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Chezy's and Manning's coefficient for Seti River

Short answer

Using bed slope 1×10⁻⁴, hydraulic radius 2.5 m and velocity 2.4 m/s, Chezy's C = V/√(RS) = 151.8 m¹/²/s. Manning's n = R¹/⁶/C = 0.0077 s/m¹/³.

The longitudinal bed slope of Seti River is 10 cm to a kilometer with hydraulic mean depth of 2.5 m. Find Chezy's coefficient and Manning's rugosity coefficient, if the velocity at the peak flood is measured to be 2.4 m/s.

Given data

Bed slope S=10 cm1 km=0.11000=1×10−4S=\dfrac{10\ \text{cm}}{1\ \text{km}}=\dfrac{0.1}{1000}=1\times10^{-4}, hydraulic mean depth R=2.5R=2.5 m, V=2.4V=2.4 m/s.

Chezy's coefficient

V=CRS  ⇒  C=VRS=2.42.5×10−4=2.40.015811=151.79 m1/2/sV=C\sqrt{RS}\;\Rightarrow\;C=\frac{V}{\sqrt{RS}}=\frac{2.4}{\sqrt{2.5\times10^{-4}}}=\frac{2.4}{0.015811}=151.79\ \text{m}^{1/2}/\text{s}

Manning's coefficient

C=R1/6n  ⇒  n=R1/6C=2.51/6151.79=1.1650151.79=0.00768C=\frac{R^{1/6}}{n}\;\Rightarrow\;n=\frac{R^{1/6}}{C}=\frac{2.5^{1/6}}{151.79}=\frac{1.1650}{151.79}=0.00768

(Check with Manning: V=1nR2/3S1/2=10.00768×1.842×0.01=2.40V=\dfrac1nR^{2/3}S^{1/2}=\dfrac{1}{0.00768}\times1.842\times0.01=2.40 m/s.)

Answer: C=151.8C=151.8 m1/2^{1/2}/s and n=0.0077n=0.0077 s/m1/3^{1/3}. The value of nn is very low for a natural river; it follows directly from the data as given.

✓ Answer checked by Claude Opus 5.5 (high effort)

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