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Topic 6 of 10 — free theory

Open-Channel Flow

Manning's equation, normal depth, critical depth, and the Froude number.

FE Civil · Hydraulics and Hydrologic Systems (8–12)PE WRE · Hydraulics—Open Channel (7–11)

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Manning's equation — the workhorse of open channels

For uniform flow in an open channel (rivers, canals, sewers flowing partly full), Manning's equation relates discharge to the channel's shape, roughness, and slope:

V = (k/n) · R2/3 · S1/2    Q = AV

Vmean velocity
Qdischarge
kunit constant: 1.486 for English units (ft/s), 1.0 for SI
nManning's roughness coefficient (concrete ≈ 0.013, earth ≈ 0.022, natural channel ≈ 0.03–0.05)
Rhydraulic radius = A/P, where P is the wetted perimeter
Slongitudinal slope of the channel (energy slope for uniform flow)

Uniform flow means depth and velocity are constant along the channel — the water surface parallels the bed. Normal depth is simply the uniform-flow depth for a given Q.

Geometry you will need on the fly

Rectangular: A = by,   P = b + 2y    Trapezoidal: A = (b + zy)y,   P = b + 2y√(1+z²)    Full circular: R = D/4

bbottom width
yflow depth
zside slope, horizontal:vertical
Do NOT use R = D/4 for a partly full pipe — recompute A and P for the actual depth.

Critical flow and the Froude number

Open-channel flow has two regimes. Subcritical (slow, deep — Fr < 1) is controlled from downstream; supercritical (fast, shallow — Fr > 1) is controlled from upstream. The boundary is critical flow:

Fr = V / √(g·Dh)   with   Dh = A/T

FrFroude number: < 1 subcritical, = 1 critical, > 1 supercritical
Dhhydraulic depth = area / top width T (for a rectangle, Dh = y)

Rectangular channels:   yc = (q²/g)1/3   and   Emin = 3yc/2

yccritical depth
q = Q/bdischarge per unit width
E = y + V²/2gspecific energy — minimised at critical depth

Specific energy E is measured from the channel bed. At a given E, two alternate depths exist (one subcritical, one supercritical) — the exam loves asking which is which.

PE depth: hydraulic jumps, culverts, and controls

A hydraulic jump converts supercritical flow to subcritical flow and dissipates energy. For a rectangular channel, conservation of momentum gives the sequent-depth relationship:

y2y1 = 0.5[√(1 + 8Fr1²) − 1]    ΔE = (y2−y1)³4y1y2

For culverts, first identify the control. Inlet control depends mainly on entrance geometry and headwater. Outlet control requires the full energy balance, including barrel friction, entrance/exit losses, tailwater, and elevation. The controlling case is the one requiring the higher headwater for the design flow.

Stormwater link: gutter, inlet, and storm-sewer design couples surface capture with closed-conduit capacity. Bypass flow from one inlet becomes approach flow to the next; never size each inlet as though it receives only local runoff.

Worked example Normal depth in a trapezoidal channel

Given:

  • Trapezoidal channel: bottom width b = 6 ft, side slopes 2H:1V (z = 2).
  • Manning's n = 0.013, bed slope S = 0.001, discharge Q = 60 cfs.

Solution:

  1. Normal depth needs trial: guess y, compute A, P, R, then Q = (1.486/n)·A·R2/3·S1/2.
  2. y = 1.50 ft → A = 13.50 ft², P = 12.71 ft, R = 1.062 ft → Q = 50.8 cfs (low).
  3. y = 1.60 ft → A = 14.72 ft², P = 13.16 ft, R = 1.119 ft → Q = 57.3 cfs (low).
  4. y = 1.65 ft → A = 15.35 ft², P = 13.38 ft, R = 1.147 ft → Q = 60.8 cfs (just high).
  5. Interpolating: yn ≈ 1.64 ft. On the exam, bracket the answer between two trials — you rarely need more than two or three guesses.

Answer: Normal depth yn ≈ 1.64 ft.

Free 5-question mini-quiz

Open-Channel Flow

Choose your answer, then check it to see the result and explanation. SI units are used unless stated otherwise.

1. A rectangular channel is 2.0 m wide, 1.2 m deep and carries water at 1.5 m/s. What is Q?

2. For the channel in Question 1, what is the hydraulic radius?

3. For V = 1.5 m/s and hydraulic depth 1.2 m, what is the flow regime?

4. A 2.0 m wide rectangular channel carries 4.0 m³/s. What is its critical depth?

5. A hydraulic jump converts:

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