Hydraulically best trapezoidal section

Hydraulically best section

A channel section is hydraulically best (most economical or most efficient) when, for a given area, bed slope and roughness, it carries the maximum discharge. Since $Q = AC\sqrt{RS}$ (or Manning), Q is maximum when the wetted perimeter $P$ is minimum, i.e. hydraulic radius $R = A/P$ is maximum. Such a section also needs the least lining and excavation.

Proof: $R = y/2$ for the best trapezoidal section

Let bottom width $b$, depth $y$, side slope $n$ H : 1 V.

   |<-------- b + 2ny -------->|
   \                          /
    \  y                     /  side slope n:1
     \_______________________/
          |<---- b ---->|

$$\begin{aligned} A &= (b + ny)y \;\Rightarrow\; b = \frac{A}{y} - ny \\ P &= b + 2y\sqrt{1+n^2} = \frac{A}{y} - ny + 2y\sqrt{1+n^2} \end{aligned}$$

For maximum discharge, $P$ must be minimum with $A$ constant:

$$\begin{aligned} \frac{dP}{dy} &= -\frac{A}{y^2} - n + 2\sqrt{1+n^2} = 0 \\ \frac{A}{y^2} &= 2\sqrt{1+n^2} - n \\ A &= y^2\left(2\sqrt{1+n^2} - n\right) \end{aligned}$$

Using $A = (b+ny)y$: $b + ny = 2y\sqrt{1+n^2} - ny$, i.e.

$$\frac{b + 2ny}{2} = y\sqrt{1+n^2}$$

(half the top width equals one sloping side).

Now the perimeter:

$$\begin{aligned} P &= \frac{A}{y} - ny + 2y\sqrt{1+n^2} \\ &= y(2\sqrt{1+n^2} - n) - ny + 2y\sqrt{1+n^2} \\ &= 2y(2\sqrt{1+n^2} - n) = \frac{2A}{y} \end{aligned}$$

Hence

$$R = \frac{A}{P} = \frac{A}{2A/y} = \frac{y}{2}$$

So for the most economical trapezoidal channel, the hydraulic mean depth equals half the depth of flow. (It is also shown that the three sides are tangent to a semicircle with centre on the water surface.)