10.3 Manning, Open Channel, and Critical Flow

Key Takeaways

  • Manning's equation, Q = (1.49/n) A R^(2/3) S^(1/2) in US units (coefficient 1.0 in SI), estimates uniform open-channel capacity.
  • Hydraulic radius R = A/P (wetted area over wetted perimeter), not flow depth, except as a rough approximation in very wide channels.
  • Normal depth is the uniform-flow depth for a given flow, slope, roughness, and shape; it need not equal the actual depth near a control.
  • Critical flow occurs at Froude number Fr = 1 and minimum specific energy; for a rectangular channel y_c = (q^2/g)^(1/3) with q = Q/b.
  • Subcritical flow (Fr < 1) is deep, slow, and downstream-controlled; supercritical flow (Fr > 1) is shallow, fast, and upstream-controlled.
Last updated: June 2026

Uniform Flow, Control, and Flow Regime

Open-channel hydraulics differs from pressure-pipe hydraulics because the water surface is exposed to the atmosphere and can adjust its depth. The NCEES WRE specification includes open-channel flow, hydraulic grade lines, energy dissipation, stormwater drainage, and subcritical and supercritical flow. The exam often asks which hydraulic model applies before it asks for a number, so identify the regime and the controlling depth first.

Manning's Equation

In US customary units, Manning's equation is Q = (1.49/n) A R^(2/3) S^(1/2); in SI the coefficient is 1.0. Here A is wetted flow area, R = A/P is hydraulic radius, P is wetted perimeter, S is the energy slope (equal to the channel slope under uniform flow), and n is the Manning roughness. Typical n values: smooth concrete ~0.012-0.015, earth ~0.022-0.030, natural channels with vegetation ~0.035-0.05+. A small n error scales Q almost linearly, so confirm the surface description.

Channel changeEffect on Q if other terms fixed
Larger area AIncreases Q
Larger hydraulic radius RIncreases Q
Steeper slope SIncreases Q (square-root sensitivity)
Larger roughness nDecreases Q
More wetted perimeter for same areaDecreases R and Q

A frequent PE mistake is substituting flow depth for hydraulic radius. For a rectangular channel A = by and P = b + 2y, so R = by/(b + 2y). Only in a very wide channel (b much greater than y) does R approach the depth y. Always build A, P, and R from the actual geometry.

Normal Depth Workflow

Normal depth is the depth that makes Manning capacity equal the given flow in a prismatic channel under uniform-flow assumptions:

  1. Identify the channel shape and write A, P, and R as functions of depth y.
  2. Insert the given n and slope S.
  3. Substitute into Manning's equation.
  4. Iterate, or test the answer choices, until computed Q matches the required Q.
  5. Confirm the depth is physically possible within the section.

Normal depth is not automatically the water depth at a culvert entrance, spillway, bridge opening, or abrupt slope break. Those are controls. A control section fixes a depth-discharge relationship and produces gradually varied (nonuniform) flow upstream or downstream.

Critical Flow and Specific Energy

Specific energy is energy per unit weight measured from the channel bottom: E = y + V^2/(2g). For a given flow and shape, critical depth occurs at the minimum specific energy. The Froude number Fr = V/sqrt(gD_h), where D_h is hydraulic depth A/T (top width T), sets the regime: Fr < 1 is subcritical (deep, slower, influenced by downstream conditions); Fr > 1 is supercritical (shallow, faster, controlled from upstream); Fr = 1 is critical.

For a rectangular channel, critical depth has a convenient closed form: y_c = (q^2/g)^(1/3), where q = Q/b is the unit discharge. This appears repeatedly in weir, spillway, culvert, and transition problems, so memorize it. At critical depth the minimum specific energy in a rectangular channel is E_min = 1.5 y_c.

Hydraulic Jumps and Energy Dissipation

A hydraulic jump occurs where supercritical flow transitions to subcritical flow. Depth rises sharply from y1 to the conjugate depth y2, velocity drops, and energy is dissipated as turbulence. Across the jump you apply the momentum equation, not Bernoulli, because energy is lost. For a rectangular channel the conjugate-depth ratio is y2/y1 = 0.5[sqrt(1 + 8 Fr1^2) - 1]. The exam often asks conceptually why a stilling basin, drop structure, plunge pool, or riprap apron is placed downstream of a culvert or spillway: to force or contain the jump and protect the channel.

Exam-Ready Checks

  • Use Manning only when uniform flow is reasonable or explicitly requested.
  • Compare normal depth to critical depth to classify the slope (mild vs steep) and the likely control.
  • Subcritical profiles respond to downstream controls; supercritical profiles respond to upstream controls.
  • Expect supercritical flow near steep outlets and energy-dissipation structures.
  • Keep units paired: cfs with ft, or m^3/s with m, never mixed.
  • Label exactly which depth is requested: normal depth, critical depth, actual depth at a control, or conjugate depth after a jump are different quantities.

Slope Classification and Profiles

Comparing normal depth (y_n) to critical depth (y_c) classifies the channel and tells you which gradually varied flow profile to expect. If y_n > y_c the slope is mild (M); if y_n < y_c the slope is steep (S); if y_n = y_c the slope is critical (C). On a mild slope, uniform flow is subcritical and the channel is downstream-controlled; on a steep slope, uniform flow is supercritical and upstream-controlled. The exam may ask you to name the profile (M1 backwater behind a dam, M2 drawdown to a free overfall, S1, S2, and so on) rather than to compute a number.

Recognizing that a backwater curve forms upstream of an obstruction on a mild slope, or that a drawdown forms approaching a free overfall, is frequently enough to pick the answer.

Worked Mini-Example

A trapezoidal channel has a 6 ft bottom width, side slopes of 2 horizontal to 1 vertical, n = 0.025, and a slope of 0.001 ft/ft carrying 200 cfs. At a trial depth y, area A = (b + zy)y = (6 + 2y)y and wetted perimeter P = b + 2y sqrt(1 + z^2) = 6 + 2y sqrt(5). At y = 3 ft: A = (6 + 6)(3) = 36 ft^2, P = 6 + 6(2.236) = 19.4 ft, R = 1.86 ft, and Q = (1.49/0.025)(36)(1.86^0.667)(0.001^0.5) = about 100 cfs, too low. Increasing the trial depth raises both A and R, so you iterate upward until computed Q matches 200 cfs (near y is about 4.6 ft).

This iteration is the heart of every normal-depth item: build A, P, and R as functions of depth, then converge. On the CBT, plugging the four answer choices in directly is often faster than blind iteration.

Test Your Knowledge

A rectangular concrete channel is 8 ft wide and carries water 2 ft deep. If n = 0.015 and the channel slope is 0.0016 ft/ft, what is the approximate uniform-flow capacity by Manning's equation?

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Test Your Knowledge

A 10-ft wide rectangular channel carries 150 cfs. What is the approximate critical depth, and how is a 3.0-ft actual depth classified?

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B
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