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Chapter 10: Problem 41

\(\mathrm{A}\) trapezoidal channel with a bottom width of \(3.0 \mathrm{m}\) and sides with a slope of 2: 1 (horizontal:vertical) is lined with fine gravel \((n=0.020)\) and is to carry \(10 \mathrm{m}^{3} / \mathrm{s}\). Can this channel be built with a slope of \(S_{0}=0.00010\) if it is necessary to keep the velocity below \(0.75 \mathrm{m} / \mathrm{s}\) to prevent scouring of the bottom? Explain.

Short Answer

Expert verified
A detailed assessment through Manning's equation is required to ascertain if the scouring can be avoided at a slope of \(S_{0}=0.00010\). Multiple iterations would be needed to estimate the depth of water and then compute the velocity through Manning's formula. If the calculated velocity doesn't exceed 0.75 m/s, then the channel won't experience scouring

Step by step solution

01

Compute Cross-Sectional Area of the Channel

The cross-sectional area (\( A \)) of a trapezoidal channel can be calculated by the formula: \( A = b \cdot d + z \cdot d^{2} \), where \( b \) is the bottom width, \( d \) is the depth and \( z \) is the side slope. As we have only the flow rate, this step needs to be solved as part of iteration process to estimate the depth.
02

Calculate the Wetted Perimeter

The wetted perimeter of a trapezoidal channel (P) is calculated by the formula \( P = b + 2d \cdot \sqrt{1+z^{2}} \). The precise value of \( d \) needs to be estimated through iteration.
03

Calculate the Hydraulic Radius

The hydraulic radius (\( R \)) is calculated by dividing the cross-sectional area (\( A \)) by the wetted parameter (\( P \)). The formula for the hydraulic radius is: \( R = A / P \). Again, the complex part is estimating the unknown water depth (\( d \)).
04

Apply Manning's Equation

With the hydraulic radius, you can compute the velocity using the Manning's equation: \( V = (1/n) \cdot R^{2/3} \cdot S_{0}^{1/2} \). Check if the calculated \( V \) is less than or equals to 0.75 m/s at reasonable estimates of \( d \). If so, the channel will not experience scouring. Otherwise, adjust the depth and repeat the process until criteria are met.

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Most popular questions from this chapter

Four sewer pipes of 0.5 -m diameter join to form one pipe of dianeter \(D\). If the Manning coefficient, \(n\), and the slope are the same for all of the pipes, and if each pipe flows half full, determine \(D\)

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?

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?

Supercritical, uniform flow of water occurs in a 5.0 -m-wide. rectangular, horizontal channel. The flow has a depth of \(1.5 \mathrm{m}\) and a flow rate of \(45.0 \mathrm{m}^{3} / \mathrm{s}\). The water flow encounters a 0.25 -m rise in the channel bottcm. Find the normal depth after the rise in the channel bottom. Is the flow after the rise subcritical, critical, or supercritical? Assume frictionless 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.
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