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Chapter 10: Problem 18

A channel has a rectangular cross section, a width of \(40 \mathrm{m}\) and a flow rale of \(4000 \mathrm{m}^{3} / \mathrm{s}\). The normal water depth is \(20 \mathrm{m}\). The flow then encounters a 4.0 -m-high dam. Find the water depth directly above the dam if the flow is critical. Assume frictionless flow.

Short Answer

Expert verified
The water depth directly above the dam when the flow is critical is 12.92 meters.

Step by step solution

01

Identifying Known Values

Identify the known values. Width of the channel \(b = 40m\); Flow rate \(Q = 4000m^3/s\); Normal water depth \(h = 20m\); Dam height \(d = 4.0m\)
02

Find the Critical Flow Speed

The formula for the critical flow speed, \(V_c\), in a rectangular channel is determined by the equation \(V_c = \sqrt{g \cdot h_c}\) where \(g\) is the gravity constant and \(h_c\) is the critical depth of water. However, before we find \(V_c\), we need to determine \(h_c\) from the given flow rate and width of the channel using the formula \(Q = b \cdot h_c \cdot V_c\)
03

Substitute For Critical Velocity

Substitute the formula for \(V_c\) into the equation found in Step 2 to derive an equation for \(h_c\): \(Q = b \cdot h_c \cdot \sqrt{g \cdot h_c}\). Simplify the equation to solve for \(h_c\): \(h_c = \left(\frac{Q^2}{g \cdot b^2}\right)^{1/3}\)
04

Find the critical depth

Substitute the known values into the equation for \(h_c\). Given that gravity \(g = 9.81 m/s^2\), flow rate \(Q = 4000m^3/s\) and width \(b = 40m\), \(h_c = \left(\frac{(4000)^2}{9.81 \cdot (40)^2}\right)^{1/3}\) = 8.92 m
05

Determine the water depth

The water depth directly above the dam is the sum of the dam's height and the critical depth. Therefore, the water depth above the dam is \(d + h_c = 4.0m + 8.92m = 12.92m\)

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Most popular questions from this chapter

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.

(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.

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?

Consider the flow down a prismatic channel having a rectangular cross section of width \(b\). The channel bottom makes an angle \(\theta\) with the horizontal. Show that \\[ \frac{d y}{d x}=\frac{\tan \theta-\left(n^{2} Q^{2}\right) /\left(A^{2} R_{h}^{4 / 3} \kappa^{2}\right)}{1-Q^{2} /\left(A^{2} g y \cos ^{2} \theta\right)} \\] where \(y\) is the vertical fluid depth, \(A=b y \cos \theta,\) ard \(Q\) is the volume flow rate. Discuss the form of the equation for small values of \(\theta\left(\theta \sim 1^{\circ}\right)\)

Do shallow waves propagate at the same speed in all fluids? Explain why or why not.
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