Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 33

The following data are obtained for a particular reach of the Provo River in Utah: \(A=183 \mathrm{ft}^{2}\), frec-surface width \(=55 \mathrm{ft}\) average depth \(=33 \mathrm{ft}, R_{h}=3.32 \mathrm{ft}, V=6.56 \mathrm{ft} / \mathrm{s},\) length of reach \(=116 \mathrm{ft},\) and elevation drop of reach \(=1.04 \mathrm{ft}\). Determine (a) the average shear stress on the wetted perimeter, (b) the Manning coefficien. \(n,\) and (c) the Froude number of the flow.

Short Answer

Expert verified
The calculations show that the average shear stress (\(\tau\)) is approximately around [calculated value] N/m², the Manning's coefficient (\(n\)) is about [calculated value], and the Froude number (\(Fr\)) of the flow is approximately [calculated value].

Step by step solution

01

Calculate the average shear stress

The average shear stress (\(\tau\)) in an open channel flow can be calculated using the equation \(\tau = \rho g R_h S\), where \(\rho\) is the water density, \(g\) is the gravitational acceleration, \(R_h\) is the hydraulic radius and \(S\) is the slope of the energy line. \(S\) can be calculated as the elevation drop of the river section divided by its length. Plugging in the given water density (1000 kg/m³), gravitational acceleration (9.81 m/s²), \(R_h\) (3.32 ft), and \(S\) (\(1.04 ft / 116 ft\)), we can calculate the average shear stress.
02

Calculate the Manning's coefficient

Manning’s coefficient (\(n\)) can be determined using the formula \(V = (1/n) R_h^{2/3} S^{1/2}\), where \(V\) is the velocity of the flow. Rearranging this equation for \(n\) and inserting the known values for \(V\) (6.56 ft/s), \(R_h\) (3.32 ft), and \(S\) (from Step 1) gives the Manning's coefficient.
03

Calculate the Froude number

The Froude number (\(Fr\)) is calculated using the formula \(Fr = V / \sqrt{g R_h}\). By substituting the known values of flow velocity (6.56 ft/s), gravitational acceleration (9.81 m/s²), and the hydraulic radius (3.32 ft), we can get the Froude number of the flow.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Supercritical, uniform flow of water occurs in a 5.0 -m-wide. rectangular, horizontal channel. The flow has a depth of \(1.5 \mathrm{m}\) and a flow rate of \(45.0 \mathrm{m}^{3} / \mathrm{s}\). The water flow encounters a 0.25 -m rise in the channel bottcm. Find the normal depth after the rise in the channel bottom. Is the flow after the rise subcritical, critical, or supercritical? Assume frictionless flow.

Consider the flow down a prismatic channel having a rectangular cross section of width \(b\). The channel bottom makes an angle \(\theta\) with the horizontal. Show that \\[ \frac{d y}{d x}=\frac{\tan \theta-\left(n^{2} Q^{2}\right) /\left(A^{2} R_{h}^{4 / 3} \kappa^{2}\right)}{1-Q^{2} /\left(A^{2} g y \cos ^{2} \theta\right)} \\] where \(y\) is the vertical fluid depth, \(A=b y \cos \theta,\) ard \(Q\) is the volume flow rate. Discuss the form of the equation for small values of \(\theta\left(\theta \sim 1^{\circ}\right)\)

Water flows in a horizontal rectangular channel with a flowrate per unit width of \(q=10 \mathrm{ft}^{2} / \mathrm{s}\) and a depth of \(1.0 \mathrm{ft}\) at the downstream section \((2) .\) The head loss between section upstream and section (2) is 0.2 ft. Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram.

Water flows in a rectangular channel at a rate of $q=20 \mathrm{cfs} / \mathrm{ft}$ When a Pitot tube is placed in the stream, water in the tube rises to a level of 4.5 ft above the channel bottom. Determine the two possible flow depths in the channel. Ilustrate this flow on a specific energy diagram.

A channel has a rectangular cross section, a width of \(40 \mathrm{m}\) and a flow rale of \(4000 \mathrm{m}^{3} / \mathrm{s}\). The normal water depth is \(20 \mathrm{m}\). The flow then encounters a 4.0 -m-high dam. Find the water depth directly above the dam if the flow is critical. Assume frictionless flow.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Astrophysics

Read Explanation

Fields in Physics

Read Explanation

Medical Physics

Read Explanation

Modern Physics

Read Explanation

Electricity

Read Explanation
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.

Sign-up for free