GATE 2024 CE (CE1) – Question 57
A standard round bottom triangular canal section as shown in the figure has a bed slope of 1 in 200. Consider the Chezy's coefficient as 150 m$^{1/2}$/s.
[Figure: a canal with straight side slopes of 1 vertical to 1.5 horizontal and a circular arc of radius y at the bottom tangent to both sides. The centre of the arc is on the water surface and the depth of flow is y.]
The normal depth of flow, y (in meters) for carrying a discharge of 20 m$^3$/s is ____________ (rounded off to 2 decimal places).
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Correct answer: 1.09 to 1.11
Explanation
The arc has its centre on the free surface and its radius equals the depth $y$. The sides have the slope 1 : 1.5, so each makes the angle $\theta = \tan^{-1}\frac{1}{1.5} = 33.69° = 0.588$ rad with the horizontal. The tangent point is at the distance $y$ from the centre, and the straight part of each side from the tangent point to the surface has the length $\frac{y}{\tan\theta} = 1.5y$. The arc subtends $2\theta$ at the centre, so its length is $2\theta y = 1.176y$. The wetted perimeter is $P = 1.176y + 3y = 4.176y$. The area is the sector $\theta y^2 = 0.588y^2$ plus two right triangles of $\frac{1}{2} \times y \times 1.5y = 0.75y^2$ each, which gives $A = 2.088y^2$. The hydraulic radius is $R = \frac{A}{P} = 0.5y$. By Chezy's formula, $Q = AC\sqrt{RS} = 2.088y^2 \times 150 \times \sqrt{0.5y/200} = 15.66y^{2.5} = 20$, so $y^{2.5} = 1.277$ and $y = 1.10$ m.