Concrete-lined canal: Q 350

Given: $Q = 350\ \text{m}^3/\text{s}$, $S = 0.000200$ (1 in 5000), $n = 0.014$, side slope 1.5H:1V, limiting velocity 2 m/s.

Most economical (best hydraulic) section

For a trapezoidal section of least perimeter: $R = D/2$ and $B = 2D(\sqrt{1+z^2} - z)$.

$$B = 2D(\sqrt{1+1.5^2} - 1.5) = 0.6056D,\qquad A = D^2\left(2\sqrt{1+z^2} - z\right) = 2.1056D^2$$

Using Manning's equation with $R = D/2$:

$$Q = \frac{1}{n}\,2.1056D^2\left(\frac{D}{2}\right)^{2/3}S^{1/2} \Rightarrow D^{8/3} = \frac{Q n\,2^{2/3}}{2.1056\sqrt{S}}$$ $$D = 8.061\ \text{m},\quad B = 4.881\ \text{m},\quad A = 136.81\ \text{m}^2,\quad R = 4.030\ \text{m},\quad V = 2.558\ \text{m/s}$$

Velocity check

$V = 2.56$ m/s is more than the limiting velocity 2 m/s, so the best section cannot be used. Fix $V = 2$ m/s and find the section from the given slope.

$$\begin{aligned} R &= \left(\frac{Vn}{\sqrt S}\right)^{3/2} = \left(\frac{2\times 0.014}{\sqrt{0.000200}}\right)^{3/2} = 2.786\ \text{m} \\ A &= \frac{Q}{V} = 175.00\ \text{m}^2,\qquad P = \frac{A}{R} = 62.82\ \text{m} \end{aligned}$$

From $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2}$:

$$\begin{aligned} 2.1056D^2 - 62.82D + 175.00 &= 0 \\ D &= 3.110\ \text{m},\qquad B = P - 3.6056D = 51.60\ \text{m} \end{aligned}$$

Answer: $D \approx 3.11$ m, $B \approx 51.6$ m, side slope 1.5H:1V, $S = 1$ in 5000, $V = 2$ m/s (the best section would give velocity above the limit).