Question 12.6: (i) Determine the most efficient section of trapezoidal chan...
(i) Determine the most efficient section of trapezoidal channel, n = 0.025, to carry 12.74 m^{3}/s. To prevent scouring, the maximum velocity is to be 0.92 m/s and the side slopes of the trapezoidal channel are 1 vertical to 2 horizontal. (ii) What slope S of the channel is required?
(i) It is known from Eq. (12.18) that for the most efficient section (the minimum wetted perimeter for a given discharge) of a trapezoidal channel
R_{h}=\frac{h^{2}(2 \operatorname{cosec} \alpha-\cot \alpha)}{h(2 \operatorname{cosec} \alpha-\cot \alpha)-h \cot \alpha+2 h \operatorname{cosec} \alpha} =\frac{h^{2}(2 \operatorname{cosec} \alpha-\cot \alpha)}{2 h(2 \operatorname{cosec} \alpha-\cot \alpha)}=\frac{h}{2} (12.18)
R_{h}=h / 2where R_{h} is the hydraulic radius and h is the depth of flow (Fig. 12.12). Hence we can write,
R_{h}=h / 2=A / P=\frac{b h+2(h / 2)(2 h)}{b+2 h(5)^{1 / 2}}Or b=2 h(5)^{1 / 2}-4 h (12.20)
= 0.472 h
where b is the width at the base (Fig. 12.12). Again, from continuity, the crosssectional area to accommodate the maximum velocity is given by
A=12.74 / 0.92=b h+2 h^{2}Or b=\left(13.85-2 h^{2}\right) / h
Equating (12.20) and (12.21), we get
h = 2.37 m and b = 1.12 m
(ii) Using Manning’s equation, i.e.
Eq., (12.9),
V=(1 / n) R_{h}^{2 / 3} S^{1 / 2}for this trapezoidal channel with b = 1.12 m, h = 2.37 m and n = 0.025, we can write
0.92=\frac{(2.37 / 2)^{2 / 3} S^{1 / 2}}{0.025}Or S = 0.00042