?
Show answer and worked solution
Answer: D. Outlet control; 7.4 ft
Compute both and use the higher headwater. The culvert can't pass the flow at less than the larger of the two, so outlet control governs at 7.4 ft.
Why the other choices are wrong
- A. Reporting the lower value understates the headwater the culvert actually needs.
- B. The 7.4 ft comes from the outlet-control calculation, so it can't be labeled inlet control.
- C. Averaging the two has no hydraulic meaning.
Basis: TxDOT Hydraulic Design Manual, Ch. 8, Sec. 3, "Headwater under Outlet Control" (both controls considered) and "Slug Flow" (assume the higher of inlet and outlet control headwater).
Question 18: Composite rational runoff
Hydrology · Multiple choice · SI
A 10-ha drainage area has 4 ha with C = 0.90 and 6 ha with C = 0.30. The design rainfall intensity, for a duration equal to the time of concentration, is 65 mm/h. Use Q = CiA / 360 (Q in m³/s, i in mm/h, A in ha).
What is the peak runoff?
Show answer and worked solution
Answer: B. 0.975 m³/s
Area-weighted C: ΣCA = 0.90 × 4 + 0.30 × 6 = 5.4 ha (C = 0.54).
Q = 65 × 5.4 / 360 = 0.975 m³/s.
Why the other choices are wrong
- A. Divides by 3,600 instead of 360.
- C. Averages the two coefficients (0.60) without weighting by area.
- D. Applies each coefficient to the full 10 ha.
Basis: FHWA HDS-4, Introduction to Highway Hydraulics (2008), §2.5.1, Eq. 2.1, Ku = 360 for SI, p. 2-5.
Question 19: Curve number runoff depth
Hydrology · Numeric entry · SI
Event rainfall is P = 85 mm on a watershed with composite CN = 78. Use S = 25,400/CN − 254 (mm), Ia = 0.2S, and Q = (P − Ia)² / (P − Ia + S) when P > Ia.
What is the direct runoff depth, in mm? Round to one decimal place.
Enter a number in mm.
Show answer and worked solution
Answer: 35.1 mm
S = 25,400/78 − 254 = 71.64 mm.
Ia = 0.2 × 71.64 = 14.33 mm.
Q = (85 − 14.33)² / (85 − 14.33 + 71.64) = 70.67² / 142.31 = 35.1 mm.
Accepted: 34.9 to 35.3.
Common mistake: Skipping the initial abstraction gives 46.1 mm. Another slip is plugging the inch form of S (1000/CN − 10) into a millimetre problem.
Basis: USACE HEC-HMS Technical Reference, SCS Curve Number Loss Model, Eqs. 1, 2, and 4 (SI form of S).
Question 20: Exceedance over a design life
Hydrology · Multiple choice
A flood level has a 2% annual exceedance probability. Assume each year is independent with the same probability.
What is the chance it is exceeded at least once in 30 years?
Show answer and worked solution
Answer: D. 45.5%
P(at least once) = 1 − P(never) = 1 − (1 − 0.02)³⁰ = 1 − 0.5455 = 45.5%.
Why the other choices are wrong
- A. That's the chance in any single year.
- B. That's the chance of no exceedance in 30 years.
- C. Multiplies 2% × 30. That overcounts years with more than one exceedance and would pass 100% for long periods.
Basis: USACE HEC-FDA Technical Reference, Long-Term Exceedance Probability, LTEP = 1 − (1 − AEP)^N.
Question 21: One storage-routing step
Hydrology · Multiple choice · SI
Route one time step through a detention pond. Δt = 600 s. At the start: storage S₀ = 5,000 m³, inflow I₀ = 3.0 m³/s, outflow O₀ = 2.0 m³/s. Inflow at the end of the step is I₁ = 5.0 m³/s. Over this range, storage and outflow are related by S = 2,500 O (S in m³, O in m³/s). Use (2S₁/Δt + O₁) = (I₀ + I₁) + (2S₀/Δt − O₀).
What is the outflow at the end of the step?
Show answer and worked solution
Answer: B. 2.43 m³/s
Right side: (3 + 5) + (2 × 5,000/600 − 2) = 8 + 16.667 − 2 = 22.667.
Left side with S₁ = 2,500 O₁: 2 × 2,500 O₁/600 + O₁ = 9.333 O₁.
O₁ = 22.667 / 9.333 = 2.43 m³/s (S₁ = 6,071 m³).
Check: storage gain = (avg inflow 4.0 − avg outflow 2.21) × 600 = 1,071 m³ = 6,071 − 5,000.
Why the other choices are wrong
- A. Holds outflow constant, ignoring the storage rise.
- C. Drops the −O₀ term from the right side.
- D. Sets outflow equal to average inflow, as if there were no storage.
Basis: USACE HEC-HMS Technical Reference, Reservoir Modeling, "Defining Routing": continuity and the (2S/Δt + O) rearrangement.
Question 22: Unit hydrograph convolution
Hydrology · Multiple choice · USCS
A watershed's 1-hour unit hydrograph (cfs per inch of runoff) has these ordinates:
| Time (h) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| UH (cfs/in.) | 0 | 100 | 300 | 200 | 100 | 0 |
A storm produces 0.5 in. of runoff in the first hour and 1.2 in. in the second hour. Ignore baseflow.
What is the peak of the direct runoff hydrograph?
Show answer and worked solution
Answer: C. 460 cfs
Scale each block's unit hydrograph by its runoff depth and lag the second block by 1 hour, then add:
| Time (h) | 0.5 × UH | 1.2 × UH (lagged 1 h) | Total |
|---|---|---|---|
| 1 | 50 | 0 | 50 |
| 2 | 150 | 120 | 270 |
| 3 | 100 | 360 | 460 |
| 4 | 50 | 240 | 290 |
| 5 | 0 | 120 | 120 |
Peak = 460 cfs at hour 3.
Why the other choices are wrong
- A. That's the unit hydrograph peak for 1 inch, not this storm.
- B. Uses only the second block (1.2 × 300).
- D. Adds both scaled peaks without the 1-hour lag (0.5 × 300 + 1.2 × 300).
Basis: NRCS NEH Part 630, Chapter 16, Hydrographs (2007), §630.1602, proportionality and superposition, p. 16-2.
Question 23: Pore-water velocity
Groundwater and Wells · Multiple choice · SI
Saturated flow in a homogeneous aquifer has a head drop of 1.8 m over 300 m. Hydraulic conductivity is 12 m/day and effective porosity is 0.24.
What is the average pore-water (seepage) velocity?
Show answer and worked solution
Answer: C. 0.300 m/day
Gradient i = 1.8 / 300 = 0.006.
Darcy flux q = Ki = 12 × 0.006 = 0.072 m/day.
Seepage velocity v = q / ne = 0.072 / 0.24 = 0.300 m/day. Water only moves through the connected pores, so it travels faster than the Darcy flux.
Why the other choices are wrong
- A. Multiplies by porosity instead of dividing.
- B. That's the Darcy flux, which spreads flow over the whole cross-section.
- D. Divides K by porosity and leaves out the gradient.
Basis: EPA/540/R-96/003, Soil Screening Guidance: Technical Background, §2.3: velocity is inversely proportional to porosity.
Question 24: Head difference between observation wells
Groundwater and Wells · Multiple choice · SI
A fully penetrating well pumps 600 m³/day from a confined aquifer 30 m thick with hydraulic conductivity 10 m/day. Flow is steady and radial, with no boundary effects. Use the Thiem equation, h₂ − h₁ = Q ln(r₂/r₁) / (2πT).
What is the head difference between observation wells at 100 m and 10 m from the pumping well?
Show answer and worked solution
Answer: B. 0.733 m
Transmissivity T = Kb = 10 × 30 = 300 m²/day.
h₂ − h₁ = 600 × ln(10) / (2π × 300) = 600 × 2.3026 / 1,885 = 0.733 m. The well closer to pumping has the lower head.
Why the other choices are wrong
- A. Uses log base 10 instead of the natural log.
- C. Uses π instead of 2π.
- D. Uses K (10 m/day) where T belongs.
Basis: EPA/540/R-96/003, Soil Screening Guidance: Technical Background, §1.3 Darcy's law as the governing flow equation; Thiem equation supplied in the stem.
Question 25: Oxygen deficit rate
Surface Water and Groundwater Quality · Multiple choice
At one point in a stream, the remaining ultimate BOD is L = 12 mg/L and the dissolved oxygen deficit is D = 2 mg/L. Use the oxygen-sag model dD/dt = k₁L − k₂D, with base-e rates k₁ = 0.20/day and k₂ = 0.50/day. Saturation DO is constant.
At this instant, what is happening to the dissolved oxygen?
Show answer and worked solution
Answer: A. DO is decreasing at 1.4 mg/L per day
dD/dt = 0.20 × 12 − 0.50 × 2 = 2.4 − 1.0 = +1.4 mg/L per day.
The deficit is growing. Since DO = DOsat − D and DOsat is constant, DO is falling at 1.4 mg/L per day. This point is upstream of the critical (lowest-DO) point.
Why the other choices are wrong
- B. Treats the deficit as the DO itself and gets the sign backward.
- C. Leaves out reaeration (k₂D).
- D. Adds reaeration instead of subtracting it.
Basis: Derived: algebra on the classic oxygen-sag deficit model given in the stem.
Question 26: TMDL allocation
Surface Water and Groundwater Quality · Multiple choice · SI
A TMDL for a pollutant is set at 120 kg/day. The load allocation for nonpoint sources and natural background is 35 kg/day. An explicit margin of safety equal to 10% of the TMDL is reserved. Nothing else is set aside.
What total wasteload allocation is left for point sources?
Show answer and worked solution
Answer: D. 73 kg/day
TMDL = ΣWLA + ΣLA + MOS.
MOS = 0.10 × 120 = 12 kg/day.
ΣWLA = 120 − 35 − 12 = 73 kg/day.
WLA covers point sources (like treatment plants); LA covers nonpoint sources and background.
Why the other choices are wrong
- A. Subtracts the margin of safety twice.
- B. Takes 10% of what's left after the LA (8.5 kg/day) instead of 10% of the total TMDL.
- C. Forgets the margin of safety.
Basis: EPA, Overview of Total Maximum Daily Loads, "What is a TMDL?": TMDL = ΣWLA + ΣLA + MOS, with definitions.
Question 27: Required log reduction
Surface Water and Groundwater Quality · Multiple choice
A source water contains 2.5 × 10⁴ organisms per 100 mL. A project's treatment target is no more than 5 organisms per 100 mL.
What minimum log reduction is needed?
Show answer and worked solution
Answer: B. 3.7-log
Log reduction = log₁₀(C_in / C_out) = log₁₀(25,000 / 5) = log₁₀(5,000) = 3.7.
That's a 99.98% reduction. Each log is a factor of 10: 3-log = 99.9%.
Why the other choices are wrong
- A. 3-log leaves 25 per 100 mL, which misses the target.
- C. That's log₁₀ of the starting count alone.
- D. More than needed; it isn't the minimum.
Basis: EPA, Disinfection Profiling and Benchmarking Technical Guidance Manual, §1.2: log inactivation definition (3-log = 99.9%), p. 2.
Question 28: Equalization storage
Drinking Water Distribution and Treatment · Multiple choice · USCS
A water system's supply is constant at 1.2 MGD. Demand repeats daily: 1.8 MGD for 6 hours, then 1.0 MGD for the other 18 hours. The tank is full when the high-demand period starts. Ignore losses, fire flow, emergency storage, and dead storage.
What is the minimum usable volume needed just for equalization?
Show answer and worked solution
Answer: A. 150,000 gal
During the 6-hour peak, demand exceeds supply by 0.6 MGD: 0.6 × 6/24 = 0.15 MG drawn from storage.
During the other 18 hours, supply exceeds demand by 0.2 MGD: 0.2 × 18/24 = 0.15 MG refilled. The cycle balances.
Required equalization volume = 150,000 gal.
Why the other choices are wrong
- B. Counts the drawdown and the refill as two separate volumes.
- C. Uses the whole 6-hour demand (1.8 × 6/24), not the shortfall.
- D. That's a full day of supply.
Basis: USACE HEC-HMS Technical Reference, Reservoir Modeling, continuity: inflow − outflow = change in storage.
Question 29: Baffling-adjusted CT
Drinking Water Distribution and Treatment · Numeric entry · SI
A disinfection contact basin has a minimum operating volume of 180 m³. Peak-hour flow is 6.0 m³/min. The baffling factor is 0.30. The free chlorine residual at the outlet is 0.80 mg/L.
What is CT, in mg·min/L? Round to one decimal place.
Enter a number in mg·min/L.
Show answer and worked solution
Answer: 7.2 mg·min/L
Theoretical detention time: TDT = V/Q = 180 / 6.0 = 30 min.
Contact time: T = TDT × BF = 30 × 0.30 = 9 min.
CT = 0.80 × 9 = 7.2 mg·min/L.
Accepted: 7.1 to 7.3. Computing CT doesn't by itself show that an inactivation requirement is met; that comparison uses CT tables for the specific organism and disinfectant, which depend on conditions such as temperature and, for free chlorine, pH and residual.
Common mistake: Skipping the baffling factor gives 24 mg·min/L. Using the applied chlorine dose instead of the measured residual also overstates CT.
Basis: EPA, Disinfection Profiling and Benchmarking Technical Guidance Manual, Eqs. 4-1 to 4-3 (CT = C × T; TDT = V/Q at peak hourly flow; T = TDT × BF), pp. 25–29.
Question 30: Filters with one out of service
Drinking Water Distribution and Treatment · Multiple choice · USCS
A water treatment plant's design capacity is 1.60 MGD. Each identical rapid-rate gravity filter can treat 0.90 MGD at the approved filtration rate. The design criterion for this project requires the plant to meet design capacity with one filter out of service.
What is the minimum number of filters?
Show answer and worked solution
Answer: B. 3
With n filters and one out, firm capacity is (n − 1) × 0.90 MGD.
- 2 filters: 0.90 MGD. Too small.
- 3 filters: 1.80 MGD ≥ 1.60. 3 filters is the minimum.
On the exam, the redundancy rule comes from the design standard NCEES lists for your test date, so look it up in that edition rather than relying on memory.
Why the other choices are wrong
- A. Two filters give 1.80 MGD total, but only 0.90 MGD with one out.
- C. Works, but it's more than the minimum.
- D. Works, but it's more than the minimum.
Basis: Derived: firm-capacity check against the design criterion given in the stem.
Question 31: Total hardness
Drinking Water Distribution and Treatment · Multiple choice
A groundwater has 64 mg/L calcium (Ca²⁺) and 18 mg/L magnesium (Mg²⁺). Equivalent weights: Ca²⁺ 20.04, Mg²⁺ 12.15, CaCO₃ 50.04 mg/meq. Classify it with the USGS scale: soft 0–60, moderately hard 61–120, hard 121–180, very hard above 180 mg/L as CaCO₃.
What are the total hardness and its classification?
Show answer and worked solution
Answer: C. 234 mg/L as CaCO₃, very hard
Convert each ion to CaCO₃ equivalents:
- Ca: 64 × 50.04/20.04 = 159.8 mg/L
- Mg: 18 × 50.04/12.15 = 74.1 mg/L
Total = 234 mg/L as CaCO₃, very hard (above 180).
Why the other choices are wrong
- A. Adds the raw ion concentrations without converting to CaCO₃.
- B. Counts calcium only.
- D. Uses CaCO₃'s molar mass (100.09) over the ions' equivalent weights, mixing a molar basis with an equivalent basis and doubling the answer.
Basis: USGS Water Science School, Hardness of Water, hardness classification ranges in mg/L as CaCO₃.
Question 32: Food-to-microorganism ratio
Wastewater Collection and Treatment · Multiple choice · SI
An aeration basin receives 5,000 m³/day with BOD₅ of 160 mg/L. The basin volume is 2,000 m³ and MLVSS is 2,500 mg/L. Define F/M as applied BOD₅ per day divided by the MLVSS mass in the basin.
What is the F/M ratio?
Show answer and worked solution
Answer: D. 0.160 /day
Food: 5,000 m³/day × 160 g/m³ = 800 kg BOD₅/day. (1 mg/L = 1 g/m³.)
Microorganisms: 2,000 m³ × 2,500 g/m³ = 5,000 kg MLVSS.
F/M = 800 / 5,000 = 0.160 /day.
Why the other choices are wrong
- A. A unit slip of 1,000 between grams and kilograms on one side only.
- B. Divides the BOD load by basin volume (a volumetric loading), not by biomass.
- C. Inverts the ratio.
Basis: Florida DEP, Wastewater Treatment Formulas (April 2024), F/M ratio formula (BOD applied ÷ MLVSS under aeration).
Question 33: Conventional nitrogen removal
Wastewater Collection and Treatment · Select all that apply
Which statements describe conventional biological nitrification followed by heterotrophic denitrification? Select all that apply.
Show answer and worked solution
Answer: A, B, D
Credit requires exactly these three; any extra or missing choice scores zero.
- A: autotrophic bacteria oxidize ammonia to nitrite, then nitrite to nitrate, with oxygen present.
- B: denitrification reduces nitrate (through intermediates) to nitrogen gas without free oxygen.
- D: heterotrophic denitrifiers need an organic carbon source such as influent BOD or methanol.
Why the other statements are false
- C. Nitrification converts ammonia to nitrate. The nitrogen is still in the water until denitrification turns it into gas.
- E. Anoxic means oxygen is present only in combined form, like nitrate. Anaerobic means neither free nor combined oxygen is available.
Basis: EPA-823-R-07-002, Biological Nutrient Removal Processes and Costs (2007), p. 1 (nitrification, denitrification, electron donors); p. 3 footnote 1 (anoxic vs anaerobic).
Question 34: Solids retention time
Wastewater Collection and Treatment · Multiple choice · SI
The total suspended solids inventory in a secondary treatment system is 12,600 kg. Solids leave only as waste activated sludge (900 kg/day) and in the final effluent (150 kg/day).
What is the solids retention time?
Show answer and worked solution
Answer: A. 12 days
SRT = solids in the system ÷ solids leaving per day = 12,600 / (900 + 150) = 12 days.
Return sludge isn't a loss; it stays inside the system.
Why the other choices are wrong
- B. Ignores effluent solids, which overstates SRT.
- C. Counts only the effluent solids as a loss.
- D. Inverts the ratio.
Basis: Florida DEP, Wastewater Treatment Formulas (April 2024), Sludge age / MCRT / SRT formula (solids in system ÷ solids lost in effluent plus solids wasted).
Question 35: What smoke testing finds
Wastewater Collection and Treatment · Multiple choice
A utility is running a sewer system evaluation to reduce wet-weather flows.
Which finding is smoke testing most directly suited to locate?
Show answer and worked solution
Answer: A. A roof downspout or area drain connected to the sanitary sewer
Smoke blown into the sewer comes out wherever the system is connected to the surface. That makes it good at finding direct inflow connections like downspouts, area drains, and catch basins (plus some indirect connections).
Inflow is surface or building water entering through connections. Infiltration is groundwater entering through defects; it rises and falls with the water table.
Why the other choices are wrong
- B. Smoke shows where it escapes; it doesn't measure a flow rate. Infiltration is quantified from flow data.
- C. That's measured with gas monitoring; it's an odor and corrosion issue.
- D. That's a pump-test and design check.
Basis: MassDEP, Guidelines for I/I Analyses and Sewer System Evaluation Surveys (2017), §IV definitions of infiltration and inflow, p. 4; §VI-6 Smoke Testing purpose, p. 26.
Question 36: Earthwork by average end areas
Project Sitework · Multiple choice · SI
Along a straight alignment, cut cross-sections are 12 m² at station 0, 20 m² at station 30 m, and 14 m² at station 70 m. Assume area changes linearly between stations; there is no fill.
What is the bank excavation volume?
Show answer and worked solution
Answer: D. 1,160 m³
Average end area, segment by segment:
- 0 to 30 m: (12 + 20)/2 × 30 = 480 m³
- 30 to 70 m: (20 + 14)/2 × 40 = 680 m³
Total = 1,160 m³ (bank measure).
Why the other choices are wrong
- A. Averages all three areas and multiplies by 70 m, ignoring the unequal spacing.
- B. Sums all three areas and multiplies by the average 35 m spacing.
- C. Forgets to halve the sum of end areas.
Basis: Derived: integrating a cross-sectional area that varies linearly between stations gives the average-end-area result.
Question 37: Compacted fill from a cut
Project Sitework · Multiple choice · SI
A cut produces 9,000 bank m³ with a dry density of 1.60 Mg/m³. Five percent of the dry soil mass is unsuitable and hauled off. The rest is compacted to a dry density of 1.80 Mg/m³ with no further loss.
What compacted fill volume does it make?
Show answer and worked solution
Answer: A. 7,600 m³
Track the dry solids, which don't change when you compact them.
Usable dry mass: 9,000 × 1.60 × 0.95 = 13,680 Mg.
Compacted volume: 13,680 / 1.80 = 7,600 m³.
Why the other choices are wrong
- B. Forgets the 5% loss.
- C. Applies the loss but not the density change.
- D. Inverts the density ratio (1.80/1.60).
Basis: FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 5, Table 5-8 unit weight–volume relationships (dry density basis).
Question 38: Sag curve low point
Project Sitework · Multiple choice · SI
A sag vertical curve has elevation z(x) = 100 − 0.020x + 0.000100x², with z and x in meters, for 0 ≤ x ≤ 200 m from the beginning of the curve.
How far from the beginning of the curve is the low point?
Show answer and worked solution
Answer: C. 100 m
The low point is where the grade is zero: dz/dx = −0.020 + 0.000200x = 0, so x = 100 m.
It's inside the curve, and d²z/dx² > 0 confirms a minimum. Elevation there: 100 − 2 + 1 = 99.0 m.
Why the other choices are wrong
- A. The start of the curve. The grade there is still −2%, so the road keeps dropping.
- B. That's the low-point elevation, not its distance.
- D. Divides 0.020 by 0.000100 and forgets the 2 from differentiating x².
Basis: Derived: calculus on the profile equation given in the stem.
Question 39: Sloped trench width
Project Sitework · Multiple choice · USCS
A 12-ft-deep trench in soil a competent person has classified as Type B will be protected by simple sloping on both sides (no benching or other protective system). The excavation stays open longer than 24 hours. The bottom width is 6 ft.
Under OSHA Subpart P Appendix B, what is the minimum top width?
Show answer and worked solution
Answer: C. 30 ft
For excavations 20 ft deep or less, Appendix B allows a maximum slope of 1H:1V in Type B soil.
Each side extends 12 ft horizontally: top width = 6 + 2 × 12 = 30 ft.
Why the other choices are wrong
- A. Slopes only one side.
- B. Uses the Type A slope (¾H:1V).
- D. Uses the Type C slope (1½H:1V).
Basis: 29 CFR 1926 Subpart P, Appendix B (eCFR), §B-1.1 to B-1.3 maximum allowable slopes (Type A ¾:1, Type B 1:1, Type C 1½:1).
Question 40: Erosion control vs sediment control
Project Sitework · Multiple choice
A disturbed slope needs temporary protection while vegetation gets established.
Which measure works directly by covering and stabilizing the exposed soil?
Show answer and worked solution
Answer: A. Anchored mulch
Mulch is erosion control: it protects the soil surface so particles don't detach in the first place.
The other three are sediment controls. They catch soil after it has already eroded and moved.
Why the other choices are wrong
- B. Traps sediment downstream; it doesn't protect the slope.
- C. Keeps sediment out of the storm drain after it's moving.
- D. Intercepts sheet flow and sediment at the perimeter.
Basis: EPA Stormwater Best Management Practice: Mulching, Description and Applicability, p. 1.
Answer key
Each worked solution sits with its question above. Use this table to mark a full attempt quickly.
| Q | Answer | Knowledge area | Topic |
|---|---|---|---|
| 1 | C. 14.0 kW | Hydraulics–Closed Conduit | Pump electrical input |
| 2 | C. 12 workdays | Project Planning | Critical path duration |
| 3 | B. System B, by about $11,300 | Project Planning | Lower present cost |
| 4 | B. 66.6 kPa | Soil Mechanics | Effective overburden stress |
| 5 | A. 2.88 kips/ft | Soil Mechanics | Rankine active thrust |
| 6 | 95.4% | Materials | Relative compaction |
| 7 | D. 0.864 | Materials | Void ratio of a saturated clay |
| 8 | A. 22 mg/L | Analysis and Design | Conservative mixing |
| 9 | C. 2,252 lb/day | Analysis and Design | Suspended solids removed |
| 10 | B. 6.77 ft³/s | Analysis and Design | Contracted rectangular weir |
| 11 | C. 5.04 m | Hydraulics–Closed Conduit | Major and minor pipe losses |
| 12 | D. 4.54 ft | Hydraulics–Closed Conduit | Force main friction loss |
| 13 | 0.100 m³/s | Hydraulics–Closed Conduit | Pump operating point |
| 14 | A. 2.22 m³/s | Hydraulics–Open Channel | Manning discharge in a trapezoid |
| 15 | C. Supercritical | Hydraulics–Open Channel | Flow regime |
| 16 | A, B, C | Hydraulics–Open Channel | What a hydraulic jump changes |
| 17 | D. Outlet control; 7.4 ft | Hydraulics–Open Channel | Culvert control |
| 18 | B. 0.975 m³/s | Hydrology | Composite rational runoff |
| 19 | 35.1 mm | Hydrology | Curve number runoff depth |
| 20 | D. 45.5% | Hydrology | Exceedance over a design life |
| 21 | B. 2.43 m³/s | Hydrology | One storage-routing step |
| 22 | C. 460 cfs | Hydrology | Unit hydrograph convolution |
| 23 | C. 0.300 m/day | Groundwater and Wells | Pore-water velocity |
| 24 | B. 0.733 m | Groundwater and Wells | Head difference between observation wells |
| 25 | A. DO is decreasing at 1.4 mg/L per day | Surface Water and Groundwater Quality | Oxygen deficit rate |
| 26 | D. 73 kg/day | Surface Water and Groundwater Quality | TMDL allocation |
| 27 | B. 3.7-log | Surface Water and Groundwater Quality | Required log reduction |
| 28 | A. 150,000 gal | Drinking Water Distribution and Treatment | Equalization storage |
| 29 | 7.2 mg·min/L | Drinking Water Distribution and Treatment | Baffling-adjusted CT |
| 30 | B. 3 | Drinking Water Distribution and Treatment | Filters with one out of service |
| 31 | C. 234 mg/L as CaCO₃, very hard | Drinking Water Distribution and Treatment | Total hardness |
| 32 | D. 0.160 /day | Wastewater Collection and Treatment | Food-to-microorganism ratio |
| 33 | A, B, D | Wastewater Collection and Treatment | Conventional nitrogen removal |
| 34 | A. 12 days | Wastewater Collection and Treatment | Solids retention time |
| 35 | A. A roof downspout or area drain connected to the sanitary sewer | Wastewater Collection and Treatment | What smoke testing finds |
| 36 | D. 1,160 m³ | Project Sitework | Earthwork by average end areas |
| 37 | A. 7,600 m³ | Project Sitework | Compacted fill from a cut |
| 38 | C. 100 m | Project Sitework | Sag curve low point |
| 39 | C. 30 ft | Project Sitework | Sloped trench width |
| 40 | A. Anchored mulch | Project Sitework | Erosion control vs sediment control |
Score your attempt
Give yourself 1 point per question. A select-all question counts only if you picked exactly the right set. A numeric entry counts if it falls inside the accepted range shown with its solution. There's no partial credit, which matches how NCEES scores every item.
| Knowledge area | Questions | Your correct |
|---|---|---|
| Project Planning | 2, 3 | ___ / 2 |
| Soil Mechanics | 4, 5 | ___ / 2 |
| Materials | 6, 7 | ___ / 2 |
| Analysis and Design | 8, 9, 10 | ___ / 3 |
| Hydraulics–Closed Conduit | 1, 11, 12, 13 | ___ / 4 |
| Hydraulics–Open Channel | 14, 15, 16, 17 | ___ / 4 |
| Hydrology | 18, 19, 20, 21, 22 | ___ / 5 |
| Groundwater and Wells | 23, 24 | ___ / 2 |
| Surface Water and Groundwater Quality | 25, 26, 27 | ___ / 3 |
| Drinking Water Distribution and Treatment | 28, 29, 30, 31 | ___ / 4 |
| Wastewater Collection and Treatment | 32, 33, 34, 35 | ___ / 4 |
| Project Sitework | 36, 37, 38, 39, 40 | ___ / 5 |
| Total | 1–40 | ___ / 40 |
Your score describes these 40 questions only. It isn't an NCEES scaled score or a prediction of passing. With two to five questions per area, one miss moves an area a lot, so use the misses to pick what to study, not to rate yourself.
NCEES reports PE results as pass or fail. It converts correct answers to a scaled score, doesn't deduct for wrong answers, and doesn't publish the passing score. Examinees who fail get a diagnostic report by major topic. (NCEES Examinee Guide, May 2026, Sections 4 and 6)
What this practice test covers
The NCEES column is the official number of questions per area on the 80-question exam. We split our 40 questions roughly in proportion to the midpoint of each range. That split is our choice, not an official weighting.
| Knowledge area | NCEES questions (of 80) | Questions here | Question numbers |
|---|---|---|---|
| 1. Project Planning | 4–6 | 2 | 2, 3 |
| 2. Soil Mechanics | 3–5 | 2 | 4, 5 |
| 3. Materials | 4–6 | 2 | 6, 7 |
| 4. Analysis and Design | 6–9 | 3 | 8, 9, 10 |
| 5. Hydraulics–Closed Conduit | 7–11 | 4 | 1, 11, 12, 13 |
| 6. Hydraulics–Open Channel | 7–11 | 4 | 14, 15, 16, 17 |
| 7. Hydrology | 8–12 | 5 | 18, 19, 20, 21, 22 |
| 8. Groundwater and Wells | 4–6 | 2 | 23, 24 |
| 9. Surface Water and Groundwater Quality | 5–8 | 3 | 25, 26, 27 |
| 10. Drinking Water Distribution and Treatment | 6–9 | 4 | 28, 29, 30, 31 |
| 11. Wastewater Collection and Treatment | 7–11 | 4 | 32, 33, 34, 35 |
| 12. Project Sitework | 9–14 | 5 | 36, 37, 38, 39, 40 |
| Total | 80 | 40 |
Source: NCEES PE Civil WRE exam specifications, effective April 2024.
This set samples every area, not every subtopic. It doesn't test bearing capacity, slope stability, concrete and piping materials, coagulation and flocculation, sedimentation basins, membranes, preliminary treatment, sludge disposal, construction layout, retaining walls, or horizontal curves. Those are gaps in this set, not signs they matter less on the exam.
The real exam also uses drag-and-drop and point-and-click items. This set uses multiple choice, numeric entry, and select-all questions.
Review your misses
- Rework every missed question on paper before you reread its solution.
- Name the cause of each miss using the table below. The fix depends on the cause.
- For a reference-lookup miss, find the method in the NCEES PE Civil Reference Handbook and note the search term that got you there. On exam day you'll search a PDF, so the search term is the useful part.
- Practice more problems from each area you missed. Two misses in an area with two questions is a reason to look closer, not proof the area is weak.
- If you want to time a full run, allow about 4 hours. That's the real exam's average pace: 8 hours for 80 questions is 6 minutes each.
| If the miss was… | Do this next |
|---|---|
| Method or concept | Restate the principle (control volume, process, or rule) before calculating again. |
| Reference lookup | Find the equation or table in the handbook or the listed standard, and confirm it's the edition for your test date. |
| Units | Write the starting and target units, and convert before you substitute. |
| Assumption | Underline what the stem gives you and what it asks, and name the assumption the model needs. |
| Arithmetic | Recompute from unrounded intermediate values and check the order of magnitude. |
Copy this log for your misses:
| Question | My method | Cause | Correct principle or reference | Next action | Retry result |
|---|---|---|---|---|---|
Exam format and references for your test date
| Row label | PE Civil: Water Resources and Environmental |
|---|---|
| Questions | 80 |
| Time | 9-hour appointment: 2-minute nondisclosure agreement, 8-minute tutorial, 8-hour exam, 50-minute scheduled break |
| Sections | Two. Once you submit the first section, you can't go back to it. |
| Item types | Multiple choice plus multiple correct, point and click, drag and drop, and fill in the blank. Each is scored right or wrong. Some unscored pretest items are mixed in and not marked. |
| Units | Both SI and U.S. customary |
| References | The NCEES PE Civil Reference Handbook and the listed design standards, on screen as searchable PDFs. Ctrl+F doesn't work; use the reference panel's search box. You can't bring your own books or notes. |
| Calculator | One NCEES-approved calculator; a TI-30XS is also available on screen |
| Where and when | Pearson VUE test centers, year-round |
| Results | Pass or fail, typically 7–10 days after the exam |
Sources: NCEES PE Civil exam page; NCEES Examinee Guide, May 2026, Sections 3, 4, and 6; NCEES computer-based testing.
Which design standards apply to your exam
The 12 knowledge areas and their question ranges stay the same in April 2027. The list of design standards supplied in the exam changes. Go by your appointment date, not the publication date of your study materials.
| Your exam date | Design standards supplied with the handbook |
|---|---|
| Before April 2027 | Recommended Standards for Wastewater Facilities, 2014; Recommended Standards for Water Works, 2018 |
| April 2027 onward | Recommended Standards for Wastewater Facilities, 2014; Recommended Standards for Water Works, 2022; plus 29 CFR 1926 (OSHA construction safety, 2024, listed subparts including fall protection and excavations); USACE EM 1110-2-1902, Slope Stability (2003); FHWA HDS-5, Hydraulic Design of Highway Culverts (3rd ed., 2012); FHWA HEC-14, Energy Dissipators for Culverts and Channels (3rd ed., 2006); UFC 3-220-05, Dewatering and Groundwater Control (2004) |
NCEES scores standard-based answers against the edition on its list, so study that edition. One trap: the wastewater standard has a newer 2026 edition, but the exam list still names 2014. (Ten States Standards editions)
Sources: specifications and standards before April 2027; specifications and standards beginning April 2027.
Other things candidates ask
Is this the PE Environmental exam? No. PE Environmental is a separate NCEES exam with its own specifications. This page covers the Water Resources and Environmental module of PE Civil.
Where do I get the handbook? Log in to MyNCEES. NCEES doesn't sell printed copies, and it doesn't provide the design standards outside the exam; those come from their publishers.
Is there an official full-length practice exam? Yes. NCEES sells an 80-question PE Civil WRE practice exam built to the April 2024 specifications, $59.95 as of October 8, 2026, through MyNCEES exam prep.
What's next after this set?
- PE exam prep: study plans and resources
- How to register for an NCEES exam
- Understanding NCEES exam results
- NCEES exam retake policy
Sources and independence
Official exam sources
- NCEES: PE Civil exam page
- NCEES: PE Civil WRE specifications and design standards, effective April 2024
- NCEES: PE Civil WRE specifications and design standards, effective April 2027
- NCEES Examinee Guide, May 2026
- NCEES: Computer-based testing and item types
Technical sources used in the solutions
- FHWA HDS-4, Introduction to Highway Hydraulics (2008)
- DOE/Hydraulic Institute, Pump Life Cycle Costs (2001)
- EPA EPANET 2.2 User Manual, EPA/600/R-20/133
- FHWA TA 5080.15, Contract Time Determination
- NIST Handbook 135 (2025), Life-Cycle Costing Manual
- FHWA LRFD Steel Bridge Design Example, Design Step P
- USACE EM 1110-2-2502, Retaining and Flood Walls (1989)
- FHWA NHI-05-037, Geotechnical Aspects of Pavements, Ch. 5
- Florida DEP, Wastewater Treatment Formulas (April 2024)
- USBR Water Measurement Manual, Ch. 7, Sec. 9
- TxDOT Hydraulic Design Manual, Ch. 8, Sec. 3
- USACE HEC-HMS Technical Reference, SCS Curve Number Loss Model
- USACE HEC-FDA Technical Reference, Long-Term Exceedance Probability
- USACE HEC-HMS Technical Reference, Reservoir Modeling
- NRCS NEH Part 630, Chapter 16, Hydrographs (2007)
- EPA/540/R-96/003, Soil Screening Guidance: Technical Background
- EPA, Overview of Total Maximum Daily Loads
- EPA, Disinfection Profiling and Benchmarking Technical Guidance Manual
- USGS Water Science School, Hardness of Water
- EPA-823-R-07-002, Biological Nutrient Removal Processes and Costs (2007)
- MassDEP, Guidelines for I/I Analyses and Sewer System Evaluation Surveys (2017)
- 29 CFR 1926 Subpart P, Appendix B (eCFR)
- EPA Stormwater Best Management Practice: Mulching
We recomputed every answer and traced each principle to the source linked with its question. That's a source and arithmetic check, not a professional engineering review.
Castleport Test Prep Editorial Team. Last verified October 8, 2026: NCEES exam format, WRE specifications, and both design-standard lists.
Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by NCEES. Exam and credential names are used to identify their subjects; trademarks belong to their respective owners. These practice questions are original and unofficial, not actual or past exam questions. Using this resource doesn't guarantee a passing result or licensure.
Drafting was AI-assisted. No licensed-engineer review has been completed.