Subject: Applied Hydraulics-II

Topic: Flow Through Open Channel(Uniform Flow)

Difficulty: Medium

Solution :

Given :

Chezy's constant (C) = 55

Bed slope = 1 in 2000

Depth of flow, $d=4.0 \space {m}$

Angle made by each side with vertical,

i.e., $\angle A B D=\angle C B D=30^{\circ}$

From Fig., we have Area,

$ \begin{aligned} A &= Area \space of A B C \\ &=2 \times \text { Area } A B D\\ &=\frac{2 \times A D \times B D}{2}\\ &=A D \times B D \\ &=B D \tan 30^{\circ} \times B D \\ &=4 \tan 30^{\circ} \times 4\\ &=9.2376 \mathrm{~m}^{2} \\ \end{aligned} \\ $

Wetted perimeter, P

$ \begin{aligned} P &=A B+B C\\ &=2 A B \\ &=2 \sqrt{B D^{2}+A D^{2}}\\ &=2 \sqrt{4.0^{2}+\left(4 \tan 30^{\circ}\right)^{2}} \\ &=2 \sqrt{16.0+5.333} \\ &=9.2375 \mathrm{~m} \end{aligned} $

Hydraulic mean Depth (m),

$ \begin{aligned} \\ m &=\frac{A}{P}=\frac{9.2376 }{9.2375} \\ &=1.0 \space m \end{aligned} $

for discharge,

$ \begin{aligned} \\ Q &=A C \sqrt{m i}\\ &=9.2376 \times 55 \times \sqrt{1 \times \frac{1}{2000}}\\ &=11.3607 \space m^3/s \end{aligned} $