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Chapter 9: Problem 44

Water discharges from a large storage reservoir via a trapezoidal channel of bottom width \(3.00 \mathrm{~m}\), side slopes 3: 1 (H:V), and longitudinal slope of \(0.5 \%\). Manning's \(n\) in the channel is estimated as \(0.025 .\) Estimate the discharge from the reservoir when the depth of the reservoir at the discharge location is \(2.00 \mathrm{~m}\).

Short Answer

Expert verified
The discharge is approximately 56.89 m鲁/s.

Step by step solution

01

Understanding the Geometry of the Channel

The channel has a trapezoidal cross-section with a bottom width \(b = 3.00\, \text{m}\) and side slopes given as 3 horizontal to 1 vertical (3:1). The depth of water \(d\) at the discharge location is \(2.00\, \text{m}\). The top width \(T\) of the channel can be calculated as \[T = b + 2 \cdot 3 \cdot d = 3 + 2 \cdot 3 \cdot 2 = 15 \, \text{m}.\]
02

Calculate the Area of the Channel

The cross-sectional area \(A\) of a trapezoidal channel is given by \[ A = \left( \frac{b + T}{2} \right) \cdot d. \] Substituting the values we have \[ A = \left( \frac{3 + 15}{2} \right) \cdot 2 = 18 \, \, \text{m}^2.\]
03

Determine the Wetted Perimeter

The wetted perimeter \(P\) in a trapezoidal channel is given by \[ P = b + 2 \sqrt{d^2 + (3d)^2}. \] Substituting the values, we find \[ P = 3 + 2 \sqrt{2^2 + (6)^2} = 3 + 2 \cdot \sqrt{4 + 36} = 3 + 12 = 15 \, \text{m}.\]
04

Calculate the Hydraulic Radius

The hydraulic radius \(R\) is defined as the cross-sectional area \(A\) divided by the wetted perimeter \(P\). Therefore, \[ R = \frac{A}{P} = \frac{18}{15} = 1.2 \, \text{m}.\]
05

Apply the Manning's Equation

Manning's equation for discharge \(Q\) is given by \[ Q = \frac{1}{n} A R^{2/3} S^{1/2}, \] where \(n = 0.025\) and \(S = 0.005\) (0.5% slope as a decimal). Substituting these into the equation, we get \[ Q = \frac{1}{0.025} \cdot 18 \cdot (1.2)^{2/3} \cdot (0.005)^{1/2}. \]
06

Solve for the Discharge

First, compute \( R^{2/3} \approx (1.2)^{2/3} \approx 1.115.\) Then compute \( S^{1/2} \approx (0.005)^{1/2} \approx 0.07071.\) Plug these into the equation: \[ Q = 40 \cdot 18 \cdot 1.115 \cdot 0.07071. \] Thus, \[ Q \approx 40 \cdot 18 \cdot 1.115 \cdot 0.07071 \approx 56.891. \] Therefore, the discharge is approximately \(56.89 \, \text{m}^3/\text{s}.\)

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Key Concepts

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

Trapezoidal Channel
A trapezoidal channel is a waterway with a cross-section shaped like a trapezoid, making it an efficient design for transporting water. This shape is widely used due to its ability to handle larger water flows with minimal construction materials. The channel's bottom width, side slopes, and depth define its geometry. Here, the channel's bottom width is 3 meters, and the sides slope at a 3:1 ratio (horizontal:vertical). This means that for every 1 meter of vertical rise, the side width increases by 3 meters. The depth of water at the discharge location is a crucial factor, here stated as 2 meters. A simple formula calculates the top width of the channel:
  • The bottom width is 3 meters.
  • Add twice the side slope length multiplied by water depth to obtain the top width.
  • Thus, the top width, \( T = 3 + 6 \cdot 2 = 15 \, \text{m} \)
These measurements are essential for further calculations related to water flow.
Hydraulic Radius
The hydraulic radius is a measure used in open channel flow, such as rivers, canals, or trapezoidal channels. It is vital in determining how easily water can flow through a channel, directly influencing the channel's discharge potential through Manning's equation. The hydraulic radius \( R \) is defined as the channel's cross-sectional area \( A \) divided by its wetted perimeter \( P \). For our trapezoidal channel example, the area is calculated as \( 18 \, \text{m}^2 \), and the wetted perimeter is \( 15 \, \text{m} \). These values result in the hydraulic radius:
  • \( R = \frac{A}{P} \)
  • Substitute the known values: \( R = \frac{18}{15} = 1.2 \, \text{m} \)
This value provides insight into how efficiently the channel can convey water based on its shape and size.
Discharge Calculation
Calculating the discharge of water in a channel involves understanding the volume of water flowing per unit time. Manning's equation is a widely used method to estimate this rate in an open channel. The formula is influenced by several factors, including the hydraulic radius, cross-sectional area, channel slope, and Manning's roughness coefficient \( n \). The discharge \( Q \) is calculated using the formula:
  • \( Q = \frac{1}{n} A R^{2/3} S^{1/2} \)
  • \( A = 18 \, \text{m}^2 \), \( R = 1.2 \, \text{m} \), \( S = 0.005 \) (slope as decimal), \( n = 0.025 \)
  • Plug these values into the formula to find \( Q \, \approx 56.89 \, \text{m}^3/\text{s} \)
This number indicates how much water exits the reservoir each second through the channel.
Wetted Perimeter
The wetted perimeter of a channel is the length of the boundary in direct contact with the water. It plays a crucial role in determining the hydraulic radius and subsequently affects discharge calculations. For a trapezoidal channel, the wetted perimeter \( P \) can be found using the geometry of the channel:
  • Base width \( + \) length of the slopes.
  • Given our channel measurements: \( P = 3 + 2 \sqrt{2^2 + (6)^2} \).
  • Simplifying gives \( 15 \, \text{m} \).
Understanding the wetted perimeter helps in computing other essential hydrodynamic properties of the channel.

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Most popular questions from this chapter

Water flows at a depth of \(4.00 \mathrm{~m}\) in a trapezoidal concrete-lined channel with a bottom width of \(4 \mathrm{~m}\) and side slopes of \(3: 1(\mathrm{H}: \mathrm{V})\). The longitudinal slope of the channel is 0.0001 , and the water temperature is \(20^{\circ} \mathrm{C}\). Assess the validity of using the Manning equation, assuming that \(n=0.013\).

A flume with a triangular cross section and side slopes of \(2: 1(\mathrm{H}: \mathrm{V})\) contains water flowing at \(0.45 \mathrm{~m}^{3} / \mathrm{s}\) at a depth of \(20 \mathrm{~cm}\). Verify that the flow is supercritical and calculate the conjugate depth.

It has been shown that in fully turbulent flow, Manning's \(n\) can be related to the height, \(d\), of the roughness projections by the relation $$ n=0.039 d^{\frac{1}{6}} $$ where \(d\) is in meters. If the estimated roughness height in a channel is \(30 \mathrm{~mm}\), determine the percentage error in \(n\) resulting from a \(70 \%\) error in estimating \(d\).

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.

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.
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