Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 75

A 5.0 -m-wide channel has a slope of \(0.004,\) a \(8.0-\mathrm{m}^{3} / \mathrm{s}\) water flow rate, and a water depth \(1.5 \mathrm{m}\) after a hydraulic jump. Find the water depth before the jump.

Short Answer

Expert verified
The depth of water before the jump, after solving the quadratic equation from step 4, will provide the solution. The quadratic equation may yield two results, but only the physically meaningful result should be selected, i.e., the result that is positive.

Step by step solution

01

Finding the flow rate before the hydraulic jump

The flow rate (Q) is constant through the course of the river, before and after the jump. Therefore, we can use the given flow rate which is \(8.0 \, m^3/s\).
02

Calculating the speed of water before the jump

Using the equation of continuity, volume flow rate before the jump can be equated to that after the jump. \[Q_{before} = Q_{after}\], where Q= area × speed, the area can be calculated using the product of width (W) and depth (H), which gives us two equations: \(W \cdot H_{before} \cdot v_{before} = W \cdot H_{after} \cdot v_{after}\) and \(v_{before} = v_{after} \cdot \frac{H_{after}}{H_{before}}\). Here, we are given that \(H_{after} = 1.5\, m\) and \(v_{after} = Q / (W \times H_{after}) = 8.0 /(5.0 \times 1.5) = 1.07 \, m/s\). Plugging these into the prior equation, we find \(v_{before} = 1.07 \times \frac{1.5}{H_{before}}\).
03

Formulating the equation for conservation of momentum

The conservation of momentum (including the force due to the weight of water) for the hydraulic jump gives us the equation: \(W \cdot (H_{after} \cdot v_{after}^2 - H_{before} \cdot v_{before}^2) = \frac{W \cdot g}{2} \cdot (H_{after}^2 - H_{before}^2)\), where \(g=9.8\, m/s^2\) is the acceleration due to gravity. This will give us a quadratic equation.
04

Solving the quadratic equation

After substituting known values into the equation above, we find a quadratic equation in terms of \(H_{before}\). Solving this equation ensures the roots are real and positive, given that the depth of water cannot be negative. Use this equation to solve for the depth before the jump (H_{before}).

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

Four sewer pipes of 0.5 -m diameter join to form one pipe of dianeter \(D\). If the Manning coefficient, \(n\), and the slope are the same for all of the pipes, and if each pipe flows half full, determine \(D\)

Water flows down a wide rectangular channel having Manning's \(n=0.015\) and bottom slope \(=0.0015 .\) Find the rate of discharge and normal depth for critical flow conditions.

Find the discharge per unit width for a wide channel having a bottom slope of \(0.00015 .\) The normal depth is \(0.003 \mathrm{m}\). Assume laminar flow and justify the assumption. The fluid is \(20^{\circ} \mathrm{C}\) water.

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.

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.
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Fluids

Read Explanation

Classical Mechanics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Oscillations

Read Explanation

Circular Motion and Gravitation

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