An open channel of trapezoidal section, \(2.5 \mathrm{~m}\) wide at the base and having sides inclined at \(60^{\circ}\) to the horizontal, has a bed slope of 1 in \(500 .\) It is found that when the rate of flow, is \(1.24 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the depth of water in the channel is \(350 \mathrm{~mm}\). Assuming the validity of Manning's formula, calculate the rate of flow when the depth is \(500 \mathrm{~mm}\).

Short Answer

Expert verified
Use Manning's formula with new depth (0.5 m) values to calculate the increased flow rate.

Step by step solution

01

- Identify the Known Parameters

The base width of the trapezoidal channel is given as 2.5 m. The side slopes are inclined at 60°. The bed slope (S) is 1/500. For the given flow rate, when the depth (d) is 350 mm (0.35 m), the flow rate (Q) is 1.24 m³/s.
02

- Calculation of Cross-Sectional Area (A) for Depth 350 mm

The area (A) is then: \[ A = \frac{1}{2} \times (\text{Base width} + \text{Top width}) \times \text{Depth} \] Substituting in the values: \[ A = \frac{1}{2} \times (2.5 + 2.5 + 2 \times 0.35 \times \tan(60^\text{°})) \times 0.35 \]
03

- Calculation of Wetted Perimeter (P) for Depth 350 mm

The formula for the wetted perimeter is: \[ P = \text{base width} + 2 \times \text{side length} \] The side length is given by: \[ \text{Side length} = \text{depth} / \text{cos(side slope angle)} \] Substituting the value: \[ \text{Side length} = 0.35 / \text{cos}(60^\text{°}) = 0.35 / 0.5 = 0.7 \] Thus, the wetted perimeter (P) is: \[ P = 2.5 + 2 \times 0.7 = 3.9 \text{ m} \]
04

- Calculation of Hydraulic Radius (R) for Depth 350 mm

The hydraulic radius (R) is given by: \[ R = \frac{A}{P} \] Using the values from the previous steps.
05

- Apply Manning's Formula to find n

Manning's formula: \[ Q = \frac{1}{n} A R^{2/3} S^{1/2} \] Rearrange to solve for n: \[ n = \frac{A R^{2/3} S^{1/2}}{Q} \] Using the values for depth 0.35 m, find n.
06

- Repeat Calculations for Depth 500 mm

Updating the depth d to 0.5 m, repeat Steps 2-4 to find new A, P, and R values.
07

- Calculate New Flow Rate (Q)

Using the new values for depth 0.5 m and previously computed n, apply Manning's formula to find the new Q.

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Key Concepts

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

Manning's formula
Manning's formula is a crucial tool in open channel flow calculations. It is used to determine the velocity or flow rate of water through a channel based on the channel's physical characteristics and surface roughness. The formula is: `Q = (1/n) * A * R^(2/3) * S^(1/2)` Here:
  • `Q` is the flow rate (m3/s)
  • `A` is the cross-sectional area of flow (m2)
  • `R` is the hydraulic radius (m)
  • `S` is the slope of the channel bed (dimensionless)
  • `n` is the Manning's roughness coefficient
Manning's formula is widely used due to its simplicity and effectiveness, especially for natural channels where flow conditions can be irregular.
Hydraulic radius
The hydraulic radius is an important concept in open channel flow as it relates to the efficiency of the channel in conveying water. It is defined as: `R = A / P` Where:
  • `R` is the hydraulic radius
  • `A` is the cross-sectional area of the flow
  • `P` is the wetted perimeter
The hydraulic radius helps to account for the shape of the channel by considering both the area of flow and the perimeter in contact with water. A higher hydraulic radius generally means a more efficient channel. For a trapezoidal channel, calculating the hydraulic radius is a bit more complex due to the shape, but it follows the same general principle.
Trapezoidal channel
A trapezoidal channel is a type of open channel that has a flat bottom and sides that slope outward. This shape is common in man-made channels because it is relatively easy to construct and maintain. The key parameters for a trapezoidal channel include:
  • Base width: The width of the bottom section.
  • Side slopes: The angle at which the sides are inclined from the base.

Calculating Areas and Perimeters


To calculate the cross-sectional area of a trapezoidal channel, the formula used is: `A = 1/2 (base width + top width) * depth` The top width can be found by considering the depth and side slopes. The wetted perimeter (P), which is the contact length between water and the channel, is calculated as: `P = base width + 2 * side length` Side length itself can be found using the trigonometric relationship involving the depth and the cosine of the slope angle.
Flow rate calculation
Flow rate calculation in open channel flow involves using Manning's formula and requires accurate determination of parameters such as area, hydraulic radius, and slope. Here’s a step-by-step approach:

Determine Cross-Sectional Area (A)


Find the cross-sectional area by considering the channel shape and depth
  • For a trapezoidal channel, use `A = 1/2 (base width + top width) * depth`

Compute Wetted Perimeter (P)


Calculate the wetted perimeter with:
  • `P = base width + 2 * side length`

Calculate Hydraulic Radius (R)


You can find the hydraulic radius using `R = A / P`
  • Ensure you have the correct hydraulic radius as it impacts the final flow rate.

Apply Manning's Formula


Finally, use the Manning's formula to compute `Q` (flow rate):
  • `Q = (1/n) * A * R^(2/3) * S^(1/2)`

This requires the Manning's roughness coefficient, which is often based on the channel material and condition.

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Most popular questions from this chapter

A long horizontal channel has a base width of \(1 \mathrm{~m}\) and sides at \(60^{\circ}\) to the horizontal. When the flow in the channel is \(0.85 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) the depth is \(500 \mathrm{~mm} .\) The discharge is suddenly reduced so that a surge wave of amplitude \(150 \mathrm{~mm}\) is propagated upstream. Determine the new rate of flow, the velocity of the wave and the Froude numbers before and after the wave.

In a long rectangular channel \(3 \mathrm{~m}\) wide the specific energy is \(1.8 \mathrm{~m}\) and the rate of flow is \(12 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). Calculate two possible depths of flow and the corresponding Froude numbers. If Manning's \(n=0.014\) what is the critical slope for this discharge?

A circular conduit is to satisfy the following conditions: capacity when flowing full, \(0.13 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\); velocity when the depth is one quarter the diameter, not less than \(0.6 \mathrm{~m} \cdot \mathrm{s}^{-1}\). Assuming uniform flow, determine the diameter and the slope if Chézy's coefficient is \(58 \mathrm{~m}^{1 / 2} \cdot \mathrm{s}^{-1}\)

Ocean waves, with a period of \(8 \mathrm{~s}\) and amplitude \(0.6 \mathrm{~m}\) in deep water, approach the shore in the normal direction. A device \(80 \mathrm{~m}\) long for extracting power from the waves is installed parallel to the shore in water \(5 \mathrm{~m}\) deep. If there is negligible dissipation of energy before the waves reach the device and its efficiency is \(50 \%\), what power is produced? What is the amplitude of the waves immediately before this position? (Density of sea-water \(\left.=1025 \mathrm{~kg} \cdot \mathrm{m}^{-3} .\right)\)

A venturi flume of rectangular section, \(1.2 \mathrm{~m}\) wide at inlet and \(600 \mathrm{~mm}\) wide at the throat, has a horizontal base. Neglecting frictional effects in the flume calculate the rate of flow if the depths at inlet and throat are \(600 \mathrm{~mm}\) and \(560 \mathrm{~mm}\) respectively. A hump of \(200 \mathrm{~mm}\) is now installed at the throat so that a standing wave is formed beyond the throat. Assuming the same rate of flow as before, show that the increase in upstream depth is about \(67.4 \mathrm{~mm}\).

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