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Chapter 10: Problem 51

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.

Short Answer

Expert verified
The diameter of the new pipe can be found by first computing the slope from the given data and Manning's equation for the old pipe, then substititating this slope into Manning's equation for the new pipe, and solving for the diameter.

Step by step solution

01

Using Manning's equation for the old pipe

Flow rate Q can be calculated using Manning's equation, \(Q=\frac{1}{n}AR^{2/3}S^{1/2}\), where \(Q\) is the flowrate, \(n\) is the Manning's coefficient, \(A\) is the cross-sectional area, \(R\) is the hydraulic radius, and \(S\) is the slope of the hydraulic grade line. Since the pipe is half full, \(A=\π D^2/8\) and \(R=D/4\). For the old pipe, \(n_{old}=0.025\) and \(Q_{old}=5.0 \, m^3/s\). Thus from Manning's equation we get, \[5.0 = \frac{1}{0.025} \cdot \frac{\π {D_{old}}^2}{8} \cdot \left(\frac{{D_{old}}}{4}\right)^{2/3} \cdot S^{1/2}\] Where \(D_{old}\) represents the diameter of the old pipe.
02

Simplifying the equation for the old pipe

If we simplify the equation from step 1 we obtain an expression for the slope \(S\) in terms of the known quantities: \[S = \left(\frac{5.0}{\frac{1}{0.025} \cdot \frac{\pi {D_{old}}^2}{8} \cdot \left(\frac{D_{old}}{4}\right)^2/3}\right)^2\] Now we can substitute \(D_{old}=2m\), and calculate the value of \(S\).
03

Using Manning's equation for the new pipe

We then apply Manning's formula for the new pipe, keeping in mind that it should flow half full at the same flow rate as the old pipe, \(Q_{new}=5.0 \, m^3/s\), and Manning's coefficient for the new pipe, \(n_{new}=0.012\). The cross-sectional area and hydraulic radius for the new pipe are \(A_{new}=\pi {D_{new}}^2/8\) and \(R_{new}=D_{new}/4\) respectively. We substitute these values and the same slope \(S\) (from Step 2) into the Manning's equation and get: \[5.0 = \frac{1}{0.012} \cdot \frac{\pi {D_{new}}^2}{8} \cdot \left(\frac{D_{new}}{4}\right)^{2/3} \cdot S^{1/2}\] where \(D_{new}\) is the diameter of the new pipe that we want to find.
04

Solve the equation for \(D_{new}\)

The equation from step 3 can be solved for \(D_{new}\) by rearranging terms and taking cube roots of both sides.\(D_{new}\) is the diameter of the new pipe.

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Most popular questions from this chapter

The following data are obtained for a particular reach of the Provo River in Utah: \(A=183 \mathrm{ft}^{2}\), frec-surface width \(=55 \mathrm{ft}\) average depth \(=33 \mathrm{ft}, R_{h}=3.32 \mathrm{ft}, V=6.56 \mathrm{ft} / \mathrm{s},\) length of reach \(=116 \mathrm{ft},\) and elevation drop of reach \(=1.04 \mathrm{ft}\). Determine (a) the average shear stress on the wetted perimeter, (b) the Manning coefficien. \(n,\) and (c) the Froude number of the flow.

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 \(E=0.6 \mathrm{m}\)

A rectangular brick-lined channel has a bottom slope of 0.0025 and is designed to carry a uniform water flow rate of \(300 \mathrm{ft}^{3} / \mathrm{s}\). Would the channel need fewer bricks if the channel were 2 ft wide, 6 ft wide, or \(10 \mathrm{ft}\) wide? Explain.

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.

Find the diameter required for reinforced concrete pipe laid at a slope of 0.001 and required to carry a uniform flow of \(19.3 \mathrm{ft}^{3} / \mathrm{sec}\) when the depth is \(75 \%\) of the diameter.
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