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Chapter 10: Problem 14

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.

Short Answer

Expert verified
The two possible flow depths can be found by solving the quadratic equation obtained in step 4. These represent the possible depth of water flow in the channel.

Step by step solution

01

Identify given data

Establish the given variables: Flow rate, \( q = 20 \, \mathrm{cfs/ft} \), and the height of water level in the Pitot tube, \( h = 4.5 \, \mathrm{ft} \).
02

Utilize the Energy Principle

According to the Bernoulli equation for energy, the specific energy, \( E \), in an open channel flow is the sum of the depth of flow, \( d \), and the kinetic head (which is the square of the velocity, \( v \), divided by twice the acceleration due to gravity, \( g \)). We can therefore state that \( E = d + \frac{v^2}{2g} \). Given that the flow rate per unit width, \( q \), is equal to the depth of flow times the velocity, i.e., \( q = dv \), we can express the velocity in terms of \( q \) and \( d \), yielding \( v = \frac{q}{d} \). Substituting this in the energy equation offers \( E = d + \frac{q^2}{2dg^2} \)
03

Solve for Energy

Solve for the specific energy \( E \) using the given height \( h \) of the water in the Pitot tube. Since the height is typically equivalent to the total energy head, it becomes \( E = h = 4.5 \, \mathrm{ft} \)
04

Solve for Flow Depth

The Equation \( E = d + \frac{q^2}{2dg^2} \) is a quadratic equation in terms of \( d \). Substituting for \( E \), \( q \) and \( g \) (approximated as \( 32.2 \, \mathrm{ft/s^2} \)) and solving for \( d \) gives the two possible flow depths.

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Most popular questions from this chapter

The depth downstream of a sluice gate in a rectangular wooden channel of width \(5 \mathrm{m}\) is \(0.60 \mathrm{m}\). If the flowrate is \(18 \mathrm{m}^{3} / \mathrm{s}\) determine the channel slope needed to maintain this depth. Will the depth increase or decrease in the flow direction if the slope is (a) \(0.02 ;\) (b) \(0.01 ?\)

(See The Wide World of Fluics article titled "Grand Canyon Rapids Building." Section \(10.6 .1 .\) ) During the flood of \(1983,\) a large hydraulic jump formed at "Crystal Rapid" on the Colorado River. People rafting the river at that time report "entering the rapid at almost 30 mph, hitting a 20 -ft-tall wall of water, and exiting at about 10 mph." Is this information (i.e., upstream and downstream velocities and change in depth) consistent with the principles of a hydraulic jump? Show calculations to support your answer.

At a given location in a 12 -ft-wide rectangular channel the flowrate is \(900 \mathrm{ft}^{3} / \mathrm{s}\) and the depth is \(4 \mathrm{ft}\). Is this location upstream or downstream of the hydraulic jump that occurs in this channel? Explain.

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 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.
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