Most Economical Trapezoidal Section

Fig shows the cross-section of a trapezoidal channel.

Let $\quad b=$ Base width of the channel,

$y=$ Depth of flow, and

$\theta=$ Angle made by the sides with horizontal.

Side slope $=1$ vertical to $n$ horizontal. Area of flow, $A=\left(\frac{A B+C D}{2}\right) \times y=\frac{b+(b+2 n y)}{2} \times y=(b+n y) y$ $\ldots(i)$ $\therefore$ $$ \frac{A}{y}=b+n y $$ or, $$ b=\frac{A}{y}-n y $$ Wetted perimeter, $P=A D+A B+B C=A B+2 B C \quad(\because A D=B C)$ $$ =b+2 \sqrt{B E^{2}+C E^{2}} $$ $$ \text { Wetted perimeter, } \begin{aligned} P &=A D+A B+B C=A B+2 B C \quad(\because A D=B C) \ &=b+2 \sqrt{B E^{2}+C E^{2}} \ &=b+2 \sqrt{n^{2} y^{2}+y^{2}} \ P &=b+2 y \sqrt{n^{2}+1} \end{aligned} $$ Substituting the value of $b$ from eqn. (ii) in eqn. (iii), we get: $$ P=\frac{A}{y}-n y+2 y \sqrt[1]{n^{2}+1} $$ The section of the channel will be most economical when its wetted perimeter $(P)$ is minimum, i.e. $\frac{d P}{d y}=0$

or, $\frac{d}{d y}\left[\frac{A}{y}-n y+2 y \sqrt{n^{2}+1}\right]=0$

or, $\frac{d}{d y}\left[\frac{A}{y}-n y+2 y \sqrt{n^{2}+1}\right]=0$

or, $\quad-\frac{A}{y^{2}}-n+2 \sqrt{n^{2}+1}=0$

$(\because n$ is c

or, $\frac{A}{y^{2}}+n=2 \sqrt{n^{2}+1}$

Substituting the value of $A$ from eqn. $(i)$, in the above equation, we get: $$ \frac{(b+n y) y}{y^{2}}+n=2 \sqrt{n^{2}+1} $$ $$ \text { or, } \quad \frac{(b+n y)}{y}+n=2 \sqrt{n^{2}+1} $$

or, $\quad \frac{b+n y+n y}{y}+2 \sqrt{n^{2}+1}$ or $\frac{b+2 n y}{y}=2 \sqrt{n^{2}+1}$

or, $\quad \frac{b+2 n y}{2}=y \sqrt{n^{2}+1}$

$\ldots(16.14)$

[i.e. Half of top width = One of the sloping sides ... Fig ]

Hydraulic radius, $R$ : Hydraulic radius, $R=\frac{A}{P}$

$$ \begin{array}{rlr} A & =(b+n y) \times y & \text { [From eqn. (i)] } \\ P & =i b+2 y \sqrt{n^{2}+1} & \text { [From eqn. (iii)] } \\ \text { But, } \quad 2 y \sqrt{n^{2}+1} & =b+2 n y & \\ \therefore \quad P & =b+(b+2 n y)=2(b+n y) & \text { [From eqn. (16.14)] } \\ \therefore \quad & \text { Hydraulic radius, } R & =\frac{(b+n y) y}{2(b+n y)}=\frac{y}{2} & \ldots(16.15) \end{array} $$ ie,hydraulic radius equals half the flow depth

$$ \begin{array}{l} O F=O C \times \frac{1}{\sqrt{n^{2}+1}}=y \sqrt{n^{2}+1} \times \frac{1}{\sqrt{n^{2}+1}}=y, \text { depth of flow } \\ {\left[\because O C=\text { Half of top width }=\frac{b+2 n y}{2}=y \sqrt{n^{2}+1} \ldots \text { Eqn. (16.14) }\right]} \end{array} $$

Thus a circle with centre $O$ and radius equal to the depth of flow will be tangential to the three sides of a most economical trapezoidal section; this condition stipulates that the most economical section of a trapezoidal channel will be a half-hexagon. Hence conditions for most economical trapezoidal section are: 1. $\frac{b+2 n y}{2}=y \sqrt{n^{2}+1}$ (i.e. Half of top width = One of the sloping sides) 2. Hydraulic radius, $R=\frac{y}{2}$ 3. A semicircle drawn from $O$ with radius equal to depth of flow will touch the three sides of the trapezoidal channel.