/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A trapezoidal drainage channel o... [FREE SOLUTION] | 91Ó°ÊÓ

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A trapezoidal drainage channel of bottom width \(5 \mathrm{~m}\), side slopes 2: 1 (H:V), Manning's \(n\) of \(0.018,\) and longitudinal slope of \(0.1 \%\) terminates at a gate where the relationship between the flow through the gate \((Q)\), headwater elevation \((H W),\) tailwater elevation \((T W),\) and gate opening \((h)\) is given by $$ Q=13.3 h \sqrt{H W-T W} $$ where \(Q\) is in \(\mathrm{m}^{3} / \mathrm{s}\) and \(H W, T W,\) and \(h\) are in meters. If the elevation of the bottom of the channel at the gate location is \(0.00 \mathrm{~m},\) the tailwater elevation is \(1.00 \mathrm{~m},\) and the flow in the channel is \(20 \mathrm{~m}^{3} / \mathrm{s}\), estimate the minimum gate opening such that the water surface elevation \(100 \mathrm{~m}\) upstream of the gate does not exceed \(2.20 \mathrm{~m}\). (Note: The depth of flow \(100 \mathrm{~m}\) upstream of the gate is not \(2.20 \mathrm{~m} .)\)

Short Answer

Expert verified
The minimum gate opening is obtained by substituting \(y\) into the equation for \(h\).

Step by step solution

01

Calculate Flow Depth 100m Upstream

First, we need to find the flow depth upstream using the given conditions. Since the channel is trapezoidal, the geometry will affect the Manning's equation. The width of the channel bottom \(b = 5 \, \text{m}\), and the side slopes are 2:1, so we will use Manning's equation:\[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \]where:- \(Q = 20 \, \text{m}^3/\text{s}\)- \(n = 0.018\)- \(S = 0.001\)Assume \(y\) is the depth of flow. The area \(A\) and the wetted perimeter \(P\) can be calculated as follows:- Trapezoidal area: \(A = b y + my^2\), where \(m = 2\) is the side slope.- Wetted perimeter: \(P = b + 2y\sqrt{1+m^2}\).Calculate the hydraulic radius \(R = \frac{A}{P}\).
02

Solve Manning's Equation

For the trapezoidal channel, using the equation from the previous step, solve for \(y\):- \(A = 5y + 2y^2\)- \(P = 5 + 2y\sqrt{5}\)Substitute these into the hydraulic radius \(R\):\[ R = \frac{5y + 2y^2}{5 + 2y\sqrt{5}} \]Now substitute \(A\), \(R\), and \(S\) back into Manning's equation and solve for \(y\) to ensure it does not exceed 2.2 m. Since this is a complex equation, use a numerical method or trial and error to find \(y\).
03

Use Gate Flow Formula

Once the appropriate \(y\) has been found, use the gate flow formula:\[ Q = 13.3 h \sqrt{HW - TW} \]Given that \(Q = 20 \, \text{m}^3/\text{s}\) and \(TW = 1.00 \, \text{m}\), solve for the minimum gate opening \(h\) needed to maintain the required flow condition upstream, ensuring \(HW = y + 1\) at elevation. Substitute the values into the equation.
04

Solve for Minimum Gate Opening

Substitute the calculated \(HW = y + 1\) from Step 2 into the gate flow equation to find \(h\):\[ 20 = 13.3 h \sqrt{(y + 1) - 1} \]Rearrange this to solve for \(h\):\[ h = \frac{20}{13.3 \sqrt{y}} \]Compute \(h\) accordingly. This calculation gives the minimum gate opening.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Manning's Equation
In open channel flow, Manning's equation is a valuable tool for engineers and hydrologists. It allows them to calculate the flow rate in a channel based on its geometry, material roughness, and slope. The equation is given by: \[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \] where:
  • \( Q \) is the discharge or flow rate in cubic meters per second (\( m^3/s \)).
  • \( n \) is the Manning's roughness coefficient, which depends on the channel's surface material.
  • \( A \) is the cross-sectional area of flow (\( m^2 \)).
  • \( R \) is the hydraulic radius (\( m \)).
  • \( S \) is the channel slope (dimensionless).
Understanding Manning's equation is crucial because it helps determine how efficiently water can flow through channels of different shapes and materials. By adjusting parameters such as the channel slope or roughness, engineers can predict the behavior of water in artificial channels, helping them design effective drainage solutions. It's important to note that finding the correct values for each parameter requires understanding the channel's physical characteristics, which can greatly influence the outcome of the flow rate.
Hydraulic Radius
The hydraulic radius \( R \) is an essential factor in calculating the flow in open channels using Manning's equation. It is defined as the ratio of the cross-sectional area of flow \( A \) to the wetted perimeter \( P \). Mathematically, it is expressed as: \[ R = \frac{A}{P} \] The hydraulic radius allows us to quantify how efficiently a channel section can convey water. The larger the hydraulic radius, the lesser the resistance to flow, resulting in a higher velocity. This concept is critical in optimizing channel design for maximum efficiency. - **Trapezoidal Channels**: In trapezoidal channels, \( A \) and \( P \) depend on the channel's base width and side slopes. For a trapezoidal channel with base width \( b \), side slope \( m \), and flow depth \( y \): - Cross-sectional area \( A \) is calculated as \( A = by + my^2 \). - Wetted perimeter \( P \) is \( P = b + 2y\sqrt{1+m^2} \). By substituting these into the hydraulic radius formula, engineers can determine \( R \), enabling precise use of Manning's equation. Understanding this helps ensure channels are designed to manage expected flow volumes, avoiding overflow or under-performance.
Trapezoidal Channel Geometry
Trapezoidal channels are a popular shape used in drainage systems due to their efficient flow characteristics and ease of construction. These channels have a flat bottom and sloped sides. Their design allows for stable flow conditions while being relatively simple to excavate. Some key aspects of trapezoidal channel geometry include:
  • **Base Width \( b \)**: This is the horizontal width at the bottom of the channel. In this exercise, it's 5 meters.
  • **Side Slopes \( m:1 \)**: The rate at which the channel walls incline upwards. A side slope of 2:1 means the channel wall rises 2 units for every 1 unit it extends horizontally.
  • **Water Depth \( y \)**: The vertical distance between the channel bottom and the water surface. This depth changes with flow conditions and must be calculated using equations like Manning's.
Channels with trapezoidal geometry are widely used in irrigation, drainage, and flood prevention. The blending of these elements ensures an efficient system that balances construction costs with performance. The process of including these elements properly is vital to the successful management of water resources.
Gate Flow Equation
The Gate Flow Equation is a formula used to calculate the flow that passes through a gate in a channel system. It is essential for managing how much water flows through controlled openings, such as sluice gates, in a canal or drainage channel. The equation provided is: \[ Q = 13.3 h \sqrt{HW - TW} \] Where:
  • \( Q \) is the flow rate through the gate in cubic meters per second (\( m^3/s \)).
  • \( h \) is the gate opening (m), which represents how high the gate is raised to allow water to pass through.
  • \( HW \) is the headwater elevation (m), indicating the height of the water surface upstream of the gate.
  • \( TW \) is the tailwater elevation (m), which is the water surface level downstream of the gate.
The Gate Flow Equation allows for precise management of water discharge, balancing upstream and downstream water levels efficiently. By adjusting the gate opening \( h \), engineers can control the flow rate \( Q \) to prevent flooding upstream while ensuring sufficient flow downstream. Understanding this relationship is vital for engineers to maintain effective water management in channels with gates.

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Most popular questions from this chapter

A trapezoidal channel is to be excavated at a site where permit restrictions require that the channel have a bottom width of \(6 \mathrm{~m}\), side slopes of \(1.5: 1(\mathrm{H}: \mathrm{V}),\) and a depth of flow of \(2 \mathrm{~m}\). If the soil material erodes when the average shear stress on the perimeter of the channel exceeds \(3 \mathrm{~Pa}\), determine the appropriate slope and corresponding flow capacity of the channel. Use the Darcy-Weisbach equation and assume that the excavated channel has an equivalent sand roughness of \(2 \mathrm{~mm}\).

Water flows in a 6 -m-wide rectangular channel that has a longitudinal slope of 0.0002 . The channel has an equivalent sand roughness of \(1.9 \mathrm{~mm} .\) Calculate the uniform flow depth in the channel when the flow rate is \(13 \mathrm{~m}^{3} / \mathrm{s}\). Use the Darcy-Weisbach equation.

Water flows at a depth of \(2.40 \mathrm{~m}\) in a trapezoidal concrete-lined section \(\left(k_{s}=1 \mathrm{~mm}\right)\) with a bottom width of \(4 \mathrm{~m}\) and side slopes of \(2: 1(\mathrm{H}: \mathrm{V})\). The longitudinal slope of the channel is \(0.0005,\) and the water temperature is \(20^{\circ} \mathrm{C}\). Assuming uniform-flow conditions, estimate the average velocity and flow rate in the channel. Use both the Darcy-Weisbach and Manning equations and compare your results.

Water flows at \(10 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(5.5 \mathrm{~m}\). The slope of the channel is \(0.15 \%,\) and the Manning roughness coefficient is 0.038 . Use the following methods to estimate the flow depth \(100 \mathrm{~m}\) upstream of a section where the flow depth is \(2.2 \mathrm{~m}:\) (a) the direct integration method and (b) the standard step method. Approximately how far upstream of this section would you expect to find uniform flow?

An open channel has a trapezoidal cross section with a bottom width of \(3 \mathrm{~m}\) and side slopes of \(2: 1(\mathrm{H}: \mathrm{V})\). If the depth of flow is \(2 \mathrm{~m}\) and the average velocity in the channel is \(1.2 \mathrm{~m} / \mathrm{s},\) calculate the discharge in the channel.

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