The flow upstream from a gate in a rectangular channel has been at a depth of Yo = 4 ft and a velocity of Vo = 5 fps for some time, when suddenly the gate is dropped so as to produce a depth of 0.8 ft downstream from it and a surge upstream from it. Determine the depth and the velocity immediately

Question

The flow upstream from a gate in a rectangular channel has been at a depth of $Y_{o}$ = 4 ft and a velocity of $V_{o}$ = 5 fps for some time, when suddenly the gate is dropped so as to produce a depth of 0.8 ft downstream from it and a surge upstream from it. Determine the depth and the velocity immediately upstream from the gate and the speed of the surge. After this initial drop, the gate is gradually closed further so as to decrease the flow past the gate at a rate dq/dt = 0.01 (ft/s)².

Use the methods of this chapter to approximately track the movement of the surge and the depth upstream from the gate.

Accepted Answer

First, a steady-state solution is called for based on the continuity and the momentum equations from the viewpoint of a moving observer, as is described in Chapter 3. Designating the steady-state upstream depth and the velocity by an s subscript, the three equations that solve [latex]Y_{2}, V_{2}[/latex], and v are

[latex]F_{1} =\left(\left| V_{2}\right|+ v\right) Y_{2}-\left(\left|V_{s}\right|+ v\right) Y_{s}=0[/latex] (1)

[latex]F_{2} =.5\left(\frac{Y_{2}}{Y_{s}}\right)\left(\frac{Y_{2}}{Y_{s}}+1\right)-\frac{\left(\left|V_{s}\right|+ v\right)^{2}}{\left(gY_{s}\right) }=0[/latex] (2)

[latex]F_{3} =Y_{2}+\frac{V^{2}_{2}}{\left(2g\right)}-Y_{d}-\frac{\left(Y_{2}V_{2}\right)^{2} }{\left(2gY^{ 2}_{d}\right)}=0[/latex] (3)

The solution gives: [latex]Y_{2}[/latex] = 4.848 ft, [latex]V_{2}[/latex] = 2.701 fps, and v = 8.140 fps, and can be obtained using any software that is designed to solve systems of nonlinear algebraic equations.

To apply the methods of this chapter to the unsteady flow assume that they are applicable to the expanding region between the moving surge and the gate and that the initial depth and the velocity solved above, i.e., the [latex]Y_{2}[/latex] and [latex]V_{2}[/latex] represent [latex]Y_{o}[/latex] = 4.848 ft and [latex]V_{o}[/latex] = −2.701 fps, i.e., the initial conditions for the unsteady solution and those that have been used in the equations of this chapter. The equations that are available are the first two equations above, i.e., the continuity and the momentum equations from a moving observer’s viewpoint, plus the equation that indicates that V − 2c is constant along negative characteristics, the equation that indicates that t =[latex] t_{1}[/latex] + Δt, that the new flow rate past the gate [latex] q_{d} = q_{o}[/latex] + (dq/dt)t, and finally again, the energy across the gate. In other words, there are six available equations for each new time step. These equations are

[latex]F_{1} =\left(\left| V_{2}\right|+ v\right) Y_{2}-\left(\left|V_{s}\right|+ v\right) Y_{s}=0[/latex] (4)

[latex]F_{2} =.5\left(\frac{Y_{2}}{Y_{s}}\right)\left(\frac{Y_{2}}{Y_{s}}+1\right)-\frac{\left(\left|V_{s}\right|+ v\right)^{2}}{\left(gY_{s}\right) }=0[/latex] (5)

[latex]F_{3} =V_{2}-2\left(gY_{2}\right)^{{1}/{2}}-V_{o}+2\left(gY_{o}\right)^{{1}/{2}}=0 [/latex] (6)

[latex]F_{4} =t-\frac{\left(V_{2}Y_{2}-q_{o} \right) }{\left|{dq}/{dt}\right| }-\frac{x}{\left\{3\left(gY_{2}\right)^{{1}/{2}}+V_{o}-2\left(gY_{o}\right)^{{1}/{2}}\right\}}=0[/latex] (7)

[latex]F_{5} =V_{o}Y_{o}-q_{d}=0 \left( in which q_{d}= q_{o}-\left| \frac{dq}{dt}\right|t\right) [/latex](8)

[latex]F_{6} =Y_{o}+\frac{V^{2}_{o}}{\left(2g\right) }-Y_{d}-\frac{\left( {q_{d}}/{Y_{d}}\right)}{\left(2g\right) }=0[/latex] (9)

These six equations can be repeatedly solved for each new time step for the following six variables: [latex]Y_{2}, V_{2}, v, Y_{o}, V_{o},[/latex] and [latex]Y_{d}.[/latex] (Note that [latex]Y_{2}[/latex] and [latex]V_{2}[/latex] from the above solution, for after the instant drop of the gate, have been renamed as [latex]Y_{o}[/latex] and [latex]V_{o}[/latex] to be consistent with the notation used for the initial uniform flow conditions, but also that these variables actually change as the gate is closed further to provide the specified dq/dt.) The position [latex]x_{s}[/latex] of the moving surge can be determined by multiplying the speed of the surge v by the time increment, or x = Σ vΔt.

The above solution procedures are implemented in the FORTRAN program SURGMO6. FOR, the listing of which is given below. This program first solves the three Equations 1, 2, and 3 that describe the beginning surge size, the velocity, etc., by the Newton method. Thereafter, for each time step, (The DO 80 K=1,NT) the six Equations 4 through 9 are solved using the Newton method giving [latex]Y_{2}, V_{2}, v, Y_{o}, V_{o},[/latex] and [latex]Y_{d}[/latex] for that time step. Note that before solving these equations that include Equations 6 and 7 from the characteristics, [latex]V_{o} [/latex] and [latex]V_{2}[/latex] are made negative to reflect that the x axis is pointing upstream. [latex]q_{o} [/latex] is also negative. The product of [latex]Y_{2}[/latex] and [latex]V_{2}[/latex] obtain from the solution of Equations 1 through 3. dq/dt is read as a positive value, and therefore to make [latex]q_{d}[/latex] less negative than [latex]q_{o} [/latex] its value is multiplied by time t which is added to [latex]q_{o} [/latex].

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