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Chapter 9: Problem 2

Water flows at \(10 \mathrm{~m}^{3} / \mathrm{s}\) through a rectangular channel \(4 \mathrm{~m}\) wide and \(3 \mathrm{~m}\) deep. If the flow velocity is \(1.4 \mathrm{~m} / \mathrm{s},\) calculate the depth of flow in the channel. If this channel expands (downstream) to a width of \(6 \mathrm{~m}\) and the depth of flow decreases by \(0.7 \mathrm{~m}\) from the upstream depth, what is the flow velocity in the expanded section?

Short Answer

Expert verified
The new flow velocity in the expanded section is approximately \(1.535 \mathrm{~m/s}\).

Step by step solution

01

Calculate the Cross-Sectional Area

In a rectangular channel, the cross-sectional area, \( A \), is the product of width \( b \) and depth \( y \). Initially, \( b = 4 \mathrm{~m} \) and \( y = 3 \mathrm{~m} \), so \( A = 4 \times 3 = 12 \mathrm{~m}^2 \).
02

Verify Initial Flow with Given Parameters

The flow volume is given as \( Q = 10 \mathrm{~m}^3/\mathrm{s} \). Using the flow equation \( Q = A \cdot v \), where \( v \) is the flow velocity, verify: \( Q = 12 \cdot 1.4 = 16.8 \mathrm{~m}^3/\mathrm{s} \). This indicates the upstream depth must allow the given flow, and the equation initially doesn't hold, suggesting a calculation or parameter reconfirm.
03

Calculate Depth of Flow in the Original Channel

Reassess using the given \( Q = 10 \mathrm{~m}^3/\mathrm{s} \) with equation \( 10 = b \cdot y \cdot v \). Solving \( 10 = 4y \cdot 1.4 \), we get \( y = 1.7857 \mathrm{~m} \).
04

Calculate New Depth After Expansion

The channel expands to \( b = 6 \mathrm{~m} \) and the depth decreases by \( 0.7 \mathrm{~m} \). The new depth = previous depth - \( 0.7 \mathrm{~m} = 1.7857 - 0.7 = 1.0857 \mathrm{~m} \).
05

Calculate New Flow Velocity in Expanded Section

Use the equation \( Q = A \cdot v \). The new cross-sectional area is \( A = 6 \times 1.0857 = 6.5142 \mathrm{~m}^2 \). Solving for \( v \), \( 10 = 6.5142 \cdot v \) gives \( v = \frac{10}{6.5142} \approx 1.535 \mathrm{~m/s} \).

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Key Concepts

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

Flow Velocity
Flow velocity is the speed at which the fluid moves through a channel. It's an important concept in open channel flow as it determines how quickly water travels from one point to another. Flow velocity is influenced by factors such as the channel鈥檚 size, shape, and slope. A key formula used to find flow velocity is based on the flow rate and cross-sectional area:
  • Flow Rate ( \( Q \) ) = Cross-Sectional Area ( \( A \) ) 脳 Flow Velocity ( \( v \) )
This means that velocity can change if there's a change in the channel's width, depth, or the flow rate. In the exercise, the flow velocity is initially given as 1.4 m/s in a rectangular channel. However, to confirm this, you must ensure that the given flow rate matches this velocity when combined with the proper cross-sectional area.
Cross-Sectional Area
The cross-sectional area is the width of the channel multiplied by its depth. This area determines how much volume flows through a channel at any point in time. In rectangular channels, the calculation is straightforward:
  • Cross-Sectional Area ( \( A \) ) = Width ( \( b \) ) 脳 Depth ( \( y \) )
For example, with an initial width of 4 m and a depth of 3 m, the cross-sectional area becomes 12 m虏. If the dimensions change, such as in the downstream section where width and depth are altered, the cross-sectional area will also change, affecting the flow characteristics.
Rectangular Channel
A rectangular channel is one where the cross-section resembles a rectangle. These channels are common in engineering due to their simplicity and ease of construction. They are defined by their width and depth, which directly affect how they handle flow.
  • Width affects how expansive the channel is, and depth affects the vertical space the water occupies.
In our problem, the channel initially has a rectangular shape with dimensions 4 m wide and 3 m deep, which impacts the initial calculations of flow depth and velocity. When it later expands, these changes are significant for recalculating flow dynamics.
Flow Rate Calculation
Flow rate is the volume of water that passes through a channel per unit time, often measured in cubic meters per second (\( ext{m}^3/ ext{s} \)). It's crucial for understanding how much water a channel can handle and is determined by:
  • Flow Rate ( \( Q \) ) = Cross-Sectional Area ( \( A \) ) 脳 Flow Velocity ( \( v \) ).
In the given exercise, the initial flow rate is assigned as 10 \( ext{m}^3/ ext{s} \). This is cross-verified with the given velocity and dimensions to ensure calculations are accurate. This formula can help diagnose if issues arise, such as disagreements between expected and calculated values, ensuring an accurate description of how the water moves through the channel. Understanding and properly applying this equation is fundamental for solving problems related to open channel flow.

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Most popular questions from this chapter

Under design conditions, flow exits at the bottom of a 10 -m-wide spillway at a rate of \(220 \mathrm{~m}^{3} / \mathrm{s}\), at a stage of \(13.5 \mathrm{~m}\), and at a depth of \(1 \mathrm{~m}\). Downstream of the spillway is a river at a stage of \(21.5 \mathrm{~m}\). The channel connecting the spillway exit to the river is to be horizontal and rectangular with a width of \(10 \mathrm{~m}\). This connecting channel (called a stilling basin) is to be constructed such that Manning's \(n\) is 0.01 and is to be designed such that any hydraulic jump would occur at the midpoint of the stilling basin. As a designer, what length of stilling basin would you specify?

Stages are measured by two recording gauges \(100 \mathrm{~m}\) apart along a constructed water supply channel. The channel has a bottom width of \(5 \mathrm{~m}\) and side slopes of \(3: 1(\mathrm{H}: \mathrm{V})\) The bottom elevations of the channel at the upstream and downstream gauge locations are \(24.01 \mathrm{~m}\) and \(23.99 \mathrm{~m}\), respectively. At a particular instance, the upstream and downstream stages are \(25.01 \mathrm{~m}\) and \(24.95 \mathrm{~m}\), respectively, and the flow is estimated as \(15 \pm 2 \mathrm{~m}^{3} / \mathrm{s}\). (a) Derive an expression for Manning's \(n\) as a function of the estimated flow rate. (b) Estimate Manning's \(n\) and the roughness height in the channel between the two measurement stations. (c) Quantitatively assess the sensitivity of the flow rate to the channel roughness.

A lined rectangular concrete drainage channel is \(10.0 \mathrm{~m}\) wide and carries a flow of \(8 \mathrm{~m}^{3} / \mathrm{s}\). To pass the flow under a roadway, the channel is contracted to a width of \(6 \mathrm{~m}\). Under design conditions, the depth of flow just upstream of the contraction is \(1.00 \mathrm{~m},\) and the contraction takes place over a distance of \(7 \mathrm{~m}\). (a) If the energy loss in the contraction is equal to \(V_{1}^{2} / 2 g,\) where \(V_{1}\) is the average velocity upstream of the contraction, what is the depth of flow in the constriction? (b) Does consideration of energy losses have a significant effect on the depth of flow in the constriction? Why or why not? (c) If the width of the constriction is reduced to \(4.50 \mathrm{~m}\) and a flow of \(8 \mathrm{~m}^{3} / \mathrm{s}\) is maintained, determine the depth of flow within the constriction (include energy losses). (d) If reducing the width of the constriction to \(4.50 \mathrm{~m}\) influences the upstream depth, determine the new upstream depth.

Water flows in a trapezoidal channel where the bottom width is \(6 \mathrm{~m}\) and side slopes are 2: 1 (H: V). The channel lining has an estimated Manning's \(n\) of \(0.045,\) and the slope of the channel is \(1.5 \%\). When the flow rate is \(80 \mathrm{~m}^{3} / \mathrm{s}\), the depth of flow at a gauging station is \(5 \mathrm{~m}\). Classify the water surface profile, state whether the depth increases or decreases in the downstream direction, and calculate the slope of the water surface at the gauging station. On the basis of this water- surface slope, estimate the depths of flow \(100 \mathrm{~m}\) downstream and \(100 \mathrm{~m}\) upstream of the gauging station.

Show that Manning's \(n\) can be expressed in terms of the Darcy friction factor, \(f,\) by the following relation: $$ n=\frac{f^{\frac{1}{2}} R^{\frac{1}{6}}}{8.86} $$ where \(R\) is the hydraulic radius of the flow. Does this relationship conclusively show that \(n\) is a function of the flow depth? Explain.
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