/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 A concrete-lined rectangular cha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A concrete-lined rectangular channel is \(8-\mathrm{m}\) wide and \(4-\mathrm{m}\) deep and has a longitudinal slope of \(0.2 \%\) and an estimated Manning's \(n\) of \(0.013 .\) A hydraulic structure controls the flow in the channel such that the depth of flow at the structure is \(3 \mathrm{~m}\) when the flow rate in the channel is \(24 \mathrm{~m}^{3} / \mathrm{s}\). Urban developers propose a localized contraction/expansion in the channel \(100 \mathrm{~m}\) upstream of the control structure to accommodate a pedestrian walkway. Taking eddy losses into account but neglecting friction losses, estimate the maximum contraction that should be allowed for the walkway. Based on your result, assess whether it is reasonable to neglect friction losses in this case.

Short Answer

Expert verified
Neglecting friction is reasonable if eddy losses significantly affect flow; otherwise, it might need reconsidering.

Step by step solution

01

Understanding the Problem

We need to find the maximum contraction allowed for a pedestrian walkway in a concrete-lined channel, considering eddy losses but neglecting friction losses. The channel is 8 meters wide, 4 meters deep, with a slope of 0.2% and Manning's n of 0.013.
02

Define Given Parameters

The given parameters are:- Channel width \(b_1 = 8\,\text{m}\)- Channel depth \(d_1 = 4\,\text{m}\)- Flow depth at structure \(d_2 = 3\,\text{m}\)- Slope \(S = 0.002\)- Manning's \(n = 0.013\)- Flow rate \(Q = 24 \,\text{m}^3/\text{s}\)
03

Initial Calculations

Calculate the flow area \(A_1\) and hydraulic radius \(R_1\) for the original section:\[A_1 = b_1 \times d_1 = 8\,\text{m} \times 4\,\text{m} = 32\,\text{m}^2\]\[R_1 = \frac{A_1}{P_1} = \frac{32}{8+4+8+4} = \frac{32}{24} = 1.33\,\text{m}\]
04

Calculating Flow Velocity

The velocity \(V_1\) in the original section can be calculated as:\[Q = A_1 \times V_1 \implies V_1 = \frac{Q}{A_1} = \frac{24}{32} = 0.75\,\text{m/s}\]
05

Apply Energy Principles

Using the energy equation to account for the change in water depth and channel width:\[Z_1 + d_1 + \frac{V_1^2}{2g} = Z_2 + d_2 + \frac{V_2^2}{2g} + h_\text{L}\]where \(h_L = k_e \times \frac{V_1^2}{2g}\) is the head loss due to contraction. Assume \(k_e \approx 0.2\) for mild contractions.
06

Compute Critical Width

Since details on total energy considerations for losses are needed, under simplifying:\[d_1 + \frac{V_1^2}{2g} = d_2 + \frac{V_2^2}{2g} + h_L\]with relation \(Q = A_2 \times V_2\) and new width \(b_2\), solve iteratively, allowing for contraction while considering the head loss.
07

Evaluating Neglected Friction Losses

Compare the energy impact of eddy losses with typical friction losses using rough comparisons:\[S_f \times L = f \times \frac{V_1^2}{2g}\] where \(S_f = 0.013\),\(f = \text{estimated} \approx 0\), see if influence high or low in energy terms.
08

Assess Conclusion

Based on estimations and energy calculations, if the contraction significantly shifts flow conditions or if friction adjustments after calculations remain minor, decision justifies neglect unless friction induces nontrivial errors.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Manning's Equation
Manning's Equation is an empirical formula used to estimate the velocity of flow in an open channel. It is a widely used tool in hydraulic engineering and fluid mechanics to calculate flow conditions in natural and artificial channels. The equation is stated as:
  • \[V = \frac{1}{n}R^{2/3}S^{1/2} \]
  • where \(V\) is the mean velocity of flow (m/s)
  • \(n\) is the Manning’s roughness coefficient, which accounts for channel surface roughness
  • \(R\) is the hydraulic radius, and \(S\) is the slope of the water surface or energy grade line
For our concrete-lined rectangular channel, this formula helps calculate the velocity and validate assumptions about neglected losses. Manning's equation also reveals how channel characteristics affect flow velocity, aiding in the design and assessment of channel modifications for projects like the proposed pedestrian walkway.
Hydraulic Radius
The hydraulic radius (\(R\)) is a key component in Manning’s equation, representing the flow efficiency of a channel. It is defined as the ratio of the flow area (\(A\)) to the wetted perimeter (\(P\)).
  • It quantifies how much water is in contact with the channel's boundaries, influencing energy dissipation.
  • For a rectangular channel, \(R\) can be calculated using: \[R = \frac{A}{P} \] where \(A\) is the cross-sectional area.
  • This aspect is crucial in fluid mechanics, as it not only affects velocity but also impacts the channel's ability to convey water efficiently.
In our exercise, we calculate the hydraulic radius to understand how the channel geometry supports the assumed flow conditions.
Energy Equation
The energy equation in fluid mechanics, particularly in open channel flow, relates the flow's energy at two sections along the channel. It considers potential energy (due to depth), kinetic energy (due to velocity), and energy losses (such as eddy losses).
  • The fundamental energy equation is: \[Z_1 + d_1 + \frac{V_1^2}{2g} = Z_2 + d_2 + \frac{V_2^2}{2g} + h_L\]
  • Here, \(Z\) is the elevation head, \(d\) is the water depth, \(V\) is velocity, and \(h_L\) represents head losses.
  • This equation allows us to explore how changes in channel width (contraction/expansion) will alter energy distribution and assess flow stability.
In this exercise, the energy equation is used to evaluate the feasibility of the channel modification, balancing energy across contractions while taking into account only eddy, not friction, losses.
Channel Flow
Channel flow refers to how water moves along a channel, influenced by factors such as gravity, channel geometry, and surface roughness. In our given scenario, understanding channel flow dynamics is essential to assessing channel changes.
  • Key considerations include:
    • The geometry of the channel, which determines volume and speed of water.
    • Flow rate: how much water passes a given point in a particular amount of time.
    • Potential changes to flow characteristics due to modifications such as contractions or expansions.
  • These factors help in predicting how different channel designs affect water flow and in evaluating the effects of different hydraulic structures or modifications.
Utilizing knowledge of channel flow, such as Manning’s equation and hydraulic radius, allows engineers to model and predict how changes will affect real-world water movement. This is crucial for designing safe and efficient infrastructure.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A trapezoidal canal has a longitudinal slope of \(1 \%,\) side slopes of \(3: 1(\mathrm{H}: \mathrm{V}),\) a bottom width of \(3.00 \mathrm{~m}\), a Manning's \(n\) of \(0.015,\) and it carries a flow of \(20 \mathrm{~m}^{3} / \mathrm{s}\). The depth of flow at a gauging station is observed to be \(1.00 \mathrm{~m}\). Respond to the following: (a) What is the normal depth of flow in the channel? (b) What is the critical depth of flow in the channel? (c) Classify the slope of the channel and the water surface profile at the gauging station. (d) How far from the gauging station is the depth of flow equal to \(1.1 \mathrm{~m}\) ? Does this depth occur upstream or downstream of the gauging station? (e) If the bottom of the channel just downstream of the gauging station is raised by \(0.20 \mathrm{~m}\), determine the resulting depth of flow at the downstream section. The bottom width of the channel remains constant at \(3 \mathrm{~m}\).

Use the Darcy-Weisbach equation to show that the head loss per unit length, \(S\), between any two sections in an open channel can be estimated by the relation $$ S=\frac{\bar{f}}{4 \bar{R}} \frac{\bar{V}^{2}}{2 g} $$ where \(\bar{f}, \bar{R},\) and \(\bar{V}\) are the average friction factor, hydraulic radius, and flow velocity, respectively, between the upstream and downstream sections.

Water flows at \(1 \mathrm{~m}^{3} / \mathrm{s}\) in a rectangular channel of width \(1 \mathrm{~m}\) and depth \(1 \mathrm{~m}\). What is the maximum contraction of the channel that will not choke the flow?

Given that hydraulically rough flow conditions occur in open channels when \(u_{*} k_{s} / \nu \geq\) 70 , show that this condition can be expressed for water in terms of Manning parameters as $$ n^{6} \sqrt{R S_{0}} \geq 7.9 \times 10^{-14} $$ If a concrete-lined rectangular channel with a bottom width of \(5 \mathrm{~m}\) is constructed on a slope of \(0.05 \%\) and Manning's \(n\) is estimated to be \(0.013,\) determine the minimum flow depth for hydraulically rough flow conditions to exist.

Determine the critical depth for a flow of \(50 \mathrm{~m}^{3} / \mathrm{s}\) in a trapezoidal channel with bottom width of \(4 \mathrm{~m}\) and side slopes of \(1.5: 1(\mathrm{H}: \mathrm{V})\). If the actual depth of flow is \(3 \mathrm{~m},\) calculate the Froude number and state whether the flow is subcritical or supercritical.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.