A rectangular channel has a bottom width of of 2 m and longitudinal slope of the channel is 0.0003. What will be the average shear stress on the channel boundary when the flow takes place at a uniform depth of 1.25 m? (Take γ = 10 kN/m3)

Concept:

The average shear stress on the channel boundary in open channel flow can be determined using the formula:

\( \tau_0 = \gamma R S \)

Where:

• \( \tau_0 \) = Shear stress on the channel boundary (N/m²)
• \( \gamma \) = Unit weight of water (10 kN/m³)
• \( R \) = Hydraulic radius (m)
• \( S \) = Slope of the channel

Calculation:

Given:

• Bottom width of the channel (\( b \)) = 2 m
• Uniform flow depth (\( y \)) = 1.25 m
• Longitudinal slope of the channel (\( S \)) = 0.0003
• Unit weight of water (\( \gamma \)) = 10 kN/m³

1. Calculate the Cross-Sectional Area (\( A \)) of the Flow:

\( A = b \times y \)

\( A = 2 \times 1.25 \)

\( A = 2.5 \, \text{m}^2 \)

2. Calculate the Wetted Perimeter (\( P \)) of the Channel:

\( P = b + 2y \)

\( P = 2 + 2 \times 1.25 \)

\( P = 4.5 \, \text{m} \)

3. Calculate the Hydraulic Radius (\( R \)):

\( R = \frac{A}{P} \)

\( R = \frac{2.5}{4.5} \)

\( R = 0.5556 \, \text{m} \)

4. Calculate the Shear Stress (\( \tau_0 \)):

\( \tau_0 = \gamma R S \)

\( \tau_0 = 10 \times 0.5556 \times 0.0003 \)

\( \tau_0 = 0.0016668 \, \text{kN/m}^2 \)

\( \tau_0 = 1.6668 \, \text{N/m}^2 \approx 1.67 \, \text{N/m}^2 \)