Question 9.2: Design a channel to carry a flow of 6.91 m³ /s. The channel ......

For stiff clay, n = 0.025, suggested side slope, s = 1 : 1 (from Table 9-2), and the permissible flow velocity (from Table 9-3) is 1.8 m/s. Hence,
\quad\quad\quad\quadA = 6.91 / 1.8 = 3.83 m²
Substituting values for V, n, and S_{o} into Manning equation, and solving for R, we get R = 0.713 m. Hence,
\quad\quad\quad\quad P=\frac{3.83}{0.713}=5.37m
\quad\quadSubstitution into expressions for P and A from Eq. 9-11 and equating them to the values computed above, we obtain

\quad\quad\quad\quad P=B_{o}+2\sqrt{1+s^{2}}y
\quad\quad\quad\quad A = (B_{o} + sy)y (9-11)
\quad\quad\quad\quad B_{o}+2.83y=5.37\\\quad\quad\quad\quad\quad(B_{o}+y)y=3.83
Elimination of B_{o} from these two equations yields
\quad\quad\quad\quad 1.83y² – 5.37y + 3.83 = 0
Solution of this equation gives y = 1.22 m. The freeboard from Eq. 9-1 may be calculated as \sqrt{0.8\times1.22} = 0.99 m, whereas the suggested value in Table 9-1 is 0.75 m. Let us select a freeboard of 0.75 m. Then, the depth of the section = 1.22 + 0.0.75 = 1.97 m. Select a depth of 2.0 m and bottom width of 1.9 m.
\quad\quad\quad\quad F_{b}=\sqrt{ky} (9-1)