Most economical triangular channel section

Economical channel section

A channel section is economical (most efficient / best hydraulic) when for a given cross-sectional area, bed slope and roughness it gives the maximum discharge. This happens when the wetted perimeter is minimum, so the cost of excavation and lining is also least.

Proof for triangular section

Let depth $y$, side slope $m$ H : 1 V on both sides, and $\theta$ the angle of each side with the vertical, so $m = \tan\theta$.

   |<------ 2my ------>|
   \                  /
    \       y        /
     \  theta|      /
      \      |     /
       \     |    /
        \    |   /
         \___|__/
             V (apex)

$$\begin{aligned} A &= \frac{1}{2}(2my)y = my^2 \;\Rightarrow\; y = \sqrt{\frac{A}{m}} \\ P &= 2y\sqrt{1+m^2} = 2\sqrt{\frac{A}{m}}\sqrt{1+m^2} = 2\sqrt{A}\sqrt{\frac{1+m^2}{m}} \end{aligned}$$

For minimum P at constant A, minimise $F = \dfrac{1+m^2}{m} = \dfrac{1}{m} + m$:

$$\begin{aligned} \frac{dF}{dm} &= -\frac{1}{m^2} + 1 = 0 \\ m^2 &= 1 \;\Rightarrow\; m = 1 \end{aligned}$$

$d^2F/dm^2 = 2/m^3 > 0$, so it is a minimum.

So the side slope is 1 : 1 (1 H : 1 V), and $\tan\theta = m = 1$ gives $\theta = 45^\circ$ with the vertical. The two sides are then at right angles to each other.

Additional results for this section: $A = y^2$, $P = 2\sqrt{2}\,y$, $R = \dfrac{y}{2\sqrt{2}}$.