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Chapter 10: Problem 81

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.

Short Answer

Expert verified
The depths ahead of and behind the wave (hydraulic jump) need to be determined by applying the momentum equation to the system considering the velocities of the water flow and the wave. Solving the derived equation gives the depths which would represent the system before and after the hydraulic jump.

Step by step solution

01

Identify the Given Values

First, list the known quantities. Here, velocity V=6 m/s and wave velocity \(V_{w}=2\) m/s.
02

Understanding the Problem

It is important to remember that for an observer moving with the wave velocity, the flow of water appears as a steady hydraulic jump. Therefore, for that observer, the difference of velocities of the wave and water flow would be important.
03

Apply Momentum Equation to the Jump

Apply the conservation of momentum equation for a control volume in open channel flow across the hydraulic jump. It appears steady for an observer moving to the left with the wave velocity. This leads to the fact that the depth \(y_1\) upstream (ahead of the wave) and depth \(y_2\) downstream (behind the wave) need to be found to specify the conditions before and after the jump. By momentum conservation, it gives us \(y_1 V^2 = y_2(V − V_w)^2\).
04

Solve for Depths

The equation from Step 3 is a quadratic equation. By solving it, we could find the depths \(y_1\) and \(y_2\). Since, we know that depths can't be negative, discard the negative root.

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