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Chapter 10: Problem 16

Water flows at a depth of \(1.2 \mathrm{~m}\) in a rectangular prismatic channel \(2.7 \mathrm{~m}\) wide. Over a smooth hump \(200 \mathrm{~mm}\) high on the channel bed a drop of \(150 \mathrm{~mm}\) in the water surface is observed. Neglecting frictional effects, calculate the rate of flow.

Short Answer

Expert verified
The rate of flow is 17.26 cubic meters per second.

Step by step solution

01

- Understand the given data

Identify the variables given in the problem: the depth of the water before the hump (\(h_1=1.2\) mewline), the width of the channel (\( b=2.7\) mewline), the height of the hump ( \(h_{hump}=0.2\) mewline), and the drop in water surface height after the hump (\(\text drop=0.15\) mewline).
02

- Calculate the depth after the hump

Use the drop height to find the depth of water after the hump: \(h_2 = h_1 - \text drop.ewline\) ewline Hence,ewline \(h_2 = 1.2 \text{ m} - 0.15 \text{ m} = 1.05 \text{ m} \).
03

- Apply the Energy Equation

Apply Bernoulli's equation for open channel flow, \( \frac{v_1^2}{2g} + h_1 = \frac{v_2^2}{2g} + h_2 + h_{hump} \), and assume that \( v_1 \) and \( v_2 \) are the velocities before and after the hump respectively.
04

- Relate velocities and rate of flow

Use continuity equation for prismatic channel, which is: \( Q = A_1v_1 = A_2v_2 \). Since \(Q=b.h_1.v_1ewline\)= \( b.h_2.v_2 \).
05

- Calculate velocities

From the continuity equation: \( v_2=\frac{b.h_1}{b.h_2}\times v_1,ewline\) which simplifies to, \( v_2=\frac{1.2}{1.05}v_1 ewline\) = \(1.14 v_1\). Subsequently: \( v_1 = h_{hump}/(2g(1-1/(1.14)^2))\) calculates to: \( v_1^2=9.8/ 0.034 = v_1=5.34 m/s.\)
06

- Compute the rate of flow

Finally, determine the rate of flow Q:\(Q = b h_1 v_1.ewline \) Plug in the numbers: \( Q= 2.7 m \times 1.2 m \times 5.34 m/s = 17.26 \text{ cubic meters per second (m}^3/\text{s) }\)

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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 refers to the movement of water in a channel with a free surface exposed to the atmosphere. In this problem, we are dealing with a rectangular prismatic channel. This type of channel has a constant width and depth, making it easier to analyze the flow. Key characteristics of open channel flow include the water depth, flow velocity, and channel dimensions. These factors are used to determine the type and behavior of the flow, such as uniform or non-uniform flow.
continuity equation
The continuity equation is essential for understanding fluid dynamics in open channels. It states that the rate of flow is constant along the channel, provided there is no addition or removal of water. Mathematically, it is expressed as \(Q = A_1 v_1 = A_2 v_2\), where \(Q\) is the discharge or flow rate, \(A_1\) and \(A_2\) are the cross-sectional areas before and after a particular section, and \(v_1\) and \(v_2\) are the flow velocities before and after. In our problem, the width of the channel is constant, so the areas can be simplified to depth times width. This allows us to express the rate of flow as \(Q = b h_1 v_1 = b h_2 v_2\).
energy equation
The energy equation, or Bernoulli's equation, is crucial to analyzing open channel flow. It describes the conservation of energy in a flowing fluid. For an open channel, it can be adapted to account for the different heights and velocities of the water. The equation is \( \frac{v_1^2}{2g} + h_1 = \frac{v_2^2}{2g} + h_2 + h_{hump} \), where \(v_1\) and \(v_2\) are velocities before and after the hump, \(h_1\) and \(h_2\) are depths before and after, and \(h_{hump}\) is the height of the hump. This equation helps us account for changes in potential and kinetic energy along the flow.
depth and velocity relationship
The depth and velocity of water in a channel are inversely related. This means that when the depth decreases, the velocity must increase to maintain the same flow rate. In our problem, we see this relationship through the continuity equation. After computing, we find that \( v_2 = \frac{h_1}{h_2} v_1 = 1.14 v_1 \). This shows that the velocity after the hump is higher than before, due to the decrease in depth from \(1.2 \text{ m}\) to \(1.05 \text{ m}\).
neglecting friction effects
In this problem, we are asked to neglect frictional effects. Friction can significantly impact flow by reducing the velocity and altering the depth. By ignoring these effects, we simplify our calculations and focus solely on the geometric and energy relationships in the channel. Although this simplification can lead to minor inaccuracies, it is often a reasonable assumption for a basic analysis, especially in smooth, unobstructed channels.

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Most popular questions from this chapter

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.

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.

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 \(58 \mathrm{~m}^{1 / 2} \cdot \mathrm{s}^{-1}\)

The cross-section of a river \(30 \mathrm{~m}\) wide is rectangular. At a point where the bed is approximately horizontal the width is restricted to \(25 \mathrm{~m}\) by the piers of a bridge. If a flood of \(450 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) is to pass the bridge with the minimum upstream depth, describe the flow past the piers and calculate the upstream depth.
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