Question 11.4: (a) A rectangular concrete channel is to convey a discharge ......

For the most efficient rectangular cross section, channel width B = 2y, A = 2y². Also, the hydraulic radius R = y/2.

From Manning’s formula,          Q=\frac{1}{n}\times R^{2/3}\times S_{0}^{1/2}\times A

or                    3.0=\frac{1}{0.012}\times\left\lgroup\frac{y}{2}\right\rgroup^{2/3}\times\left(0.001\right)^{1/2}\times\left(2  y^{2}\right)=3.322~y^{8/3}

Therefore, y = 0.962 m, and B = 1.924 m. A freeboard of say 0.15 m could be provided.

(b) In problem (a) if a trapezoidal section without any constraint is provided, the most efficient cross section will be half-hexagon. Thus, tan θ = tan 30° = 0.577. Furthermore, hydraulic radius R = y/2, and the area A = \sqrt3 y² . Therefore, from Manning’s equation

Therefore, y = 1.016 m, and B={\frac{2}{\sqrt{3}}}y=1.173{\mathrm{~m}}

(c) For the specifications of problem (a), the sides slopes of a trapezoidal section are restricted to 1.5 H : 1V because of the prevailing site conditions. Find the most efficient cross section of the channel.

For the most efficient section, the hydraulic radius R = y/2.

From Eq. (11.53), the area of cross section A = y² (2 sec θ – tan θ)

{\frac{A}{y^{2}}}=2\sec\theta-\tan\theta           (11.53)

Given, tan θ = m = 1.5.    A = y² (2 \sqrt{1+1.5^2} – 1.5) = 2.10 y²

From Manning’s equation, 3.0=\frac{1}{0.012}\times\left\lgroup\frac{y}{2}\right\rgroup^{2/3}\times\left(0.001\right)^{1/2}\times\left(2.10  y^{2}\right)=3.487~y^{8/3}

Therefore, y = 0.945 m

From Eq. (11.54), B = 2y (sec θ – tan θ) = 2 × 0.945 × (\sqrt{1+1.5^2} – 1.5) = 0.572 m

The three efficient sections are shown in Fig. 11.13. The area of cross section and the perimeter for the three cases are (a) 1.85 m², 3.85 m, (b) 1.78 m², 3.52 m, (c) 1.88 m² and 3.96 m, respectively. The area of cross section is nearly the same for all the shapes. However, the perimeter is a minimum for the half-hexagon case.