Question
Download Solution PDFA 3.0 m wide rectangular channel flowing at its normal depth of 0.8 m carries a discharge of 9.5 cumec. The bed slope of the channel is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
To determine the type of bed slope of a rectangular channel, we need to compare the normal depth (\( y_n \)) with the critical depth (\( y_c \)). The type of slope is categorized based on whether the normal depth is greater than, less than, or equal to the critical depth.
Calculation:
Given:
- Width of the rectangular channel (\( b \)) = 3.0 m
- Normal depth (\( y_n \)) = 0.8 m
- Discharge (\( Q \)) = 9.5 cumec (cubic meters per second)
1. Calculate the Cross-Sectional Area (\( A \)) of Flow:
\( A = b \times y_n \)
\( A = 3.0 \times 0.8 \)
\( A = 2.4 \, \text{m}^2 \)
2. Calculate the Hydraulic Radius (\( R \)):
The hydraulic radius \( R \) is given by:
\( R = \frac{A}{P} \), where \( P \) is the wetted perimeter.
\( P = b + 2y_n \)
\( P = 3.0 + 2 \times 0.8 \)
\( P = 4.6 \, \text{m} \)
\( R = \frac{2.4}{4.6} \)
\( R \approx 0.5217 \, \text{m} \)
3. Calculate the Critical Depth (\( y_c \)):
The critical depth for a rectangular channel is calculated using the formula:
\( y_c = \left( \frac{Q^2}{g \times b^2} \right)^{1/3} \)
\( y_c = \left( \frac{9.5^2}{9.81 \times 3^2} \right)^{1/3} \)
\( y_c = \left( \frac{90.25}{88.29} \right)^{1/3} \approx 1 \, \text{m} \)
4. Determine the Slope Type:
- **Mild Slope**: \( y_n > y_c \)
- **Steep Slope**: \( y_n < y_c \)
- **Critical Slope**: \( y_n = y_c \)
Given:
- \( y_n = 0.8 \, \text{m} \)
- \( y_c = 1 \, \text{m} \)
Since \( y_n < y_c \), the slope of the channel is considered steep.
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