Chapt10

1051 approaches zero so small step sizes are required

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Unformatted text preview: Is the 4-ft depth position upstream or downstream in Fig. E10.8a? Solution In Example 10.4 we computed yc 4.27 ft. Since our initial depth y 3 ft is less than yc, we know the flow is supercritical. Let us also compute the normal depth for the given slope S0 by setting q 50 ft3/(s ft) in the Chézy formula (10.19) with Rh yn: q | v v Solve for: | e-Text Main Menu n 2/3 1/2 ARh S0 1.486 2/3 [yn(1 ft)]yn (0.0048)1/2 0.022 yn | 50 ft3/(s ft) 4.14 ft Textbook Table of Contents | Study Guide Chapter 10 Open-Channel Flow y= c 4.27 ft y= n 4.14 ft y0 = 3 ft y L = 4 ft S0 = 0.0048 L=? x=0 E10.8a x=L Thus both y(0) 3 ft and y(L) 4 ft are less than yn, which is less than yc, so we must be on an S-3 curve, as in Fig. 10.14a. For a wide channel, Eq. (10.51) reduces to dy dx S0 n2q2/( 2y10/3) 1 q2/(gy3) (0.022)2(50)2/(2.208y10/3) 1 (50)2/(32.2y3) 0.0048 with y(0) 3 ft The initial slope is y (0) 0.00494, and a step size x 5 ft would cause a change (0.00494)(5 ft) 0.025 ft, less than 1 percent. We therefore integ...
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