Can you please show how to do this problem? I’m confused on this homework question and please box answer and pleas write clear thank you. ### Problem StatementA 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 StepsTo 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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Answered: A 2.1-m wide rectangular channel with a…

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).

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Highway Drainage

Highway drainage is the process of extracting and regulating the excess surface water on the pavement by giving the desired slope. The construction of a highway drain helps to divert the water present on the surface, and sub-grade by the means of slope. Highway drainage systems are a necessary part of smooth functioning of traffic, since the highway drainage systems provide proper disposal of unwanted substances from the highways.

Question

Can you please show how to do this problem? I’m confused on this homework question and please box answer and pleas write clear thank you.

### 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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