/*! 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 11 A hydraulic jump (see Video \(\v... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 11

A hydraulic jump (see Video \(\vee 10.11\) ) is in place downstream from a spillway as indicated in Fig. P5.11. Upstream of the jump, the depth of the stream is \(0.6 \mathrm{ft}\) and the average stream velocity is \(18 \mathrm{ft} / \mathrm{s}\), Just downstream of the jump, the average stream velocity is \(3.4 \mathrm{ft} / \mathrm{s}\). Calculate the depth of the stream, \(h\), just downstreum of the jump.

Short Answer

Expert verified
The downstream depth of the stream is approximately 3.176 ft.

Step by step solution

01

Identify Known Values

We are given the upstream depth of the stream, denoted as \(h_1 = 0.6 \text{ ft}\), the upstream average stream velocity, \(v_1 = 18 \text{ ft/s}\), and the downstream average stream velocity, \(v_2 = 3.4 \text{ ft/s}\). We need to find the downstream depth of the stream, \(h_2\).
02

Use Continuity Equation

According to the continuity equation for incompressible fluids, the product of cross-sectional area and velocity is constant. For flow in a channel, this becomes \(A_1v_1 = A_2v_2\). Assuming constant width, this simplifies to \(h_1v_1 = h_2v_2\).
03

Solve for Downstream Depth

Rearrange the simplified continuity equation, \(h_1v_1 = h_2v_2\), to solve for \(h_2\): \[ h_2 = \frac{h_1v_1}{v_2} \]Substitute the known values:\[ h_2 = \frac{0.6 \text{ ft} \times 18 \text{ ft/s}}{3.4 \text{ ft/s}} \]
04

Calculate Numerical Result

Carry out the division to find \(h_2\): \[ h_2 = \frac{10.8 \text{ ft}^2/ ext{s}}{3.4 \text{ ft/s}} = 3.176 \text{ ft}\] Thus, the depth of the stream just downstream of the jump is approximately \(3.176 \text{ ft}\).

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 91影视!

Key Concepts

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

Continuity Equation
In fluid dynamics, the continuity equation is fundamental in describing mass conservation within a given system. It asserts that the mass of fluid entering a system equals the mass of fluid exiting the system, provided no sources or sinks are present. This principle is particularly useful for analyzing incompressible flow conditions, such as those seen in hydraulic jumps.
For a stream where the width remains constant, the continuity equation states that the product of the depth (h) and velocity (v) remains constant. Therefore, we can express it algebraically as:
  • \(A_1v_1 = A_2v_2\)
  • Simplifying due to constant width: \(h_1v_1 = h_2v_2\)
This approach allows us to solve for the unknown downstream depth in hydraulic structures like spillways by utilizing known upstream values. It's essential to ensure that units are consistent when making these calculations, to maintain accuracy.
Incompressible Flow
Incompressible flow is a common assumption for liquids traveling through channels or pipes, where the density remains constant. This means that the fluid does not experience any significant volume change when subjected to varying pressures while flowing.
In the context of hydraulic jumps, the assumption of incompressibility simplifies the analysis, allowing us to apply the continuity equation and fundamental fluid properties:
  • Densities of the fluid are constant across the stream, leading to simpler calculations.
  • Flow volumes remain uniform, so the dynamic changes are easier to predict and solve for.
Understanding incompressible flow can significantly enhance our ability to model and predict stream behaviors in controlled environments, leading to improved fluid management and engineering calculations.
Stream Velocity
Stream velocity plays a crucial role in hydraulic systems, affecting both upstream and downstream conditions of flow. This measure represents the speed at which water is moving through a channel at any given point.
In hydraulic jump scenarios, varying velocities will lead to changes in the stream's depth. The notable decrease in velocity post-jump often corresponds with an increase in depth, providing valuable insights into fluid dynamics:
  • Velocity determines the momentum of the water and influences changes in depth across the hydraulic jump.
  • Knowing both the initial and final velocities allows for the application of the continuity equation to find unknown parameters.
By carefully analyzing stream velocity, engineers can predict how changes upstream will affect downstream flow characteristics, ensuring the safe and efficient design of water management systems.

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

Describe a few examples (include photographs/images) of pumps where a fluid is forced to move by "blades" mounted on a rotating shaft.

Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per hour. The basement floor area is \(1500 \mathrm{ft}^{2} .\) What capacity (gal/min) pump would you rent to (a) keep the water accumulated in your basement at a constant level until the storm sewer is blocked off, and (b) reduce the water accumulation in your basement at a rate of 3 in./hr even while the backup problem exists?

How long would it take to fill a cylindrical-shaped swimming pool having a diameter of \(8 \mathrm{m}\) to a depth of \(1.5 \mathrm{m}\) with water from a garden hose if the flowrate is 1.0 liter/s?

The results of a wind tunnel test to determine the drag on a body (see Fig. \(\mathrm{P} 5.69\) ) are summarized below. The upstream [section (1)] velocity is uniform at 100 fUs. The static pressures are given by \(p_{1}=p_{2}=14.7\) psia. The downstream velocity distribution, which is symmetrical about the centerline, is given by \\[ \begin{array}{ll} u=100-30\left(1-\frac{|y|}{3}\right) & |y| \leq 3 \mathrm{ft} \\ u=100 & |y|>3 \mathrm{ft} \end{array} \\] where \(u\) is the velocity in \(\mathrm{ft} / \mathrm{s}\) and \(y\) is the distance on either side of the centerline in feet (see Fig. P5.69). Assume that the body shape does not change in the direction normal to the paper. Calculate the drag force (reaction force in \(x\) direction) exerted on the air by the body per unit length normal to the plane of the sketch.

Fresh water flows steadily into an open 55 -gal drum initially filled with seawater. The fresh water mixes thoroughly with the seawater, and the mixture overflows out of the drum. If the fresh water flowrate is 10 gal/min, estimate the time in seconds required to decrease the difference between the density of the mixture and the density of fresh water by \(50 \%\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Wave Optics

Read Explanation

Medical Physics

Read Explanation

Mechanics and Materials

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.