Question
Download Solution PDFA 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)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
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 (\( \tau_0 \)):
\( \tau_0 = \gamma R S \)
\( \tau_0 = 10 \times 0.5556 \times 0.0003 \)
\( \tau_0 = 0.0016668 \, \text{kN/m}^2 \)
\( \tau_0 = 1.6668 \, \text{N/m}^2 \approx 1.67 \, \text{N/m}^2 \)
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