Question 12.6: (i) Determine the most efficient section of trapezoidal chan...

(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)

where $R_{h}$ is the hydraulic radius and h is the depth of flow (Fig. 12.12). Hence we can write,

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

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),

for this trapezoidal channel with b = 1.12 m, h = 2.37 m and n = 0.025, we can write

Or      S = 0.00042