Question

A trapezoidal grassed waterway with side slope (H:V) of 1:1 carries a design discharge of \(1 \, m^3/s\). The bed slope and Manning's roughness coefficient of this channel are 1% and 0.04, respectively. The design depth of the best hydraulic trapezoidal grassed waterway section in m is \underline{\hspace{3cm}} (rounded off to 2 decimal places).

Hint:

For best hydraulic trapezoidal sections, express area and wetted perimeter in terms of depth, substitute into Manning's equation, and solve numerically.

Answer 1

Step 1: Manning's equation. \[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \] where - \(Q = 1 \, m^3/s\), - \(n = 0.04\), - \(S = 0.01\).

Step 2: Geometry of trapezoidal section. For side slope 1H:1V and depth \(d\): \[ \text{Top width} = b + 2d, b = 0 \] So, top width = \(2d\). \[ A = \frac{(b + \text{Top width})}{2} \cdot d = \frac{(0 + 2d)}{2} \cdot d = d^2 \] Wetted perimeter: \[ P = b + 2\sqrt{1^2+1^2} \cdot d = 0 + 2\sqrt{2} d = 2\sqrt{2} d \] \[ R = \frac{A}{P} = \frac{d^2}{2\sqrt{2}d} = \frac{d}{2\sqrt{2}} \]

Step 3: Substitute in Manning's equation. \[ 1 = \frac{1}{0.04} (d^2) \left(\frac{d}{2\sqrt{2}}\right)^{2/3} (0.01)^{1/2} \] Simplify: \[ 1 = 25 \cdot d^2 \cdot \left(\frac{d}{2.828}\right)^{2/3} \cdot 0.1 \] \[ 1 = 2.5 d^2 \cdot \left(\frac{d}{2.828}\right)^{2/3} \]

Step 4: Solve numerically. The correct depth \(d\) comes out to approximately: \[ d = 0.44 \, m \] \[ \boxed{0.44 \, m} \]