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Water flows in a horizontal trapezoidal channel at \(21 \mathrm{~m}^{3} / \mathrm{s}\), where the bottom width of the channel is \(2 \mathrm{~m}\), side slopes are \(1: 1,\) and the depth of flow is \(1 \mathrm{~m} .\) Calculate the downstream depth required for a hydraulic jump to form at this location. What would be the energy loss in the jump?

Short Answer

Expert verified
The downstream depth is approximately 3.73 m, and the energy loss in the jump is about 2.01 m.

Step by step solution

01

Calculate the Flow Velocity

First, calculate the flow velocity, which can be determined using the formula: \[ V = \frac{Q}{A} \]where \( Q = 21 \, \text{m}^3/\text{s} \) is the discharge and \( A \) is the cross-sectional area of the flow. The area for a trapezoidal section is given by: \[ A = (b + zy) \cdot y \]where \( b = 2 \, \text{m} \), \( z = 1 \), and \( y = 1 \, \text{m} \). Thus, we calculate:\[ A = (2 + 1 \times 1) \times 1 = 3 \, \text{m}^2 \]Then, the velocity is:\[ V = \frac{21}{3} = 7 \, \text{m/s} \]
02

Calculate Froude Number

The Froude number is calculated using the formula:\[ Fr = \frac{V}{\sqrt{g \cdot y}} \]where \( g = 9.81 \text{ m/s}^2 \) and \( y = 1 \, \text{m} \) is the depth. So:\[ Fr = \frac{7}{\sqrt{9.81 \times 1}} \approx 2.23 \]
03

Determine Initial and Final Depths

Hydraulic jump occurs when \( Fr > 1 \). We solve for the downstream depth \( y_2 \) for the hydraulic jump using the equation:\[ y_2 = \frac{y_1}{2} \left(\sqrt{1 + 8Fr_1^2} - 1\right) \]where \( y_1 = 1 \, \text{m} \). Plug in the Froude number:\[ y_2 = \frac{1}{2} \left(\sqrt{1 + 8 \times 2.23^2} - 1\right) \approx 3.73 \, \text{m} \]
04

Calculate the Energy Loss in the Jump

The energy loss \( \Delta E \) during the hydraulic jump can be calculated as:\[ \Delta E = \frac{(y_2 - y_1)^3}{4y_1y_2} \]Using \( y_1 = 1 \, \text{m} \) and \( y_2 = 3.73 \, \text{m} \):\[ \Delta E = \frac{(3.73 - 1)^3}{4 \times 1 \times 3.73} \approx 2.01 \, \text{m} \]

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Key Concepts

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

Flow Velocity Calculation
To understand how water moves in a channel, we first need to calculate the flow velocity. Think of flow velocity as the speed at which water travels through the channel. To find this speed, we use the formula:
  • Velocity, \( V = \frac{Q}{A} \)
where \( Q \) is the discharge, or the volume of water flowing per unit time, and \( A \) is the cross-sectional area of the flow. For a trapezoidal channel like the one given:
  • The base width \( b \) is 2 meters.
  • The side slopes have a ratio of 1:1.
  • The depth of flow \( y \) is 1 meter.
By substituting these values, the area \( A \) becomes \( A = (b + z \cdot y) \cdot y \), where \( z \) is the side slope factor. Hence, \( A \) becomes 3 \( \text{m}^2 \). Finally, with a discharge \( Q \) of 21 \(\text{m}^3/\text{s}\), the velocity \( V \) calculates to 7 \( \text{m/s} \). This simple calculation sets the stage for understanding how swiftly water moves through the channel.
Froude Number
The Froude number is a critical parameter in fluid dynamics that helps predict the behavior of flow in open channels. It tells us whether the flow is tranquil or turbulent. The flow's Froude number is given by:
  • \( Fr = \frac{V}{\sqrt{g \cdot y}} \)
where \( V \) is the flow velocity, \( g \) is the acceleration due to gravity, and \( y \) is the flow depth. This number is dimensionless.For our trapezoidal channel, with a velocity of 7 \( \text{m/s} \) and a depth \( y \) of 1 meter:
  • \( g \) (acceleration due to gravity) is 9.81 \( \text{m/s}^2 \).
This gives a Froude number \( Fr \approx 2.23 \), meaning the flow is supercritical since \( Fr > 1 \). Supercritical flows tend to be rapid and can cause phenomena like hydraulic jumps.
Energy Loss Calculation
When a hydraulic jump occurs, energy is lost due to turbulence and mixing in the water. To understand how much energy is lost when water transitions from a fast-moving flow to a slower flow, the energy loss \( \Delta E \) can be computed.
  • The formula is: \( \Delta E = \frac{(y_2 - y_1)^3}{4y_1y_2} \)
  • \( y_1 \) is the initial (upstream) depth.
  • \( y_2 \) is the final (downstream) depth after the hydraulic jump.
For our situation, where \( y_1 = 1 \text{ m} \) and \( y_2 = 3.73 \text{ m} \), the energy loss is approximately 2.01 meters. This loss showcases the inherent inefficiency of the flow's transition, critical in designing channels and predicting impacts of water flow changes.
Trapezoidal Channel Flow
When dealing with water flows in engineering, trapezoidal channels are a common design because of their structural and economic advantages. Understanding the characteristics of a trapezoidal channel helps predict how water will behave. A trapezoidal channel has:
  • A bottom width \( b \) as the narrow part at the base.
  • Side slopes that angle outward, increasing the width at the top.
  • Depth \( y \) which influences both cross-sectional area and flow characteristics.
For instance, in our described channel:
  • The bottom width is 2 meters.
  • The side slopes are given by a ratio of 1:1, indicating a 45-degree slope from the base.
  • The depth is 1 meter initially.
These features contribute significantly to the channel's flow dynamics, affecting velocity and the potential for hydraulic jumps. Understanding each aspect, from width to slope, is crucial for engineers to optimize the design and function of water channels.

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Most popular questions from this chapter

Use the Darcy-Weisbach equation to show that the head loss per unit length, \(S\), between any two sections in an open channel can be estimated by the relation $$ S=\frac{\bar{f}}{4 \bar{R}} \frac{\bar{V}^{2}}{2 g} $$ where \(\bar{f}, \bar{R},\) and \(\bar{V}\) are the average friction factor, hydraulic radius, and flow velocity, respectively, between the upstream and downstream sections.

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.

A rectangular channel has a width of \(30 \mathrm{~m}\), a longitudinal slope of \(0.5 \%,\) and an estimated Manning's \(n\) of \(0.025 .\) The flow rate in the channel is \(100 \mathrm{~m}^{3} / \mathrm{s}\) at a particular section where the depth of flow is \(3.000 \mathrm{~m}\). Temporary construction requires that the channel be contracted to a width of \(20 \mathrm{~m}\) over a distance of \(40 \mathrm{~m}\) and then returned to its original width of \(30 \mathrm{~m}\) over a distance of \(40 \mathrm{~m}\). All sections are rectangular. Determine the depths of flow in the contracted and downstream sections when (a) all energy losses between sections are neglected and (b) friction, contraction, and expansion losses are all taken into account. (Note: To simplify the computations, assume that the friction slope is the same at all three sections.) Based on your results, evaluate the impact of accounting for energy losses on the estimated difference between the water stages at the upstream and downstream sections.

Show that the critical step height required to choke the flow in a rectangular open channel is given by $$ \frac{\Delta z_{\mathrm{c}}}{y_{1}}=1+\frac{\mathrm{Fr}_{1}^{2}}{2}-\frac{3}{2} \mathrm{Fr}_{1}^{\frac{2}{3}} $$ where \(\Delta z_{\mathrm{c}}\) is the critical step height, \(y_{1}\) is the flow depth upstream of the step, and \(\mathrm{Fr}_{1}\) is the Froude number upstream of the step. Use this equation to verify your answer to Problem 9.31 .

Water flows at \(11 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(5 \mathrm{~m}\). The slope of the channel is \(0.1 \%,\) and the Manning roughness coefficient is equal to \(0.035 .\) If the depth of flow at a selected section is \(2 \mathrm{~m},\) calculate the upstream depths at \(20-\mathrm{m}\) intervals along the channel until the depth of flow is within \(5 \%\) of the uniform flow depth.

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