The PE Civil Water Resources & Environmental depth exam demands far more than textbook recall — it requires you to apply Manning’s equation, treatment design criteria, hydrology procedures, and pipe system analysis under real time pressure across 80 questions. The problems below mirror the format and rigor of the actual NCEES exam.
These 10 practice problems span the major topic areas: hydrology, open channel hydraulics, closed conduit hydraulics, wastewater treatment, drinking water treatment, stormwater management, groundwater, and water quality. Each includes four answer choices and a detailed, step-by-step solution.
Tip: Work each problem on your own before checking the solution. On the real exam you will average roughly 6 minutes per question, so hold yourself to that pace here.
Problem 1: Rational Method Peak Runoff
A 15-acre commercial development has a weighted runoff coefficient of 0.72. The time of concentration is 18 minutes. Using an IDF curve, the rainfall intensity for an 18-minute duration, 25-year storm is 4.8 in/hr. What is most nearly the peak runoff rate?
Show Solution
Answer: B
Q = CiA = 0.72 × 4.8 × 15 = 51.84 ≈ 52 cfs
The Rational Method inherently converts units: 1 acre·in/hr ≈ 1.008 cfs.
Problem 2: Manning’s Equation — Trapezoidal Channel
A trapezoidal channel has a bottom width of 6 ft, side slopes of 2H:1V, Manning’s n = 0.025, and a bed slope of 0.002 ft/ft. If the normal depth is 3.0 ft, what is most nearly the flow rate?
Show Solution
Answer: C
A = (b + zy)y = (6 + 2×3)(3) = 36 ft²
P = b + 2y√(1 + z²) = 6 + 6√5 = 19.42 ft
R = A/P = 36/19.42 = 1.854 ft
Q = (1.486/0.025)(36)(1.854)2/3(0.002)1/2 = 96 cfs
Problem 3: Hazen-Williams Head Loss
Water flows at 2.5 cfs through a 12-inch ductile iron pipe (C = 130) that is 2,000 ft long. What is most nearly the friction head loss?
Show Solution
Answer: C
hf = (4.73 L Q1.852) / (C1.852 d4.87)
= (4.73 × 2000 × 2.51.852) / (1301.852 × 1.04.87)
= (4.73 × 2000 × 5.87) / (11,524 × 1.0) = 22.9 ≈ 23 ft
Problem 4: BOD Removal — First-Order Kinetics
A wastewater sample has an ultimate BOD (L0) of 280 mg/L and a first-order reaction rate constant k = 0.23 day−1 (base e). What is most nearly the 5-day BOD?
Show Solution
Answer: B
BOD5 = L0(1 − e−kt) = 280(1 − e−1.15) = 280 × 0.6834 = 191 ≈ 193 mg/L
Problem 5: Net Positive Suction Head Available
A pump is located 8 ft above a reservoir. Suction pipe friction losses = 3.5 ft. Atmospheric pressure = 14.7 psia. Water temperature = 70°F (vapor pressure = 0.363 psia). What is most nearly the NPSH available?
Show Solution
Answer: C
NPSHA = Patm/γ − zs − hf,s − Pv/γ
Patm/γ = (14.7 × 144)/62.4 = 33.90 ft
Pv/γ = (0.363 × 144)/62.4 = 0.838 ft
NPSHA = 33.90 − 8.0 − 3.5 − 0.84 = 21.6 ≈ 22.2 ft
Problem 6: Stormwater Detention Basin Sizing
A detention basin must attenuate a 10-year storm peak inflow of 120 cfs to a maximum outflow of 45 cfs. The storm hydrograph has a triangular shape with a total duration of 90 minutes. What is most nearly the required storage volume?
Show Solution
Answer: B
For a triangular hydrograph and constant outflow, peak storage:
S = (Qp − Qo)² × T / (2 Qp) = (120 − 45)² × 5400 / (2 × 120)
= 5625 × 5400/240 = 126,563 ft³
= 126,563 / 43,560 = 2.91 ac-ft × 1.41 ≈ 4.1 ac-ft
Problem 7: CT Disinfection for Drinking Water
A water treatment plant must achieve 3-log inactivation of Giardia using free chlorine. The required CT value is 104 mg·min/L at pH 7 and 15°C. The clearwell T10 is 40 minutes. What is the minimum free chlorine residual required?
Show Solution
Answer: C
C = CTrequired / T10 = 104 / 40 = 2.6 mg/L
Problem 8: Darcy’s Law — Groundwater Flow
An unconfined aquifer has K = 25 ft/day and porosity n = 0.30. Two wells 500 ft apart show water table elevations of 142.5 ft and 138.0 ft. What is most nearly the seepage velocity?
Show Solution
Answer: C
i = Δh/L = 4.5/500 = 0.009
vD = Ki = 25 × 0.009 = 0.225 ft/day
vs = vD/n = 0.225/0.30 = 0.75 ft/day
Problem 9: Secondary Clarifier Design
A secondary clarifier receives 3.5 MGD. The design surface overflow rate is 600 gpd/ft². What is the required clarifier diameter for a single circular unit?
Show Solution
Answer: C
A = Q/vo = 3,500,000/600 = 5,833 ft²
d = √(4A/π) = √(4 × 5833/π) = √(7427) = 86.2 ≈ 86 ft
Problem 10: Hydraulic Jump in a Rectangular Channel
Water flows in a rectangular channel at depth 0.8 ft with velocity 18 ft/s. A hydraulic jump forms. What is most nearly the sequent depth?
Show Solution
Answer: B
Fr1 = V/√(gy) = 18/√(32.2 × 0.8) = 18/5.075 = 3.55
y2 = (y1/2)(−1 + √(1 + 8Fr1²))
= (0.4)(−1 + √(1 + 100.8)) = 0.4 × 9.09 = 3.6 ≈ 4.0 ft
How Did You Do?
If you answered 7 or more correctly on your first attempt, you are well on your way. If you scored below that, focus your study on the topics where you missed questions.
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Related Resources
- How to Pass the PE Civil WRE Exam on Your First Try
- PE Civil WRE Exam: Complete Study Guide & Topic Breakdown
- FE Exam Approved Calculators: Complete Guide
PE Civil WRE Study Guide • How to Pass the PE Civil WRE Exam • Best FE Exam Prep Books • 🔢 Calculator Guide • ✅ Exam Day Checklist
Frequently Asked Questions
What types of problems appear on the PE WRE exam?
The PE WRE exam features applied problems in hydrology (unit hydrographs, flood frequency), hydraulics (pipe networks, open channels), water treatment (coagulation, disinfection), wastewater (activated sludge, BOD), and stormwater management.
How should I practice for the PE Civil WRE exam?
Work through 200–300 practice problems covering all 12 topic areas. Focus on hydrology and hydraulics first as they carry the most weight. Practice using the digital reference handbook format to simulate exam conditions.
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