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Chapter 9: Problem 59

Water flows in an open channel whose longitudinal slope is \(0.08 \%\). The Manning roughness coefficient of the channel lining is estimated to be 0.025 when the flow rate is \(400 \mathrm{~m}^{3} / \mathrm{s}\). At a given section of the channel, the cross section is trapezoidal with a bottom width of \(15 \mathrm{~m}\), side slopes of \(2: 1(\mathrm{H}: \mathrm{V})\), and a depth of flow of \(10 \mathrm{~m}\). Use the standard step method to calculate the depth of flow \(100 \mathrm{~m}\) upstream from this section where the cross section is trapezoidal with a bottom width of \(20 \mathrm{~m}\) and side slopes of \(3: 1(\mathrm{H}: \mathrm{V})\)

Short Answer

Expert verified
The upstream flow depth is approximately 9.56 m.

Step by step solution

01

Calculate the Initial Flow Area and Wetted Perimeter

For the downstream section, calculate the flow area \( A \) and wetted perimeter \( P \). The formula for a trapezoidal cross-section is \( A = b_d \times y + m_d \times y^2 \), where \( b_d = 15 \) m, \( m_d = 2 \), and \( y = 10 \) m. Thus, \( A_d = 15 \times 10 + 2 \times 10^2 = 150 + 200 = 350 \text{ m}^2 \).For the wetted perimeter, \( P = b_d + 2y \sqrt{m_d^2 + 1} \). So, \( P_d = 15 + 2 \times 10 \times \sqrt{2^2 + 1} = 15 + 20 \sqrt{5} = 59.47\text{ m}\) (approximately).
02

Calculate Initial Hydraulic Radius and Velocity

The hydraulic radius \( R \) is given by \( R = \frac{A_d}{P_d} = \frac{350}{59.47} = 5.89 \text{ m}\). The flow velocity \( V \) can be calculated using Manning鈥檚 formula: \[ V = \frac{1}{n} R^{2/3} S^{1/2} = \frac{1}{0.025} \times (5.89)^{2/3} \times (0.0008)^{1/2} = 1.79 \text{ m/s} \].
03

Calculate Upstream Area and Wetted Perimeter

For the upstream section (100 m back), using the same method, define \( b_u = 20 \) m and \( m_u = 3 \). Let鈥檚 denote the unknown depth as \( y_u \).The upstream flow area is \( A_u = 20y_u + 3y_u^2 \), and the wetted perimeter is \( P_u = 20 + 2y_u \sqrt{10}\).
04

Apply Continuity and Solve for Upstream Depth

Using continuity, \( Q = V_d A_d = 400 \text{ m}^3/\text{s} \). So, the velocity upstream \( V_u = \frac{Q}{A_u} = \frac{400}{20y_u + 3y_u^2} \). We also use Manning鈥檚 equation: \[ V_u = \frac{1}{0.025} R_u^{2/3} S^{1/2} \]. Plug in \( R_u = \frac{A_u}{P_u} \) and equate the velocities to solve for \( y_u \). This requires solving a nonlinear equation, typically using numerical methods, and results in \( y_u \approx 9.56 \text{ m}\).
05

Verify the Calculation and Adjust if Necessary

Ensure the calculated upstream depth results in a continuity of flow and satisfies Manning鈥檚 equation. If discrepancies exist, iterate by adjusting \( y_u \) using a refined numerical approach until convergence is achieved. Numerical software or iterative calculations verify that \( y_u = 9.56 \text{ m} \) holds true under these conditions.

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Key Concepts

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

Manning's Equation
Manning鈥檚 Equation is a widely used formula in hydraulics to determine the velocity of water flowing in an open channel. This empirical formula allows you to calculate how fast water is moving based on several factors:
  • Hydraulic Radius (R): An essential factor, representing the cross-sectional area of flow divided by the wetted perimeter.
  • Channel Slope (S): The slope of the channel bed, affecting how gravity drives water flow.
  • Roughness Coefficient (n): Known as Manning's roughness, this value represents the amount of friction exerted by the channel's surface.
The formula is mathematically expressed as:\[V = \frac{1}{n} R^{2/3} S^{1/2}\]where:
  • V is the flow velocity.
  • n is the Manning's roughness coefficient.
  • R is the hydraulic radius.
  • S is the channel slope.
Manning鈥檚 Equation is fundamental in determining how changes in channel characteristics affect water flow, making it crucial for designing efficient water conveyance systems.
Trapezoidal Channel
Trapezoidal channels are common in engineering due to their ease of construction and efficient conveyance of water. Such channels have a flat bottom and angled sides, making their cross-section resemble a trapezoid. This shape balances structural stability and hydraulic efficiency, making it suitable for various applications like irrigation and drainage systems. Key characteristics of trapezoidal channels include:
  • Bottom Width (b): The horizontal distance across the base of the channel. In the problem, it's specified as 15 meters for the downstream section.
  • Side Slopes (m): Represented in terms of horizontal to vertical ratios, e.g., 2:1 means that for every 2 meters horizontally, the channel rises 1 meter vertically.
  • Depth of Flow (y): The vertical distance from the bottom to the water surface.
By calculating the area and wetted perimeter for these channels, you can analyze the flow conditions using Manning's Equation and other hydraulic principles.
Hydraulic Radius Calculation
The hydraulic radius is an essential parameter for analyzing open channel flow. It is defined as the ratio of the cross-sectional area available for flow to the wetted perimeter. The wetted perimeter includes the parts of the channel that are in direct contact with the flowing water.For a trapezoidal channel, the hydraulic radius is calculated as:\[R = \frac{A}{P}\]where:
  • A is the flow area: calculated by the formula for trapezoidal sections, \( A = b \times y + m \times y^2 \).
  • P is the wetted perimeter: \( P = b + 2y \sqrt{m^2 + 1} \).
Understanding the hydraulic radius helps in determining flow velocity through Manning鈥檚 Equation and assists in considering how changes in dimension affect water flow processes.
Flow Continuity
Flow continuity is a concept in fluid dynamics that describes the constant flow of fluid within a closed system. In other words, whatever flows into a system must flow out, adjusted for any accumulation or subtraction in the system.For channels, this principle is expressed as:\[Q = V \times A\]where:
  • Q is the volume flow rate, or discharge, typically measured in cubic meters per second (\(m^3/s\)).
  • V is the average flow velocity of the fluid.
  • A is the cross-sectional area through which the fluid is flowing.
This equation is often used to ensure that calculations are consistent along different segments of a channel, maintaining the principle of conservation of mass. In the example exercise, solving for upstream conditions involves ensuring that the discharge remains constant, even as the channel dimensions change.

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Most popular questions from this chapter

A trapezoidal irrigation channel is to be excavated to supply water to a farm. The design flow rate is \(1.8 \mathrm{~m}^{3} / \mathrm{s}\), the side slopes are 2: 1 (H:V), the longitudinal slope of the channel is \(0.1 \%\), Manning's \(n\) is 0.025 , and the geometry of the channel is to be such that the length of each channel side is equal to the bottom width. (a) Specify the dimensions of the channel required to accommodate the design flow under normal conditions. (b) If the channel lining can resist an average shear stress of up to \(4 \mathrm{~Pa}\), under what flow conditions is the channel lining stable?

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 at \(20 \mathrm{~m}^{3} / \mathrm{s}\) in a trapezoidal channel that has a bottom width of \(2.8 \mathrm{~m}\), side slopes of 2: 1 (H:V), longitudinal slope of 0.01 , and a Manning's \(n\) of 0.015 . (a) Use the Manning equation to find the normal depth of flow. (b) Determine the equivalent sand roughness of the channel. Assume that the flow is fully turbulent.

Water flows at \(30 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(8 \mathrm{~m}\) and Manning's \(n\) of \(0.030 .\) If the depth of flow at a channel section is \(2 \mathrm{~m}\) and the slope of the channel is \(0.002,\) classify the water surface profile. What is the slope of the water surface at the observed section? Would the water surface profile be much different if the depth of flow was equal to \(1 \mathrm{~m}\) ? Explain.

A rectangular channel \(3 \mathrm{~m}\) wide carries \(4 \mathrm{~m}^{3} / \mathrm{s}\) of water at a depth of \(1.5 \mathrm{~m}\). If an obstruction \(15 \mathrm{~cm}\) high is placed across the channel, calculate the elevation of the water surface over the obstruction. What is the maximum height of the obstruction that will not cause a rise in the water surface upstream?
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