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Chapter 10: Problem 69

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) \(0.01 ?\)

Short Answer

Expert verified
The slope needed to maintain a depth of 0.60 m in the rectangular wooden channel is determined using Manning's equation. Compare the determined slope with the given slopes \(0.02, 0.01\). If the given slope is larger than the determined slope, it will decrease the flow depth downstream, and vice versa.

Step by step solution

01

STEP 1: Compute the Hydraulic Radius & Cross-Sectional Area

Recall that the hydraulic radius \( R \) for a rectangular channel is the cross-sectional area of the flow \( A \) divided by the wetted perimeter \( P \). For a rectangular channel, \( A = b \cdot h \) and \( P = b + 2h \) where \( b \) is the width of the channel, and \( h \) is the depth of the flow. Thus: \[ R = \frac{bh} {b+2h} = \frac{0.60 \cdot 5} {5+2 \cdot 0.60} \] and \[ A = b \cdot h = 5 \cdot 0.60 \]
02

STEP 2: Apply Manning's equation

Manning's equation relates the flow rate \( Q \), the hydraulic radius \( R \), the slope of the energy line \( S \), the cross-sectional area \( A \) and the Manning's roughness coefficient \( n \). In this exercise, \( Q \) and \( n \) are provided (with \( n = 0.012 \) for wooden channels), and \( A \) and \( R \) were calculated in step 1. Rearranging Manning's equation for \( S \), we get: \[ S = \frac{n^2 \cdot Q^2} {A^2 \cdot R^{4/3}} \] Substitute \( n, Q, A \) and \( R \) into this formula to find \( S \).
03

STEP 3: Determine the channel slope

Substituting the known values into the formula from Step 2 will give \( S \), the channel slope needed to maintain the given flow depth. Solve to find the value of \( S \).
04

STEP 4: Examine given slopes' effect on flow

If the given slope is larger than \( S \), the channel depth will decrease downstream considering it leads to faster flow than required for current depth. Conversely, a lower slope will slow the flow and increase depth. Comparing the slopes \( S \) from Step 3 and given slopes will determine whether the depth will increase or decrease for given slopes. If \( S_1 > S \) then the depth decreases downstream and if \( S_2 < S \) then the depth increases downstream.

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Most popular questions from this chapter

Water flows at \(150 \mathrm{ft}^{3} / \mathrm{s}\) in a 3 -ft-wide rectangular cleanearth irrigation canal. The canal slope is \(0.275^{\circ} .\) Al one point, the water depth is \(3 \mathrm{ft}\) (a) Accurately compute the water depth at a location 200 fit downstream. (b) Is the flow at the upstream location subcritical or supercritical? At the downstream location? (c) Sketch the canal and the water surface profile.

An old, rough-surfaced, 2 -m-diameter concrete pipe with a Manning coefficient of 0.025 carries water at a rate of \(5.0 \mathrm{m}^{3} / \mathrm{s}\) when it is half full. It is to be replaced by a new pipe with a Manning coefficient of 0.012 that is also to flow half full at the same flowrate. Determine the diameter of the new pipe.

Water flows in a horizontal rectangular channel with a flowrate per unit width of \(q=10 \mathrm{ft}^{2} / \mathrm{s}\) and a depth of \(1.0 \mathrm{ft}\) at the downstream section \((2) .\) The head loss between section upstream and section (2) is 0.2 ft. Plot the specific energy diagram for this flow and locate states (1) and (2) on this diagram.

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 \(\theta\left(\theta \sim 1^{\circ}\right)\)

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.
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