Economical trapezoidal channel side length

A trapezoidal channel is most economical when its wetted perimeter is minimum for a given area, which gives maximum discharge for given $A$, $S$ and roughness.

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

   |<-------- T = b + 2ny ------->|
   \                             /
    \ sloping side = y*sqrt(1+n^2)
     \__________________________/
          |<----- b ----->|

Area and wetted perimeter

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

Condition for minimum P (A constant)

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

Substitute $A/y = b + ny$:

$$\begin{aligned} b + ny &= 2y\sqrt{1+n^2} - ny \\ b + 2ny &= 2y\sqrt{1+n^2} \\ \frac{b + 2ny}{2} &= y\sqrt{1+n^2} \end{aligned}$$

Interpretation

• Top width $T = b + 2ny$, so the left side is half the top width.
• Length of one sloping side $= \sqrt{(ny)^2 + y^2} = y\sqrt{1+n^2}$, which is the right side.

Therefore, for the most economical trapezoidal section, length of one sloping side = half the top width. Hence verified.

Also $d^2P/dy^2 = 2A/y^3 > 0$, so this is indeed a minimum of $P$. A related result is $R = y/2$.