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Hydraulics — Open Channel

10% of exam

Manning uniform flow, specific energy and critical depth, sub-/supercritical flow and hydraulic jumps, stormwater collection, gutter/inlet flow, and culverts.

4 concepts

A. Open-channel flow

Manning Uniform Flow and Normal Depth

Manning's equation in both SI and USCS forms, hydraulic radius and conveyance, finding normal depth by iteration, best hydraulic sections, and partially full circular pipes.

Open-channel flow — water moving with a free surface in a ditch, gutter, storm sewer, culvert, or natural stream — is the single most heavily weighted topic on the PE Civil WRE exam, and almost every problem in it starts from one equation: Manning's. The handbook places it in §6.4 Open-Channel Flow, but it reappears in hydrology, stormwater, and wastewater collection. The points that slip away on exam day are rarely the physics; they are the units constant (

1.486

in USCS,

1.0

in SI), confusing the bed slope

S_0

Hydrology

Groundwater and Wells

Surface Water & Groundwater Quality

Drinking Water Distribution & Treatment

Wastewater Collection & Treatment

Project Sitework

with the friction slope

S_f

, mis-computing the hydraulic radius

R_H = A/P

, and — above all — not recognizing that finding normal depth from a known discharge requires iteration because

y

appears in both

A

and

R_H

. Master that loop and a third of the open-channel questions become bookkeeping.

Manning's equation and the units constant

Manning's equation gives the discharge for steady uniform flow as a function of cross-sectional geometry (

A

,

R_H

), boundary roughness

n

, and the energy-grade-line slope

S

. For uniform flow the energy slope equals the bed slope,

S = S_0

. The constant

k

is the trap: it is

1.486

in US customary units (feet, cfs) and exactly

1.0

in SI (metres, m³/s). The roughness

n

is numerically the same in both systems — about

0.013

for concrete pipe,

0.024

for corrugated metal,

0.030

for an earth channel with rubble sides. Because

Q \propto S^{1/2}

, doubling the slope raises capacity only

41\%

; because

Q \propto R_H^{2/3}

, channel shape matters less than raw area.

Q = \dfrac{k}{n}\,A\,R_H^{2/3}\,S^{1/2}, \qquad v = \dfrac{k}{n}\,R_H^{2/3}\,S^{1/2}, \qquad k = \begin{cases}1.486 & \text{USCS}\\[2pt]1.0 & \text{SI}\end{cases}

Hydraulic radius and section geometry

The hydraulic radius

R_H = A/P

— flow area over wetted perimeter — is the geometric heart of the equation and where arithmetic errors breed. The wetted perimeter is only the channel boundary in contact with water; the free surface (

T

, the top width) is never part of

P

. For a rectangle of width

b

at depth

y

:

A = by

,

P = b + 2y

. For a trapezoid of bottom width

b

and side slope

z

(horizontal:vertical):

A = (b + zy)y

and

P = b + 2y\sqrt{1 + z^2}

. For a pipe flowing exactly full,

R_H = D/4

. Carry

R_H

to three significant figures — the

2/3

power magnifies rounding.

R_H = \dfrac{A}{P}, \qquad \text{trapezoid: } A = (b + zy)\,y,\quad P = b + 2y\sqrt{1 + z^2}

Conveyance: separating geometry from slope

Conveyance

K

bundles everything in Manning's equation except the slope, so

Q = K\,S^{1/2}

. It is a convenience worth knowing because it isolates the geometry-and-roughness factor that does not change when the slope does — useful for comparing reaches, for compound (overbank) channels where you sum conveyances, and for backwater computations. Note this

K

(units of cfs because it already carries

1.486

) is distinct from the

k

units constant; the exam expects you to keep them apart.

K = \dfrac{k}{n}\,A\,R_H^{2/3}, \qquad Q = K\,S^{1/2}

Normal depth by iteration

Normal depth

y_n

is the uniform-flow depth a long prismatic channel settles to for a given

Q

,

n

,

S

, and shape. Except for the widest rectangles you cannot solve it in closed form because

y

sits inside both

A

and

R_H

. The reliable method is to write

Q\,n/k = A\,R_H^{2/3}\,S^{1/2}

(a known constant on the left), then guess

y

, compute the right side, and adjust: too small a discharge means raise

y

. Two or three rounds bracket it; a half-interval (bisection) search converges fast. Always finish by checking the Froude number — a mild channel returns

Fr < 1

(subcritical), confirming the depth is physically the normal one and not a critical or supercritical artifact.

\dfrac{Q\,n}{k} = A(y_n)\,R_H(y_n)^{2/3}\,S^{1/2} \quad\Longrightarrow\quad \text{iterate for } y_n

Best hydraulic section

For a fixed area, the section that maximizes

R_H

(minimizes

P

) carries the most flow — the 'best hydraulic section.' Among all shapes it is the semicircle; among trapezoids it is the half-hexagon (side slope

z = 1/\sqrt{3}\approx 0.577

, i.e.

60^\circ

sides), for which

R_H = y/2

and

A = \sqrt{3}\,y^2

. The best rectangle is one half as deep as it is wide (

b = 2y

), again giving

R_H = y/2

. These are efficiency optima, not always economic optima — excavation, lining, and freeboard usually push real designs away from them — but the exam tests whether you know the half-hexagon and the

b = 2y

rectangle.

\text{Half-hexagon: } z = \tfrac{1}{\sqrt3},\ R_H = \tfrac{y}{2}; \qquad \text{Best rectangle: } b = 2y,\ R_H = \tfrac{y}{2}

Partially full circular pipes

Storm and sanitary sewers run partly full, and their hydraulic elements do not vary linearly with depth. A standard result: a pipe flowing exactly half full has the same hydraulic radius as flowing full (

R_H = D/4

), so it carries the same velocity but exactly half the discharge. Peculiarly, maximum velocity occurs near

y/D \approx 0.81

and maximum discharge near

y/D \approx 0.94

— a pipe actually carries slightly more than its full-flow

Q_{full}

just before sealing. The handbook's Hydraulic Elements Graph plots

A

,

R_H

,

v

, and

Q

ratios against

y/D

; read

v/v_{full}

and

Q/Q_{full}

off it rather than re-deriving the circular-segment geometry under time pressure. For full-flow design the minimum diameter follows directly from Manning.

R_H = \dfrac{D}{4}\ (\text{full}), \qquad D_{min} = \left(\dfrac{2.16\,Q\,n}{S^{1/2}}\right)^{3/8}\ (\text{USCS, full flow})

Exam strategy

Write the units constant first and box it —

1.486

for cfs/ft,

1.0

for m³/s/m — because the most common single error here is dropping or misplacing it. For a 'find the depth' problem, recognize immediately that it is iterative; set up

Qn/k = AR_H^{2/3}S^{1/2}

and bisect. For a 'find the capacity' problem it is one direct substitution — just nail

R_H

. For partly full pipes, default to the hydraulic-elements chart and remember the half-full = half-

Q

, same-

v

anchor. Finish every uniform-flow answer with a Froude check so you know whether the slope is mild or steep before the next part of the problem asks.

Key equations

Manning's equation (discharge)

Q = \dfrac{k}{n}\,A\,R_H^{2/3}\,S^{1/2}

Steady uniform flow. $k = 1.486$ USCS (ft, cfs), $1.0$ SI (m, m³/s); $n$ = roughness; $A$ = flow area; $R_H$

Manning's equation (velocity)

v = \dfrac{k}{n}\,R_H^{2/3}\,S^{1/2}

Hydraulic radius

R_H = \dfrac{A}{P}

Flow area $A$ over wetted perimeter $P$

Trapezoidal geometry

A = (b + zy)\,y, \quad P = b + 2y\sqrt{1 + z^2}, \quad T = b + 2zy

Conveyance

K = \dfrac{k}{n}\,A\,R_H^{2/3}, \qquad Q = K\,S^{1/2}

Normal-depth iteration target

\dfrac{Q\,n}{k} = A\,R_H^{2/3}\,S^{1/2}

Best hydraulic sections

\text{Half-hexagon: } R_H = \tfrac{y}{2},\ A=\sqrt3\,y^2; \quad \text{Best rectangle: } b=2y,\ R_H=\tfrac{y}{2}

Minimum full-flow diameter

D_{min} = \left(\dfrac{2.16\,Q\,n}{S^{1/2}}\right)^{3/8}

Worked examples

Normal depth in a trapezoidal channel (iteration)

Problem. A trapezoidal drainage channel has bottom width $b = 10\,\text{ft}$ , side slopes $z = 2$ (2H:1V), Manning $n = 0.025$ , and bed slope $S = 0.0009\,\text{ft/ft}$ . Find the normal depth that carries $Q = 400\,\text{cfs}$ , and classify the slope.

Solution. Target:

\dfrac{Qn}{k} = \dfrac{400(0.025)}{1.486} = 6.729 = A\,R_H^{2/3}\,S^{1/2}

y_n \approx 5.14\ \text{ft}, \qquad Fr = 0.366\ (\text{subcritical})

Full-flow capacity of a circular storm sewer

Problem. A $30\text{-in}$ ( $D = 2.5\,\text{ft}$ ) concrete storm sewer ( $n = 0.013$ ) is laid at $S = 0.004\,\text{ft/ft}$

Common pitfalls

•Dropping or misplacing the units constant: USCS Manning uses $1.486$ (for ft and cfs); SI uses $1.0$ . Using $1.486$ with metric inputs inflates $Q$ by ~ $49\%$ .
•Treating normal depth as a closed-form solve. Because $y$ is inside both $A$ and $R_H$ , you must iterate (bisection or trial) on $Qn/k = AR_H^{2/3}S^{1/2}$ .
•Including the free surface (top width $T$ ) in the wetted perimeter $P$ . Only the boundary touching water counts; $P$ never contains $T$ .
•Confusing bed slope $S_0$ with friction slope $S_f$ . They are equal ONLY in uniform flow; in gradually varied flow they differ and Manning gives $S_f$
•Assuming a circular pipe carries maximum $Q$ when full. Peak discharge occurs near $y/D \approx 0.94$ and peak velocity near $0.81$ — a partly full pipe can beat $Q_{full}$
•Rounding $R_H$ too early. The $2/3$ exponent amplifies small errors; carry $R_H$ to at least three significant figures before powering it.
•Using a 'best hydraulic section' result ( $R_H = y/2$ ) for an arbitrary trapezoid. That shortcut holds only for the half-hexagon ( $z = 0.577$ ) and the $b = 2y$

References

NCEES PE Civil Reference Handbook — §6.4 Open-Channel Flow
NCEES PE Civil Reference Handbook — §6.4.5 Steady Uniform Flow (Manning's Equation)
Chow, Open-Channel Hydraulics — uniform flow and best hydraulic sections

B. Hydraulic grade lines and energy dissipation

Hydraulic Jump, GVF Profiles and Energy Dissipation

The momentum (sequent-depth) equation, energy loss across a jump, gradually varied flow M/S profile classification, and stilling-basin energy dissipation.

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C. Stormwater collection and drainage

Gutter Flow, Inlets and Culverts

Triangular gutter spread by modified Manning, inlet interception and bypass, storm-sewer pipe sizing, and culvert hydraulics under inlet versus outlet control.

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D. Sub- and supercritical flow

Specific Energy, Critical Depth and the Froude Number

The specific-energy diagram, critical depth and minimum specific energy, the Froude number, alternate depths, and flow over a hump or through a contraction.

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= hydraulic radius;

S = S_0

= bed slope (ft/ft or m/m).

Mean velocity (ft/s or m/s). Equal to $Q/A$ ; same $k$ convention.

(the boundary in contact with water, excluding the free surface). For a full circular pipe,

R_H = D/4

.

Bottom width $b$ , depth $y$ , side slope $z$ (H:V). $T$ = top width. Set $z = 0$ for a rectangle.

Geometry-and-roughness factor (cfs) independent of slope; useful for compound channels (sum $K_i$ ) and backwater work.

Left side is a known constant; iterate $y$ until the right side matches. The solution is the normal depth $y_n$ .

Sections that minimize wetted perimeter for a given area (maximum efficiency). Semicircle is the overall optimum.

USCS; smallest circular pipe (ft) that conveys $Q$ (cfs) just full at slope $S$ . Round up to the next commercial size.

, with

\sqrt{S} = 0.03

. Guess

y = 5.0\,\text{ft}

:

A = (10 + 2(5))5 = 100\,\text{ft}^2

,

P = 10 + 2(5)\sqrt{5} = 32.36\,\text{ft}

,

R_H = 3.090

, so

Q = \frac{1.486}{0.025}(100)(3.090)^{2/3}(0.03) = 378\,\text{cfs}

— low. Guess

y = 5.2\,\text{ft}

:

A = 106.1

,

P = 33.26

,

R_H = 3.19

,

Q = 410\,\text{cfs}

— high. Interpolate to

y \approx 5.14\,\text{ft}

:

A = 104.2\,\text{ft}^2

,

P = 32.98\,\text{ft}

,

R_H = 3.16\,\text{ft}

, giving

Q = 400\,\text{cfs}

. So

y_n = 5.14\,\text{ft}

. Froude check:

v = Q/A = 3.84\,\text{ft/s}

,

T = 10 + 2(2)(5.14) = 30.6\,\text{ft}

, hydraulic depth

D = A/T = 3.41\,\text{ft}

,

Fr = v/\sqrt{gD} = 3.84/\sqrt{32.2(3.41)} = 0.366 < 1

. The answer is

y_n = 5.14\,\text{ft}

on a mild (subcritical) slope. Sanity:

Fr < 1

confirms a subcritical normal depth, consistent with the gentle

0.0009

slope.

. Find its full-flow capacity and velocity, and the discharge when it runs half full.

Solution. Full flow:

A = \pi D^2/4 = \pi(2.5)^2/4 = 4.909\,\text{ft}^2

,

R_H = D/4 = 0.625\,\text{ft}

.

Q_{full} = \dfrac{1.486}{0.013}(4.909)(0.625)^{2/3}(0.004)^{1/2}

. With

(0.625)^{2/3} = 0.731

and

\sqrt{0.004} = 0.0632

:

Q_{full} = 114.3 \times 4.909 \times 0.731 \times 0.0632 = 25.9\,\text{cfs}

.

v_{full} = Q_{full}/A = 25.9/4.909 = 5.28\,\text{ft/s}

. Half full: the hydraulic radius is the same as full (

R_H = D/4

), so velocity is unchanged at

5.29\,\text{ft/s}

, but the area is halved, so

Q_{half} = 0.5\,Q_{full} = 13.0\,\text{cfs}

. Final:

Q_{full} = 25.9\,\text{cfs}

at

5.29\,\text{ft/s}

;

Q_{half} = 13.0\,\text{cfs}

. Sanity:

v_{full} > 2\,\text{ft/s}

meets the usual self-cleansing minimum, and the half-full equals-velocity result is the classic circular-pipe anchor.

Q_{full} = 25.9\ \text{cfs}, \quad v_{full} = 5.28\ \text{ft/s}, \quad Q_{half} = 13.0\ \text{cfs}

, not

S_0

.

rectangle.