Initially, the depth upstream and downstream from a gate is 8 and 2 ft respectively. Suddenly, the gate is raised so as to produce a depth of Y2 = 3.5 ft downstream from the gate. Assuming that initially, a uniform flow exists upstream and downstream from the gate, use the characteristic to

Question

Initially, the depth upstream and downstream from a gate is 8 and 2 ft respectively. Suddenly, the gate is raised so as to produce a depth of [latex]Y_{2} [/latex]= 3.5 ft downstream from the gate. Assuming that initially, a uniform flow exists upstream and downstream from the gate, use the characteristic to determine the new depth [latex]Y_{1} [/latex]and the velocity [latex]V_{1} [/latex] immediately upstream from the gate, the velocity [latex]V_{2} [/latex] immediately downstream from the gate, the depth [latex]Y_{3}[/latex], and the velocity [latex]V_{3}[/latex] at the wave that will move downstream from the gate and the velocity of this wave w.

Accepted Answer

First, the energy equation 8 + [latex]q^{2}_{1}[/latex]/(2g8²)= 2 + [latex]q^{2}_{1}[/latex]/(2g2²) across the gate needs to be solved for [latex]q_{1}[/latex] = 40.603 cfs/ft. Next, solve [latex]V_{o1}[/latex] =[latex]q_{1}[/latex] /8 = 5.075 fps, and [latex]V_{o2}[/latex] = [latex]q_{1}[/latex]/2 = 20.302 fps. Now, the six equations can be solved for the six unknowns. Three separate TK-Solver models are given below to solve this problem. The first solves all six equations simultaneously; the second solves the three equations that govern the flow upstream from the gate; and the third, the three equations that govern the flow downstream from the gate. Note, the two approaches yield the same solution: [latex]Y_{1}[/latex] = 6.144 ft, [latex]V_{1}[/latex] = 9.045 fps, [latex]V_{2}[/latex] = 15.877 fps, [latex]V_{3}[/latex] = 20.890 fps, [latex]Y_{3}[/latex] = 2.096 ft, and w = 28.613 fps. The flow rate after the gate is raised is [latex]q_{2} = V_{1}Y_{1}[/latex] = (9.045) (6.144) = 55.569 cfs/ft or [latex]q_{2} = V_{2}Y_{2}[/latex] = (15.877)(3.5) = 55.569 cfs/ft, or the increased flow rate is Δq = 14.966 cfs/ft. To the moving observer, the flow is supercritical downstream from the wave and the subcritical upstream therefrom, i.e., [latex]F^{2}_{ro2}=[/latex](w-[latex]V_{o2}[/latex])²/(g[latex]Y_{o2}[/latex])=1.0727 and [latex]F^{2}_{r3}[/latex]=(w-[latex]V_{3}[/latex])²/(g[latex]Y_{3}[/latex])=0.9327. (See Homework Problem 6.103 for a Fortran Program to solve this problem.)

Question 6.21: Initially, the depth upstream and downstream from a gate is ...

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