A rectangular channel 22 feet wide carries a flow of 25,000 cfs on a slope of 0.01. Manning's n for the channel is 0.012. Find the depth of flow.
A rectangular channel 22 feet wide carries a flow of 25,000 cfs on a slope of 0.01. Manning's n for the channel is 0.012. Find the depth of flow.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![A rectangular channel 22 feet wide carries a flow of 25,000 cfs on a slope of 0.01. Manning’s n for the channel is 0.012. Find the depth of flow.
**Explanation:**
This problem involves calculating the depth of flow in a rectangular channel using given parameters:
- **Width of the channel**: 22 feet
- **Flow rate**: 25,000 cubic feet per second (cfs)
- **Slope of the channel**: 0.01
- **Manning’s roughness coefficient (n)**: 0.012
To solve for the depth of flow, you can use Manning’s equation:
\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \]
Where:
- \( Q \) is the flow rate (cfs)
- \( A \) is the cross-sectional area of flow (square feet)
- \( R \) is the hydraulic radius (feet)
- \( S \) is the slope of the energy grade line (bed slope in this case)
For a rectangular channel, \( A = b \cdot y \) and \( R = \frac{b \cdot y}{b + 2y} \), where:
- \( b \) is the width of the channel (22 feet)
- \( y \) is the depth of flow (feet)
The task is to solve for \( y \) using these expressions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ebcf202-e71c-4e7a-b7be-45ee67012454%2F365e814f-ffec-4df8-8844-8ea61a00ae8e%2F8wkikhv_processed.png&w=3840&q=75)
Transcribed Image Text:A rectangular channel 22 feet wide carries a flow of 25,000 cfs on a slope of 0.01. Manning’s n for the channel is 0.012. Find the depth of flow.
**Explanation:**
This problem involves calculating the depth of flow in a rectangular channel using given parameters:
- **Width of the channel**: 22 feet
- **Flow rate**: 25,000 cubic feet per second (cfs)
- **Slope of the channel**: 0.01
- **Manning’s roughness coefficient (n)**: 0.012
To solve for the depth of flow, you can use Manning’s equation:
\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \]
Where:
- \( Q \) is the flow rate (cfs)
- \( A \) is the cross-sectional area of flow (square feet)
- \( R \) is the hydraulic radius (feet)
- \( S \) is the slope of the energy grade line (bed slope in this case)
For a rectangular channel, \( A = b \cdot y \) and \( R = \frac{b \cdot y}{b + 2y} \), where:
- \( b \) is the width of the channel (22 feet)
- \( y \) is the depth of flow (feet)
The task is to solve for \( y \) using these expressions.
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