Geometric properties from channel offsets

Plotting the section

Take distance as $x$ and depth as measured below the water surface (the depth is $0$ at both banks, so the water surface is level with the ground at $x=0$ and $x=10$ m). Top width $T=10$ m, maximum depth $5$ m at $x=5$ m.

 0    2    4    6    8   10  (m)
 ~~~~~~~~~~~~~~~~~~~~~~~~~~  water surface
 \                        /
  \__ 3 m ___      ___   /
          5 m  \  /   /
                \/

(points: (0,0), (1,3), (3,3), (5,5), (7,3), (10,0))

Area and wetted perimeter by segments

Area of each strip = width × mean depth; length = $\sqrt{\Delta x^2+\Delta y^2}$.

(i) Area

$A=28.00$ m².

(ii) Equivalent hydraulic radius

$$R=\frac{A}{P}=\frac{28.00}{15.062}=1.8590\ \text{m}$$

(iii) Conveyance factor

$$K=\frac{1}{n}AR^{2/3}=\frac{1}{0.02}\times28.00\times(1.8590)^{2/3}=2116.7\ \text{m}^3/\text{s}$$

(so that $Q=K\sqrt{S_0}$; the section factor is $AR^{2/3}=42.333$ m$^{8/3}$.)

(iv) Hydraulic depth

$$D_h=\frac{A}{T}=\frac{28.00}{10}=2.80\ \text{m}$$

(v) Section factor for uniform flow

$$AR^{2/3}=28.00\times(1.8590)^{2/3}=42.333\ \text{m}^{8/3}$$

Answer: $A=28.0$ m², $K=2117$, $R=1.859$ m, $D_h=2.8$ m, $AR^{2/3}=42.33$ m$^{8/3}$.