A trapezoidal channel with a bottom width b = 10 ft, and a side slope of m = 1.0 initially has a uniform depth of Yo = 4.2 ft, and a velocity Vo = −3.517 fps. (The velocity is negative because the origin for the unsteady solution is at the downstream end of the channel.) The gate at the downstream

Question

A trapezoidal channel with a bottom width b = 10 ft, and a side slope of m = 1.0 initially has a uniform depth of [latex]Y_{o} [/latex]= 4.2 ft, and a velocity [latex]V_{o} [/latex] = −3.517 fps. (The velocity is negative because the origin for the unsteady solution is at the downstream end of the channel.) The gate at the downstream end of this channel is suddenly opened allowing the channel to discharge freely into a reservoir whose water surface is well below the water in the channel. Determine the flow rate into the reservoir.

Accepted Answer

First, compute [latex]Y^{\prime }_{o}[/latex]= (1)(4.2)/10 = 0.42 and entering Table 6.1 with this value gives [latex]w^{\prime }_{o}[/latex] = 1.304, and [latex]w_{o}[/latex](gb/ m)[latex]^{{1}/{2}}[/latex] [latex]w^{\prime }_{o}[/latex] =17.944(1.304)=23.399fps. The dimensionless initial velocity is [latex]V^{\prime }_{o}[/latex]= [latex]V_{o}[/latex]/(gb/ m)[latex]^{{1}/{2}}[/latex] = −3.517/17.944 = −0.196 , or [latex]w^{\prime }_{o}− V^{\prime }_{o}[/latex] = 1.304 + 0.196 = 1.5 . Thus, we seek a dimensionless depth Y′ so that the corresponding w′ + c′ = 1.5. Trying several values of Y′ in Table 6.1 results in the following:

Thus, the minimum depth when the maximum flow rate occurs is [latex]Y_{min} [/latex]= (b/m)[latex]Y^{\prime }_{min}[/latex] = 2.74 ft, the corresponding velocity is [latex]V_{max}[/latex]= w + [latex]V_{o} − w_{o}[/latex] = 17.944(1.025) − 3.517 − 23.399 = −8.523 fps and the maximum flow rate is Q = VA = −8.523(10 + 2.74)2.74 = −8.523(34.908) = 297.5 cfs. Note that if 1.5 is entered on the abscissa on the first graph of Figure 6.3a Y′ = .27 can be read. As a useful learning exercise, you should program Equation 6.42 (or Equation 6.43) into an equation solver such as that available in MathCad, TK-Solver or MATLAB and by iterative providing values of w by using Table 6.1, find the values that result in this equation equaling zero.

c+(w+[latex]V_{o}-w_{o}[/latex]){1}=0 or

w+c = [latex]w_{o}-V_{o}[/latex] (6.42)

w′+c′=[latex]w^{\prime }_{o}-V^{\prime }_{o}[/latex] (6.43)

Question 6.15: A trapezoidal channel with a bottom width b = 10 ft, and a s...

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