Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 3

A rectangular channel \(3 \mathrm{m}\) wide carries \(10 \mathrm{m}^{3} / \mathrm{s}\) at a depth of \(2 \mathrm{m} .\) Is the flow subcritical or supercritical? For the same flowrate, what depth will give critical flow?

Short Answer

Expert verified
The flow is subcritical. The depth that would provide a critical flow is approximately 1.34 meters.

Step by step solution

01

Determine the Velocity

The velocity (V) of the water flow can be calculated using the formula V = Q/A, where Q is the flowrate and A is the cross-sectional area of the flow (A = width * depth). Here, Q = 10 cubic meters per second, width = 3 m, and depth = 2 m. Thus, V = (10 cubic meters per second) / (3 m * 2 m) = 1.67 m/s.
02

Calculate the Froude Number

The Froude number (Fr) is a key dimensionless number in fluid mechanics and is used to define the flow regime. It is given by the formula Fr = V / sqrt(g * D), where g is the acceleration due to gravity and D is the hydraulic depth. In this case, g = 9.81 m/s^2 and D = depth = 2 m. Hence, Fr = 1.67 m/s / sqrt(9.81 m/s^2 * 2 m) = 0.6.
03

Determine the Flow Regime

The Froude number gives an indication of the flow regime of the fluid. If Fr < 1, the flow is deemed subcritical. If Fr = 1, the flow is critical, and if Fr > 1, the flow is supercritical. As computed in Step 2, Fr = 0.6, which is less than 1. Thus, it can be concluded that the flow is subcritical.
04

Find the Critical Depth

In hydrology, critical depth (Dc) is the depth of flow in an open channel at which the specific energy is a minimum. For a rectangular channel, it is calculated by the formula Dc = (Q^2 / g)^(1/3), where Q and g are as defined previously. Substituting the given values into the formula, the critical depth is given as Dc = (10 cubic meters per second^2 / 9.81 m/s^2)^(1/3) = 1.34 m.

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

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 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.

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 \(E=0.6 \mathrm{m}\)

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 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.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Quantum Physics

Read Explanation

Radiation

Read Explanation

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Electromagnetism

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