Most economical rectangular channel section

A most economical (best hydraulic) section of an open channel is the section which, for a given area, bed slope and roughness, carries the maximum discharge. Since $Q = A C\sqrt{RS}$ (Chezy) or $Q = \frac{1}{n}AR^{2/3}S^{1/2}$ (Manning), $Q$ is maximum when $R = A/P$ is maximum, i.e. when the wetted perimeter $P$ is minimum for the given area. Such a section also needs the least lining and excavation.

Proof for a rectangular section

   |<------- b ------->|
   |                   |  ^
   |      water        |  y
   |___________________|  v

Let width $= b$, depth $= y$.

$$A = by \quad \Rightarrow \quad b = \frac{A}{y}$$ $$P = b + 2y = \frac{A}{y} + 2y$$

For given $A$, $P$ is minimum when $dP/dy = 0$:

$$\frac{dP}{dy} = -\frac{A}{y^2} + 2 = 0 \;\Rightarrow\; A = 2y^2$$

Since $A = by$: $by = 2y^2 \Rightarrow b = 2y$. (Check: $d^2P/dy^2 = 2A/y^3 > 0$, so it is a minimum.)

Hydraulic mean depth:

$$\begin{aligned} R &= \frac{A}{P} = \frac{by}{b + 2y} \\ &= \frac{2y\cdot y}{2y + 2y} = \frac{2y^2}{4y} = \frac{y}{2} \end{aligned}$$

Result

For the most economical rectangular channel:

• width is twice the depth, $b = 2y$;
• hydraulic mean depth is half the depth, $R = y/2$.

Example: if $y = 1.5$ m, then $b = 3$ m and $R = 0.75$ m.