CivilHydraulicshardFE ExamCourseworkPE / Advanced

Discharge in a Rectangular Channel by Manning's Equation

A rectangular open channel is $b = 3.0\ \text{m}$ wide and flows at a uniform depth of $y = 1.5\ \text{m}$ .

The bed slope is $S = 0.001$ and Manning's roughness coefficient is $n = 0.013$ .

Find the discharge $Q$ (SI form of Manning's equation).

Given

b	3 m	Channel width
y	1.5 m	Flow depth
S	0.001 -	Bed slope
n	0.013 -	Manning's n

Hint 1

Manning gives a velocity; you need area as well to get discharge.

Hint 2

The hydraulic radius is area over wetted perimeter — and the free surface does not count as wetted.

Hint 3

P = b + 2y = 6 m, so R = 0.75 m. Then V = (1/n)R^(2/3)S^(1/2).

Worked solution — try the problem first

$V = \frac{1}{n}R^{2/3}S^{1/2}, \qquad Q = VA$

Flow area: $A = by = 3.0 \times 1.5 = 4.50\ \text{m}^2$

Wetted perimeter. For an open channel this is the bed plus two sides — the free surface is not wetted perimeter:

$P = b + 2y = 3.0 + 2(1.5) = 6.00\ \text{m}$

Hydraulic radius: $R = \frac{A}{P} = \frac{4.50}{6.00} = 0.750\ \text{m}$

Velocity: $R^{2/3} = 0.750^{0.6667} = 0.8255, \qquad S^{1/2} = \sqrt{0.001} = 0.03162$

$V = \frac{1}{0.013}(0.8255)(0.03162) = 76.92 \times 0.02611 = 2.008\ \text{m/s}$

Discharge: $Q = 2.008 \times 4.50 = 9.04\ \text{m}^3/\text{s}$

$\boxed{Q = 9.04\ \text{m}^3/\text{s}}$

Careful with units systems. This is the SI form with the $1/n$ coefficient. US customary units use $1.486/n$ — mixing the two produces a 49% error.

Concepts:Manning's equationOpen channel flowHydraulic radiusUniform flow

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