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Chapter 10: Problem 8

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.

Short Answer

Expert verified
New discharge rate: 0.85 m³/s. Wave velocity: 2.53 m/s. Froude numbers: 0.598 (initial) and 0.374 (new).

Step by step solution

01

Calculate Initial Flow Depth

Convert the initial depth from millimeters to meters: 500 mm = 0.5 m.
02

Calculate Cross-Sectional Area

The cross-sectional area (A) of the trapezoidal channel is given by: \[ A = b \times y + z \times y^2 \]where: - b is the base width (1 m)- y is the flow depth (0.5 m)- z is the side slope (cot(60°) = 1 / \sqrt{3})Substitute the values to get: \[ A = 1 \times 0.5 + \frac{1}{\sqrt{3}} \times 0.5^2 = 0.5 + \frac{0.25}{\sqrt{3}} \approx 0.5 + 0.144 = 0.644 m^2 \]
03

Calculate Velocity

Velocity (V) can be calculated using the flow rate (Q) and area (A): \[ V = \frac{Q}{A} \]Given Q = 0.85 m^3/s and A = 0.644 m^2, then: \[ V = \frac{0.85}{0.644} \approx 1.32 m/s \]
04

Calculate Initial Froude Number

The Froude number (Fr) is calculated as: \[ Fr = \frac{V}{\sqrt{g y}} \]Substituting the values: \[ Fr = \frac{1.32}{\sqrt{9.81 \times 0.5}} \approx 0.598 \]
05

Determine New Flow Depth

The new flow depth is the initial depth plus the surge wave amplitude: \[ y_{new} = 0.5 + 0.15 = 0.65 m \]
06

Calculate New Cross-Sectional Area

Using the revised depth: \[ A_{new} = 1 \times 0.65 + \frac{1}{\sqrt{3}} \times 0.65^2 \approx 0.65 + 0.243 = 0.893 m^2 \]
07

New Flow Rate

Using continuity (Q before surge assumed to equal Q after surge): \[ Q_{new} = 0.893 \times V_{new} \]Solving for new velocity (V): \[ Q_{new} = 0.85 = 0.893 \times V_{new} \implies V_{new} \approx 0.952 m/s \]
08

Calculate Velocity of the Wave

The wave velocity (c) can be calculated as: \[ c = \sqrt{g (y + \Delta y)} \]Substitute the values: \[ c = \sqrt{9.81 \times 0.65} \approx 2.53 m/s \]
09

Calculate New Froude Number

Using the new velocity: \[ Fr_{new} = \frac{V_{new}}{\sqrt{g y_{new}}} \approx \frac{0.952}{\sqrt{9.81 \times 0.65}} \approx 0.374 \]

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Key Concepts

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

open channel flow
Open channel flow involves the movement of a fluid, such as water, in a conduit with a free surface open to the atmosphere. Unlike pipe flow, open channel flow is influenced by gravity and the shape of the channel. Factors like slope, channel shape, flow depth, and fluid velocity are crucial. Understanding these aspects helps in predicting how water will behave in natural and artificial channels.
trapezoidal channel
A trapezoidal channel is a type of open channel with a flat bottom and sides that slope outwards. The side slopes are often described by the angle they make with the horizontal. Given an angle of 60°, the slope of the sides can be used to calculate the cross-sectional area of the channel, which is vital for determining flow characteristics. The cross-sectional area is calculated using the formula: \[ A = b \times y + z \times y^2 \] where \( b \) is the channel's base width, \( y \) is the flow depth, and \( z = \frac{1}{\sqrt{3}} \) is related to the side slope.
surge wave propagation
Surge waves occur when there is a sudden change in flow conditions, such as when discharge is rapidly changed. These waves can travel upstream or downstream and affect water depth and velocity. In the given problem, a surge wave increases the water depth from 0.5 m to 0.65 m. The speed of the wave can be calculated using the formula: \[ c = \sqrt{g (y + \Delta y)} \] This represents the wave velocity, where \( g \) is the gravitational acceleration, \( y \) is the initial depth, and \( \Delta y \) is the wave amplitude.
fluid mechanics equations
Fluid mechanics involves various equations to describe the behavior of fluids. Essential equations include the continuity equation, which states that the discharge remains constant if the flow conditions are steady. This is written as: \[ Q = A \times V \] The Froude number is another critical concept, which compares the flow velocity to the wave velocity of the fluid. It is given by: \[ Fr = \frac{V}{\sqrt{g y}} \] This helps in classifying the flow as subcritical (Fr<1), critical (Fr=1), or supercritical (Fr>1), which affects how energy is dissipated and how waves move through the channel.

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Most popular questions from this chapter

Uniform flow occurs with a depth of \(900 \mathrm{~mm}\) in a rectangular prismatic channel \(2.5 \mathrm{~m}\) wide. If Manning's \(n=0.015\) and the bed slope is 1 in 1200 what is the minimum height of hump in the bed over which critical flow will be produced?

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 \(67.4 \mathrm{~mm}\).

In a rectangular wave-tank \(4.5 \mathrm{~m}\) wide containing fresh water, waves of total height (trough to crest) \(0.5 \mathrm{~m}\) and period \(5 \mathrm{~s}\) are generated over a still-water depth of \(4 \mathrm{~m}\). Verifying the assumptions made, determine the wave speed, wavelength, group velocity and total power. For a position midway between a trough and a crest and at half the still-water depth, determine the velocity of a particle and the static pressure. What are the semi-axes of a particle orbit at this position?

A venturi flume installed in a horizontal rectangular channel \(700 \mathrm{~mm}\) wide has a uniform throat width of \(280 \mathrm{~mm}\). When water flows through the channel at \(0.140 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\), the depth at a section upstream of the flume is \(430 \mathrm{~mm}\). Neglecting friction, calculate the depth of flow at the throat, the depth at a section just downstream of the flume where the width is again \(700 \mathrm{~mm}\), and the force exerted on the stream in passing through the flume.

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 \(\left.=1025 \mathrm{~kg} \cdot \mathrm{m}^{-3} .\right)\)
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