10. An open channel has a bottom width of 4.5m. The velocity of flow is 1.2 m/s. Determine the discharge in cubic meters per second under the following conditions:

a. Most efficient trapezoidal section.

b. Trapezoidal section with one side vertical and the other sloping 45 degrees and depth of flow of 2/3 the base width.

c. Minimum seepage with side slope of 65 degrees.

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(a) Most efficiest \left\{R=\frac{y}{2}\right\}\begin{array}{l} V=\frac{1}{h} R^{\frac{2}{3}} s^{1 / 2} \\1.2=0.015\left(\frac{y}{2}\right)^{2 / 3} \sqrt{0.0015} \\y=0.633 \mathrm{~m} \quad A=\sqrt{3} y^{2} \\\theta=A V=\sqrt{3} \times y^{2} \times y .2 \\\theta=0.832 \mathrm{~m}^{3} / \mathrm{s}\end{array}\begin{array}{l}\text { (B) } \quad V=1.2 \mathrm{~m} / \mathrm{s} \quad n=0.015 \quad S=0.0015 \quad B=4.5 \\\begin{array}{ll}\underbrace{\cdots}_{+-4.5 \rightarrow 0} & v=\frac{1}{h} R^{2 / 3} s^{1 / 2} \\y_{f} & 1.2=\frac{1}{0.015}\left(\frac{A}{P}\right)^{2 / 3} \sqrt{0.0015}\end{array} \\A=4.5 y+\left(\frac{y}{2} \times y\right) \\\left(\frac{A}{P}\right)^{2 / 3}=0.464 \\A=4 \cdot 5 y+\frac{y^{2}}{2} \\A=0.3168 \mathrm{P} \\p=y+4 \cdot 5+y \sqrt{2} \\p=2.414 y+4.5 \\4.5 y+0.5 y^{2}=0.3168(2.414 y+4.5) \\0.5 y^{2}+3.735 y-7.4256=0 \\y=0.36 \mathrm{~m} \\\theta=A V=\left(4.5 y+0.5 y^{2}\right) \times 1.2=2.02 \mathrm{~m}^{3} / \mathrm{s} \\\end{array}Now(C)\begin{array}{l} A=(B+m y) y \\=(4.5+0.466 \times 2.25) \times 2.25 \\A=12.48 \mathrm{~h}^{2} \\P=B+2 y \sqrt{1+\mathrm{m}^{2}} \\=4.5+4.5 .5 \sqrt{1+0.466^{2}} \\P=9.46 \mathrm{~m} \\\therefore \quad R=\frac{A}{P}=1.318 \quad V=1.2 \mathrm{~L} / \mathrm{s} \\\theta \quad A V=12.48 \times 1.2=14.976 \mathrm{~m}^{3} / \mathrm{s}\end{array}I Hope this answer will be helpful to you.Please like this answer don't give dislike.Thank you. ...