PE Civil: Water Resources & Environmental

PE Civil WRE Exam: 10 Practice Problems with Detailed Solutions

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PE Civil: Water Resources & Environmental — complete guide set: Study guide & 16-week plan How to pass (first try) Reference & codes in the exam
Quick answer: These 10 problems span the major WRE topic areas, from Rational Method runoff and Manning’s flow to BOD kinetics, CT disinfection, detention storage and hydraulic jumps. Work each one before checking the solution, at about 6 minutes apiece. If you get fewer than 7 right on the first try, focus your study on the topics you missed.
Source note: These are FE Test Prep practice problems, not official NCEES questions. Verify current PE Civil WRE specifications and reference materials on the official NCEES PE Civil page, and confirm scheduling details through Pearson VUE for NCEES.

The PE Civil Water Resources & Environmental 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 discipline-specific questions. The problems below are written for NCEES-style practice, not as official NCEES questions.

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: Solve each problem before you open the solution. The real exam averages roughly 6 minutes per question; hold that pace here and note which reference each problem sent you to.

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?

  • A) 42 cfs
  • B) 52 cfs
  • C) 62 cfs
  • D) 72 cfs
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 0.0009 ft/ft. If the normal depth is 3.0 ft, what is most nearly the flow rate?

  • A) 68 cfs
  • B) 82 cfs
  • C) 96 cfs
  • D) 112 cfs
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(0.0009)1/2 = 96.9 ≈ 96 cfs

Problem 3: Hazen-Williams Head Loss

Water flows at 5.0 cfs through a 12-inch ductile iron pipe (C = 130) that is 2,000 ft long. What is most nearly the friction head loss?

  • A) 14 ft
  • B) 18 ft
  • C) 23 ft
  • D) 29 ft
Show Solution

Answer: C

hf = (4.73 L Q1.852) / (C1.852 d4.87)

= (4.73 × 2000 × 5.01.852) / (1301.852 × 1.04.87)

= (4.73 × 2000 × 19.70) / (8,223 × 1.0) = 22.7 ≈ 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?

  • A) 168 mg/L
  • B) 191 mg/L
  • C) 210 mg/L
  • D) 235 mg/L
Show Solution

Answer: B

BOD5 = L0(1 − e−kt) = 280(1 − e−1.15) = 280 × 0.6834 = 191 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?

  • A) 18.3 ft
  • B) 21.5 ft
  • C) 22.2 ft
  • D) 25.8 ft
Show Solution

Answer: B

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 ≈ 21.5 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?

  • A) 2.9 ac-ft
  • B) 4.1 ac-ft
  • C) 5.2 ac-ft
  • D) 6.4 ac-ft
Show Solution

Answer: A

For a symmetric triangular hydrograph with constant outflow, storage is the area between the inflow hydrograph and the allowable outflow line.

The inflow exceeds 45 cfs after 45/120 = 0.375 of the rising limb and drops below 45 cfs at the same fraction of the falling limb. The excess-flow triangle therefore has a base of 90(1 − 0.375) = 56.25 min and a height of 120 − 45 = 75 cfs.

S = 0.5 × 75 × (56.25 × 60) = 126,563 ft³

S = 126,563 / 43,560 = 2.91 ≈ 2.9 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?

  • A) 1.8 mg/L
  • B) 2.3 mg/L
  • C) 2.6 mg/L
  • D) 3.1 mg/L
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?

  • A) 0.23 ft/day
  • B) 0.38 ft/day
  • C) 0.75 ft/day
  • D) 1.13 ft/day
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?

  • A) 72 ft
  • B) 82 ft
  • C) 86 ft
  • D) 95 ft
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?

  • A) 3.2 ft
  • B) 3.6 ft
  • C) 4.6 ft
  • D) 5.3 ft
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 = 3.64 ≈ 3.6 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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These 10 problems are just a sample. FE Test Prep offers 450+ PE Civil WRE practice questions mapped to the current PE Civil WRE exam specification with detailed, step-by-step solutions.

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Continue your PE Civil WRE preparation:
Use the matching PE Civil WRE reference/code guide Rehearse the section and subsection lanes before timed practice; the clarifier and CT disinfection problems above, for example, depend on knowing which supplied reference to open before you search. Pair it with free PE Civil WRE practice.
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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?

Plan on about 450–600 problems from our bank covering all 12 topic areas, about a quarter of them spaced re-attempts of earlier misses; PE passers report about 500–900 problems at typical vendor length, and ours are shorter, about 9 minutes each including review. 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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