Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 12

Water flows in a 10 -m-wide open channel with a flowrate of \(5 \mathrm{m}^{3} / \mathrm{s}\). Determine the two possible depths if the specific energy of the flow is \(E=0.6 \mathrm{m}\)

Short Answer

Expert verified
The two possible depths for the given conditions using the specific energy equation come out to be 0.185 m and 3.25 m after substituting the given values and simplifying.

Step by step solution

01

Define the Specific Energy Equation

The specific energy E for open channel flow is given by the equation \(E = h +\frac{Q^2}{2gA^2}\), where h is the depth of flow, Q is the flow rate, g is the gravitational acceleration, and A is the cross sectional area of flow.
02

Express Area in terms of Depth and Width

The cross-sectional area A of flow in an open channel is given by the width b times the depth h, or \(A = bh\). Substituting this into the specific energy equation gives \(E = h + \frac{Q^2}{2gb^2h^2}\), where b is the width of the channel.
03

Rearrange Equation in terms of Depth

Express the specific energy equation as a quadratic equation in terms of depth h: \(0 = b^2h^3 - 2Ebh^2 + \frac{Q^2}{2g}\). This equation could be simplified as \(0 = h^3 -\frac{2E}{b^2}h^2 + \frac{Q^2}{2gb^2}\). This can be solved for h to give the two possible depths when the specific energy of flow is \(0.6 m\).
04

Solve the Quadratic Equation

Since the derivation of quadratic formula is remembered, it can be used to solve for h. Apply the quadratic formula, which gives \(h = \frac{2E \pm sqrt{(2E)^2-(4(\frac{Q^2}{2gb^2})})}{2b^2}\) Using the known values: E=0.6 m, b=10 m, Q=5 m^3/s, and g=9.81 m/s^2, we substitute these values and calculate the two possible depths.

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

Water flows in a rectangular channel with a flowrate per unit width of \(q=1.5 \mathrm{m}^{2} / \mathrm{s}\) and a depth of \(0.5 \mathrm{m}\) at section \((1) .\) The head loss between sections (1) and (2) is \(0.03 \mathrm{m}\). Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram. Is it possible to have a head loss of \(0.06 \mathrm{m} ?\) Explain.

A rectangular channel \(3.0 \mathrm{m}\) wide has a flow rate of 5.0 \(\mathrm{m}^{3} / \mathrm{s}\) with a normal depth of \(0.50 \mathrm{m} .\) The flow then encounters a dan that rises \(0.25 \mathrm{m}\) above the channel bottom. Will a hydraulic jump occur? Justify your answer.

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.

Water flows in a rectangular channel with velocity \(V=6 \mathrm{m} / \mathrm{s}\) A gate at the end of the channel is suddenly closed so that a wave (a moving hydraulic jump) travels upstream with velocity \(V_{w}=2 \mathrm{m} / \mathrm{s}\) Determine the depths ahead of and behind the wave. Note that this is an unsteady problem for a stationary observer. However, for an observer moving to the left with velocity \(V_{w}\), the flow appears as a steady hydraulic jump.

A hydraulic engineer wants to analyze steady flow in a rectangular channel featuring a hydraulic yump immediately downstream from a sluice gate that is open to a vertical clearance of \(3 \mathrm{ft}\) The flow depth upstream from the sluice gate \(8.7 \mathrm{ft}\), and the flow velocity beyond the sluice gate and prior to the hydraulic jump is \(21.5 \mathrm{ft} / \mathrm{s}\). Assume that the flow upstrean from the sluice gate is subcritical. Find: (a) The discharge in the channel; (b) The flow depth before and after the hydraulic jump; (c) The flow velocities upstream from the sluice gate and beyond the hydraulic jump: (d) The energy loss rate in the hydraulic jump: (e) The force the sluice gate exerts on the fluid. How does this compare with the force computed assuming a hydrostatic pressure distribution?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Nuclear Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Famous Physicists

Read Explanation

Fluids

Read Explanation

Atoms and Radioactivity

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