A 2.1-m wide rectangular channel with a bed slope of 0.043 has a depth of flow of 9.29 in m. Manning's roughness coefficient is 0.015. Determine the steady uniform discharge in the channel (cms).
A 2.1-m wide rectangular channel with a bed slope of 0.043 has a depth of flow of 9.29 in m. Manning's roughness coefficient is 0.015. Determine the steady uniform discharge in the channel (cms).
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN:9781305970939
Author:Braja M. Das, Khaled Sobhan
Publisher:Braja M. Das, Khaled Sobhan
Chapter8: Seepage
Section: Chapter Questions
Problem 8.11P
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![### Problem Statement
A 2.1-m wide rectangular channel with a bed slope of 0.043 has a depth of flow of 9.29 m. Manning's roughness coefficient is 0.015. Determine the steady uniform discharge in the channel (cms).
### Solution Steps
To calculate the steady uniform discharge in the channel, we use Manning's equation:
\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \]
where
- \( Q \) is the discharge (cubic meters per second, cms)
- \( n \) is Manning's roughness coefficient
- \( A \) is the cross-sectional area of flow (square meters)
- \( R \) is the hydraulic radius (meters)
- \( S \) is the slope of the channel bed
#### Step-by-Step Calculation:
1. **Calculate the cross-sectional area (A):**
For a rectangular channel,
\[ A = b \cdot h \]
where \( b \) is the width of the channel and \( h \) is the depth of flow.
\[ A = 2.1 \, \text{m} \times 9.29 \, \text{m} = 19.509 \, \text{m}^2 \]
2. **Calculate the wetted perimeter (P):**
For a rectangular channel,
\[ P = b + 2h \]
\[ P = 2.1 \, \text{m} + 2 \times 9.29 \, \text{m} = 20.68 \, \text{m} \]
3. **Calculate the hydraulic radius (R):**
\[ R = \frac{A}{P} \]
\[ R = \frac{19.509 \, \text{m}^2}{20.68 \, \text{m}} = 0.943 \, \text{m} \]
4. **Apply Manning's equation:**
\[ Q = \frac{1}{0.015} \times 19.509 \, \text{m}^2 \times (0.943 \, \text{m})^{2/3} \times (0.043)^{1/2} \]
5. **Perform the calculations:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b7dd8f1-dbab-4607-8b33-bde9974f18c9%2Fe6d8f0fd-8464-4662-8508-cbf49f7df3fe%2F0a9gh4n_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A 2.1-m wide rectangular channel with a bed slope of 0.043 has a depth of flow of 9.29 m. Manning's roughness coefficient is 0.015. Determine the steady uniform discharge in the channel (cms).
### Solution Steps
To calculate the steady uniform discharge in the channel, we use Manning's equation:
\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \]
where
- \( Q \) is the discharge (cubic meters per second, cms)
- \( n \) is Manning's roughness coefficient
- \( A \) is the cross-sectional area of flow (square meters)
- \( R \) is the hydraulic radius (meters)
- \( S \) is the slope of the channel bed
#### Step-by-Step Calculation:
1. **Calculate the cross-sectional area (A):**
For a rectangular channel,
\[ A = b \cdot h \]
where \( b \) is the width of the channel and \( h \) is the depth of flow.
\[ A = 2.1 \, \text{m} \times 9.29 \, \text{m} = 19.509 \, \text{m}^2 \]
2. **Calculate the wetted perimeter (P):**
For a rectangular channel,
\[ P = b + 2h \]
\[ P = 2.1 \, \text{m} + 2 \times 9.29 \, \text{m} = 20.68 \, \text{m} \]
3. **Calculate the hydraulic radius (R):**
\[ R = \frac{A}{P} \]
\[ R = \frac{19.509 \, \text{m}^2}{20.68 \, \text{m}} = 0.943 \, \text{m} \]
4. **Apply Manning's equation:**
\[ Q = \frac{1}{0.015} \times 19.509 \, \text{m}^2 \times (0.943 \, \text{m})^{2/3} \times (0.043)^{1/2} \]
5. **Perform the calculations:
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