Question 9.1: Design a trapezoidal channel to carry a discharge of 10 m³ /......

For the blasted rock surface, n = 0.030 and the side slopes may be almost vertical. Let us select a value for the side slope s as 1 horizontal to 4 vertical. The substitution of these values into the Manning equation yields
\quad\quad\quad\quad AR^{\frac{2}{3}}=\frac{nQ}{C_{o}S^{0.5}_{o}}\\\quad\quad\quad\quad\quad\quad =\frac{0.030\times 10}{(0.00025)^{\frac{1}{2}}}\\\quad\quad\quad\quad\quad\quad=18.97
\quad\quadSince the channel section is almost rectangular, let us select B_{o} = 2y. Then,
A=(B_{o}+\frac{1}{4}y)=2.25y^{2};P=B_{o}+\frac{1}{2}\sqrt{17}y=4.06y;R=(2.25y^{2}/(4.06y))=0.55y.Hence,
AR^{\frac{2}{3}}=(2.25y^{2})(0.55y)^{\frac{2}{3}}\\\quad\quad=1.518y^{2.67}\\\quad\quad=18.97
Solving this equation for y, we get y = 2.57 m. Then, B_{o} = 2 × 2.57 = 5.14 m. For ease of construction, let us use B_{o} = 5 m. Then, the corresponding value of y for which AR^{\frac{2}{3}} = 18.97 is determined by trial and error as 2.64 m. Based on Eq. 9-1, the freeboard =\sqrt{0.8\times 2.64} = 1.45 m. As compared to this value, a freeboard of 0.75 m selected from Table 9-1 appears to be more appropriate.
Therefore, total depth = 2.64 + 0.75 = 3.39 \simeq 3.4 m.
\quad\quadThe flow area for a flow depth of 2.64 m is 14.94 m². Therefore, the flow velocity = 10/14.94 = 0.67 m/s. This is close to the minimum allowable flow velocity; thus, a bottom width of 5 m and a cross section depth of 3.4 m is satisfactory.
\quad\quad\quad\quad F_{b}=\sqrt{ky} (9-1)