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Open Channel Flow Practice Problems — Manning's, Critical Depth & Hydraulic Jumps

Nine original open-channel problems, every step shown, every wrong answer traced to the mistake that produces it. Open-channel flow is 7–11 of the 80 questions on the PE Civil: WRE exam and a staple of the FE Civil hydraulics section — and the arithmetic here is where confident candidates quietly bleed points.

Last reviewed: 2026-10-03

V = (1.486/n) R2/3 S1/2  ·  yc = (q²/g)1/3 (rect.)  ·  E = y + V²/2g  ·  y2 = (y1/2)(√(1 + 8Fr1²) − 1)

1.486US-units Manning constant (1.0 for SI) — the classic unit-trap; R = A/P, never the depth
yccritical depth; the cube-root form is for rectangular channels only — other shapes need Fr = 1 by trial
y2sequent depth after a hydraulic jump; energy is lost, so E2 < E1 always

Normal Flow — Manning's Equation

Problem 1 — Trapezoidal channel discharge

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A trapezoidal channel has bottom width 8 ft, side slopes 2H:1V, flow depth 3 ft, Manning's n = 0.025, and bed slope 0.0016. Compute the discharge under uniform flow.

  1. 156 cfs
  2. 105 cfs
  3. 208 cfs
  4. 101 cfs
Full solution
  1. A = (b + zy)y = (8 + 2×3)(3) = 42 ft². P = b + 2y√(1+z²) = 8 + 6√5 = 21.42 ft (free surface excluded). R = A/P = 1.961 ft.
  2. V = (1.486/n)R2/3S1/2 = (1.486/0.025)(1.961)2/3(√0.0016) = 59.44 × 1.562 × 0.04 = 3.71 ft/s.
  3. Q = AV = 42 × 3.71 = 156 cfs.

Answer: A — 156 cfs.

Why the wrong answers are wrong: B (105 cfs) uses the SI constant 1.0 with foot units — velocity comes out a third low. C (208 cfs) plugs the depth y = 3 ft in as R. D (101 cfs) adds the top width (20 ft) into the wetted perimeter, a mistake the exam counts on.

Problem 2 — Rectangular channel velocity

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A rectangular concrete channel is 10 ft wide, carries a normal depth of 2.5 ft on a slope of 0.0009 with n = 0.015. Find the mean velocity.

  1. 4.18 ft/s
  2. 2.81 ft/s
  3. 5.47 ft/s
  4. 2.97 ft/s
Full solution
  1. A = 10 × 2.5 = 25 ft²; P = 10 + 2(2.5) = 15 ft; R = 25/15 = 1.667 ft.
  2. V = (1.486/0.015)(1.667)2/3(√0.0009) = 99.07 × 1.406 × 0.03 = 4.18 ft/s.

Answer: A — 4.18 ft/s.

Why the wrong answers are wrong: B (2.81) — SI constant again. C (5.47) — R replaced by depth. D (2.97) — top width counted in the wetted perimeter (P = 25 ft). Three traps, one problem.

Problem 3 — Normal depth by trial

PE WRE Open Channel Flow (7–11 of 80) Target pace: ~6 min

A trapezoidal channel (b = 6 ft, z = 1.5, n = 0.022, S = 0.002) must carry 120 cfs under uniform flow. Estimate the normal depth.

  1. 2.73 ft
  2. 2.35 ft
  3. 3.34 ft
  4. 2.50 ft
Full solution
  1. Normal depth needs trial: guess y, compute A, P, R, then Q = (1.486/n)·A·R2/3·S1/2.
  2. y = 2.50 ft → A = 24.38 ft², P = 15.01 ft, R = 1.624 ft → Q = 101.7 cfs (low).
  3. y = 2.75 ft → A = 27.84 ft², P = 15.92 ft, R = 1.749 ft → Q = 122.1 cfs (high).
  4. Interpolate between 101.7 and 122.1: yn ≈ 2.50 + 0.25 × (120 − 101.7)/(122.1 − 101.7) = 2.73 ft. (Direct solve: 2.725 ft.)

Answer: A — 2.73 ft.

Why the wrong answers are wrong: B (2.35 ft) uses the R ≈ y shortcut, which is only fair for wide channels — this one isn't wide. C (3.34 ft) runs the whole trial with k = 1.0. D (2.50 ft) stops after the first trial without checking that Q ≠ 120 cfs.

Critical Depth & Specific Energy

Problem 4 — Critical depth, rectangular channel

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A wide rectangular channel carries 8 ft²/s per foot of width. Compute the critical depth.

  1. 1.26 ft
  2. 1.99 ft
  3. 1.87 ft
  4. 1.41 ft
Full solution
  1. For a rectangular channel: yc = (q²/g)1/3 = (8²/32.2)1/3 = (1.9876)1/3 = 1.26 ft.

Answer: A — 1.26 ft.

Why the wrong answers are wrong: B (1.99) forgets the cube root. C (1.87) uses g = 9.81 m/s² with foot units. D (1.41) takes a square root instead of a cube root.

Problem 5 — Critical depth in a trapezoidal channel

PE WRE Open Channel Flow (7–11 of 80) Target pace: ~6 min

A trapezoidal channel (b = 5 ft, z = 2) carries 60 cfs. Estimate the critical depth.

  1. 1.36 ft
  2. 1.65 ft
  3. 1.87 ft
  4. 4.82 ft
Full solution
  1. The rectangular shortcut doesn't apply — use the critical-flow condition Q²T/(gA³) = 1 by trial, with T = b + 2zy.
  2. y = 1.30 ft → A = 9.88 ft², T = 10.20 ft → Q²T/(gA³) = 1.18 (high).
  3. y = 1.40 ft → A = 10.92 ft², T = 10.60 ft → ratio = 0.95 (low).
  4. Interpolate: yc ≈ 1.36 ft. Check: A = 10.49 ft², V = 5.72 ft/s, Fr = V/√(g·A/T) = 1.00 ✓.

Answer: A — 1.36 ft.

Why the wrong answers are wrong: B (1.65) forces the rectangular formula with q = Q/b = 12 — wrong section, wrong answer. C (1.87) is the specific energy at critical depth (Ec = 1.36 + 0.51), not the depth. D (4.82) plugs the total Q = 60 cfs in as the unit discharge q.

Problem 6 — Alternate depths on the E–y curve

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A rectangular channel carries q = 8 ft²/s per foot of width with specific energy E = 3.00 ft. Find the subcritical alternate depth.

  1. 2.88 ft
  2. 0.65 ft
  3. 1.26 ft
  4. 3.00 ft
Full solution
  1. E = y + q²/(2gy²) = y + 64/(64.4y²) = 3.00. Solve by trial.
  2. y = 2.90 ft → E = 2.90 + 0.118 = 3.018 (high); y = 2.85 ft → E = 2.85 + 0.122 = 2.972 (low).
  3. Interpolate: y ≈ 2.88 ft (subcritical — the deeper root).

Answer: A — 2.88 ft.

Why the wrong answers are wrong: B (0.65) is the supercritical alternate depth — right equation, wrong branch. C (1.26) is the critical depth, where the two branches meet; alternate depths are never critical. D (3.00) reports the specific energy as a depth.

Hydraulic Jump

Problem 7 — Sequent depth and energy loss ★ shown in full

PE WRE Open Channel Flow (7–11 of 80) Target pace: ~6 min

One problem on each page is shown worked in full, so you can judge the quality before trusting the rest. This is that problem — and it includes the energy-loss step, where arithmetic slips are most common.

A hydraulic jump occurs in a rectangular channel with upstream depth y1 = 0.80 ft and upstream velocity V1 = 12 ft/s. Compute the energy dissipated in the jump.

  1. 0.46 ft
  2. 1.85 ft
  3. 2.57 ft
  4. 0.23 ft
Full solution
  1. Froude number upstream: Fr1 = V1/√(gy1) = 12/√(32.2 × 0.80) = 12/5.076 = 2.364 (supercritical, so a jump is possible).
  2. Sequent depth: y2 = (y1/2)(√(1 + 8Fr1²) − 1) = 0.40 × (√(1 + 44.72) − 1) = 0.40 × 5.762 = 2.30 ft.
  3. Energy by the long method: q = V1y1 = 9.6 ft²/s. E1 = 0.80 + 12²/(2 × 32.2) = 0.80 + 2.236 = 3.036 ft. V2 = q/y2 = 4.17 ft/s; E2 = 2.30 + 4.17²/64.4 = 2.30 + 0.270 = 2.574 ft. ΔE = 3.036 − 2.574 = 0.462 ft.
  4. Cross-check with the shortcut: ΔE = (y2 − y1)³/(4y1y2) = (1.505)³/(4 × 0.80 × 2.30) = 3.407/7.375 = 0.462 ft ✓ — both methods agree.

Answer: A — 0.46 ft.

Why the wrong answers are wrong: B (1.85) drops the 4 in the shortcut denominator — exactly the slip this problem double-checks against. C (2.57) reports E2, the energy remaining, not the energy lost. D (0.23) halves the loss (an 8 in place of the 4 in the denominator).

Problem 8 — Sequent depth, rectangular channel

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A rectangular channel carries q = 15 ft²/s per foot of width. Just upstream of a hydraulic jump the depth is y1 = 1.0 ft. Find the sequent depth y2.

  1. 3.27 ft
  2. 6.54 ft
  3. 6.29 ft
  4. 2.64 ft
Full solution
  1. Fr1 = q/(y1√(gy1)) = 15/(1.0 × √32.2) = 15/5.676 = 2.643.
  2. y2 = (y1/2)(√(1 + 8Fr1²) − 1) = 0.5 × (√(1 + 55.88) − 1) = 0.5 × 6.543 = 3.27 ft.

Answer: A — 3.27 ft.

Why the wrong answers are wrong: B (6.54) forgets the ÷2 outside the brackets. C (6.29) runs the whole thing with g = 9.81. D (2.64) reports the Froude number as a depth.

Culvert Hydraulics

Problem 9 — Outlet-control headwater

PE WRE Open Channel Flow (7–11 of 80) Target pace: ~6 min

A 5-ft × 5-ft concrete box culvert, 180 ft long (n = 0.013), carries 120 cfs under outlet control. Entrance loss coefficient Ke = 0.5, tailwater is 3.0 ft above the outlet invert, and the culvert slope is 0.005. Find the headwater depth above the inlet invert.

  1. 4.67 ft
  2. 3.77 ft
  3. 4.31 ft
  4. 3.41 ft
Full solution
  1. Full-flow velocity: A = 25 ft², V = 120/25 = 4.8 ft/s; V²/2g = 23.04/64.4 = 0.358 ft.
  2. Outlet-control losses: hL = [1 + Ke + 29.1n²L/R4/3] · V²/2g. R = 25/20 = 1.25 ft; friction term = 29.1(0.013)²(180)/1.254/3 = 0.657. Bracket = 1 + 0.5 + 0.657 = 2.157; hL = 2.157 × 0.358 = 0.772 ft.
  3. HW above the outlet invert = TW + hL = 3.0 + 0.772 = 3.772 ft. Refer back to the inlet: HW = 3.772 + S·L = 3.772 + 0.005 × 180 = 4.67 ft.

Answer: A — 4.67 ft.

Why the wrong answers are wrong: B (3.77) reports headwater above the outlet invert — the slope add-back is the step everyone skips. C (4.31) drops the exit loss (the leading 1 in the bracket). D (3.41) drops both.

How normal depth, critical depth, and the jumps connect

Normal depth ynUniform flow; from Manning's by trial. Depends on slope and roughness.
Critical depth ycMinimum specific energy; from Fr = 1. Depends only on discharge and section shape.
Alternate depthsTwo depths, same energy, one E–y curve (one subcritical, one supercritical).
Sequent depthsBefore/after a hydraulic jump — energy is lost, so they are not alternates.
Reading the flowyn > yc → mild slope, subcritical normal flow; yn < yc → steep slope, supercritical — a jump waits downstream.

Frequently asked questions

What is the difference between normal depth and critical depth?

Normal depth is the depth of uniform flow for a given discharge, slope, and roughness — found from Manning's equation, usually by trial. Critical depth is the depth of minimum specific energy for a given discharge — found from Fr = 1. They are equal only when the channel happens to be on a critical slope.

Do alternate depths and sequent depths mean the same thing?

No. Alternate depths share the same specific energy on one E–y curve (one subcritical, one supercritical). Sequent depths are the before-and-after depths of a hydraulic jump — they have different specific energies because the jump dissipates energy. You cannot read sequent depth off the E–y curve.

Which depth goes in the Froude number for a trapezoidal channel?

The hydraulic depth D = A/T (area over top width), not the flow depth y. Using y in a non-rectangular section is one of the most common exam traps on this topic.

How is headwater found for a culvert under outlet control?

HW = TW + hL + S0·L, where hL = (1 + Ke + friction term)·V²/2g covers exit, entrance, and friction losses, TW is the tailwater depth above the outlet invert, and S0·L converts the headwater back to the inlet invert.

What is the fastest way to get energy loss across a hydraulic jump?

For a rectangular channel, ΔE = (y2 − y1)³/(4·y1·y2) gives the loss directly from the sequent depths. Compute E1 − E2 as a cross-check — if the two disagree, the arithmetic slipped somewhere.

Keep practicing

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