Section 4: Soil and Water Conservation Engineering
Fluid Mechanics
Subtopic: Flow in open channels
Question Statement:
A triangular channel has side slopes of $1\text{ horizontal to } 1\text{ vertical}$ ($z = 1$). If the flow depth is $0.80\text{ m}$, the hydraulic radius of the section in meters is ________ (round off to 3 decimal places).
Answer Key:
0.283
Step-by-Step Solution & Explanation
For a triangular channel with side slope $z:1$:
1. Area $A = z y^2 = 1 \times (0.80)^2 = 0.64\text{ m}^2$.
2. Wetted perimeter $P = 2 y \sqrt{1 + z^2} = 2 \times 0.80 \times \sqrt{1 + 1^2} = 1.60 \sqrt{2} = 2.2627\text{ m}$.
3. Hydraulic radius $R$:
$R = \frac{A}{P} = \frac{0.64}{2.2627} = \frac{y}{2\sqrt{2}} = \frac{0.80}{2.8284} \approx 0.2828\text{ m} \approx 0.283\text{ m}$