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Chapter 10: Problem 44

Find the discharge per unit width for a wide channel having a bottom slope of \(0.00015 .\) The normal depth is \(0.003 \mathrm{m}\). Assume laminar flow and justify the assumption. The fluid is \(20^{\circ} \mathrm{C}\) water.

Short Answer

Expert verified
The discharge per unit width is \(0.000064 m^2/s\). Verification of the Reynolds number, which is 63.38, justifies the assumption of laminar flow since its value is less than 2000.

Step by step solution

01

Determine the Hydraulic Radius

The hydraulic radius (R) for a wide channel, where the width (b) is much greater than the depth (y), can be approximated by using the normal depth. Hence, \(R = y = 0.003\) m.
02

Calculate the Reynolds Number

The Reynolds number (Re) can be calculated using the formula \(Re = \frac{{V * d}}{{\nu}}\), where V is the average velocity, d is the hydraulic diameter and \(\nu\) is the kinematic viscosity. But since we do not have V, we initially cannot validate the laminar flow assumption. So move to calculating the discharge first.
03

Calculate the Roughness Coefficient

The roughness coefficient \(n\) for water at \(20^{\circ} \mathrm{C}\) can be assumed to be \(0.01\).
04

Calculate the Discharge per Unit Width

The discharge (Q) per unit width for a wide channel is given by the Manning's equation \(Q = \frac{1}{n} R^{2/3} S^{1/2}\), where \(S\) is the bottom slope and \(n\) is the roughness coefficient. Substituting for \(R\), \(S\) and \(n\), we get \(Q = \frac{1}{0.01} * (0.003)^{2/3} * (0.00015)^{1/2} = 0.000064\) m/s.
05

Recalculate the Reynolds Number

Now that we have the discharge per unit width (Q), we can infer the average velocity (V) since \(Q= V * y\). Thus, \(V=Q/y = 0.000064 m/s / 0.003 m = 0.0213 m/s \). Now we can calculate Reynolds number \(Re = \frac{{V * d}}{{\nu}}\). Generally, for water at \(20^{\circ} \mathrm{C}\), \(\nu = 1.004 * 10^{-6} m^2/s\). Hence, \(Re = \frac{{0.0213 * 0.003}}{{1.004 * 10^{-6}}} = 63.38\). Since Re is less than 2000, the assumption of laminar flow is valid.

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Most popular questions from this chapter

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?

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.

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 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.

A rectangular channel \(3.0 \mathrm{m}\) wide has a flow rate of 5.0 \(\mathrm{m}^{3} / \mathrm{s}\) with a normal depth of \(0.50 \mathrm{m} .\) The flow then encounters a dan that rises \(0.25 \mathrm{m}\) above the channel bottom. Will a hydraulic jump occur? Justify your answer.
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