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Chapter 10: Problem 22

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.

Short Answer

Expert verified
The velocity at section 2 is \(10 ft/s\). The specific energy at the downstream (\(E_{2}\)) is \(2.55 ft\) and at the upstream (\(E_{1}\)) is \(2.75 ft\). The respective points are marked on the specific energy curve.

Step by step solution

01

Calculation of velocity at section 2

Using the flow-rate definition which is the product of area and velocity, one can find the velocity as \(v_{2} = \frac{q}{h_{2}} = \frac{10}{1} = 10 \: ft/s\)
02

Calculation of specific energy at section 2

The specific energy at a cross section of an open channel flow is defined as the total energy head with respect to the channel bottom. From the energy equation, we know that \(E = h + \frac{v^{2}}{2g}\). We substitute the known values to find the specific energy at section 2 (\(E_{2}\)) \(E_{2} = h_{2} + \frac{v_{2}^2}{2g} = 1+ \frac{(10)^{2}}{2*32.2} = 2.55 \: ft\)
03

Calculation of specific energy at section 1

The head loss is the difference in the total energy between two sections, expressed as \(h_{L} = E_{1} - E_{2}\). So, \(E_{1} = h_{L} + E_{2} = 0.2 + 2.55 = 2.75 \:ft\)
04

Plot the specific energy diagram

The specific energy curve is typically a parabola with energy on the y-axis and depth on the x-axis. Following this approach, plot the curve. On the curve, identify points for \(E_{1}\) and \(E_{2}\).

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Most popular questions from this chapter

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.

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.

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.

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.

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.
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