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?

  • 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 0.002 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(0.002)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?

  • 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 × 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?

  • A) 168 mg/L
  • B) 193 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 ≈ 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?

  • A) 18.3 ft
  • B) 21.5 ft
  • C) 22.2 ft
  • D) 25.8 ft
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?

  • A) 3.1 ac-ft
  • B) 4.1 ac-ft
  • C) 5.2 ac-ft
  • D) 6.4 ac-ft
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?

  • 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) 4.0 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 = 3.6 ≈ 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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