Previously, it has been shown that, if a point outflow takes place at a intermediate position in a rectangular channel with an initially uniform flow, the depth at this point is uniquely determined by the magnitude of this outflow at any given time. Prove that for a nonrectangular channel the same

Question

Previously, it has been shown that, if a point outflow takes place at a intermediate position in a rectangular channel with an initially uniform flow, the depth at this point is uniquely determined by the magnitude of this outflow at any given time. Prove that for a nonrectangular channel the same is true, i.e., the relationship ΔQ = 2A([latex]w_{o}[/latex] − w) holds in which ΔQ is the point outflow. For a trapezoidal channel with b = 3 m, and m = 1.2, with an initial uniform flow of [latex]Y_{o}[/latex] = 2 m and [latex]Q_{o}[/latex] = 12 m³/s, make up a table that gives the depth, the velocity and the flow rate at this point, as well as these variables at positions x = 100 m, x = 200 m, x = −50 m, and x = −100 m for several time steps if the rate of change of the point outflow is d(ΔQ)/dt = 0.1 (m/s)² for 50 s and remains constant at Q = 5 m³/s, thereafter.

Accepted Answer

The solution of the requested flow variables involves three parts: (1) at the origin where the outflow takes place, (2) at the downstream positions, and (3) at the upstream positions.

At the origin, x = 0: Here, the point outflow ΔQ equals the difference between the upstream and the downstream flow rates, or ΔQ = [latex]Q_{u} − Q_{d}[/latex] = A([latex]V_{u} − V_{d}[/latex]) = A[[latex]V_{o} + w_{o}[/latex] − w − (w + [latex]V_{o} − w_{o}[/latex])]= 2A([latex]w_{o}[/latex] − w). (Note A and w do not have subscripts since the upstream and the downstream depths are the same at x = 0.) Thus, we have proven that for a given initial uniform flow in a nonrectangular channel that produces [latex]w_{o}[/latex] there is a depth that produces w that satisfies the equation F = .5ΔQ + A(w − [latex]w_{o}[/latex]) = 0. Therefore, at x = 0, this equation must be solved for a depth Y with ΔQ = {d(ΔQ)/dt}t computed for any time step until ΔQ not longer changes. Once Y is known, the other variables are computed from [latex]V_{d} = w + V_{o} - w_{o}[/latex], [latex]Q_{d} = AV_{d},[/latex] [latex]V_{u} = V_{o} + w_{o} − w, Q_{u}= AV_{u}.[/latex] The first table below gives these computed values for times using t = 10 s. For example at t = 40 s: Q=.1(40)= 4 m³/s; the solution of the above implicit equation gives Y = 1.927 m, and A=8.310 m², [latex]V_{d} =[/latex] 0.916 m/s, [latex]V_{u}[/latex] = 1.306 m/s, [latex]Q_{d}[/latex] = 9.4 m³/s, and [latex]Q_{u}[/latex] = 13.4 m³/s. (See a home work problem to write a computer program to carry out these computations.)

At downstream positions where x is positive: At these positions we seek the [latex]C^{+}[/latex] characteristic along which [latex]t_{1}[/latex] + Δt = t, in which [latex]t_{1}[/latex]= ΔQ/{d(ΔQ)/dt} and Δt = x/([latex]V_{d} + c_{d}[/latex]). Since, as shown above, there is a unique relationship between ΔQ and the depth, this equation is solved for Y, or the Newton Method obtains Y that satisfies: F = ΔQ/{d(ΔQ)/dt} + x/([latex]V_{d} + c_{d}[/latex]) − t = 0. Note, ΔQ in this equation is the outflow associated with the [latex]C^{+}[/latex] characteristics through the point (x, t) and is not the ΔQ at time t, thus in the tables below it is denoted by Δ[latex]Q_{c}.[/latex] For example, at t = 40 s and x = 100 m, the solution of the above implicit equation produces Y = 1.967 m, from which A = 8.577 m², V=1.023 m/s, Q=11.4 m³/s. The constant outflow along the [latex]C^{+}[/latex] characteristic through point (100, 40) of the xt plane is ΔQ = 1.865 m³/s. The second table below provides these values for t = 0, 10, 20,…, 70 s. Note that the initial uniform flow conditions exist prior to the time when the wave reaches the position x = 100 or x = 200 m. The position of the wave is shown in the last column of this table.

At upstream positions, where x is negative: At these positions we seek the [latex]C^{−}[/latex] characteristic (that is a straight line) along which Y, [latex]V_{u} , Q_{u},[/latex] and w are constant such that [latex]t_{1}[/latex] + Δt = t, in which [latex]t_{1}[/latex] = ΔQ/{d(ΔQ)/dt} and Δt = x/([latex]V_{u} − c_{u}[/latex]). Note that Δt is not the same as for downstream positions at time t, i.e., the depth that satisfies the equation F = ΔQ/{d(ΔQ)/dt} + x/([latex]V_{u} − c_{u}[/latex]) − t = 0 will not equal the depth that satisfies the downstream implicit equation for the same time t. This means that, the [latex]C^{−}[/latex] characteristic we are now seeking originates at a different time t from x = 0. For example, at t = 40 s and x = −50 m, the solution of the above implicit equation produces: Y = 1.965 m, from which [latex]V_{u}[/latex]= 1.204 m/s, [latex]Q_{u}[/latex] = 12.7 m³/s, [latex]c_{u}[/latex] = 3.66 m/s, and ΔQc = 1.963 m³/s. It should be noted that when the upstream flow becomes critical, or when |[latex]V_{u}[/latex]| = c, or [latex]Q^{2}_{u}[/latex]T /(gA³) = 1, it limits the amount of the point outflow.

[latex]Y_{o} = 2.000, Q_{o} = 12.0, V_{o}= 1.111, A_{o}= 10.800, c_{o} = 3.686, w_{o}[/latex] = 8.996

At the origin, x = 0

Note: While, it is not possible to integrate the stage variable w (or w′) in the closed form to obtain an algebraic equation that gives w′ as a function of y′, it is possible to take the data in Table 6.1 and fit it to a model by a least squared regression analysis. Once such an equation is available, it can be used instead of looking up values for w′ corresponding to y′, or by finding y′ corresponding to w′ from the table, to implicitly solve y′ for a given w′. Such a regression fit using a 6 degree polynomial using the data in Table 6.1 for a trapezoidal channel gives

w′ = 0.140785 + 5.254419y′−11.42496y′² +19.398805y′³ −18.48886y′[latex]^{4}[/latex]
+ 8.998029y′[latex]^{5}[/latex] −1.7405045y′[latex]^{6}[/latex]
Average absolute difference = 0.003468

The multiple regression correlation coefficient for this fit is R² = 0.9999, with the major deviations occurring for small values of y′, which would seldom be used. Thus, the above equation might be used to replace Table 6.1 especially in light of the approximations used in the theory and if changes in w′ are relative small.

More accurate estimates of w′ might be obtained by dividing Table 6.1 into several parts and performing a regression fit for each part of this table. Using three parts, Part 1 with y′ from 0.01 to 0.5080, Part 2 with y′ from 0.5100 to 1.008, and Part 3 with y′ from 1.010 to 1.508 gives the following three sixth order polynomial equations (with the average absolute difference between the w′ in the table and that given after each equation.)

Part 1 (y′ = 0.01 to y′ = 0.508)

w′ =0.07378655 + 8.18346421 y′ -48.52012353 y′² + 228.8879673y′³- 618.51160685y′[latex]^{4}[/latex]+864.136452 y′[latex]^{5}[/latex] - 483.52952655 y′[latex]^{6}[/latex]

Average absolute difference = 0.0012065 = 1.2065 × 10[latex]^{−3}[/latex]

Part 2 (y′ = 0.51 to y′ = 1.008)

w′ = 0.32126859 + 3.25239439y′ − 3.24689267y′² + 3.396349093y′³ - 2.739827y′[latex]^{4}[/latex] + 0.95440249y′[latex]^{5}[/latex] - 0.1664955 y′[latex]^{6}[/latex]

Average absolute difference = 0.000003748 = 3.7485 × 10[latex]^{−6}[/latex]

Part 3 (y′ = 1.01 to y′ = 1.508)

w′ = 0.39850342 + 2.74650447y′ − 1.85174341y′² +1.32995644y′³ − 0.64462275y′ [latex]^{4}[/latex] + 0.18036973y′[latex]^{5}[/latex] - 0.02192338 y′[latex]^{6}[/latex]

Average absolute difference = 0.00001001 = 1.0009 × 10[latex]^{−5}[/latex]

Fitting the data in Table 6.2, which follows, for a circular channel with a fourth degree polynomial equation produces a multiple regression correlation coefficient equally close to unity. This equation is

w′ = -0.3047431+ 6.0278815y′- 10.9985247y′² +11.9441034y′³ − 4.97205264y′ [latex]^{4}[/latex]

Average absolute difference = 0.01116496 = 1.116496 × 10[latex]^{−2}[/latex]

Using three parts: Part #1 y′ = 0 to y′ = .3206

w′ = 0.1951268 + 10.8143864y′- 61.1991232y′² + 206.7081028y′³− 261.8789472y′ [latex]^{4}[/latex]

Average absolute difference = 0.0067979 = 6.79793 × 10[latex]^{−2}[/latex]

Using three parts: Part #2 y′ = 0.322 to y′ = .6566

w′= 0.5187091 +3.9167468y′- 4.332116y′²+ 3.6181778y′³- 1.4055572y′ [latex]^{4}[/latex]

Average absolute difference = 0.0000079713 = 7.9713 × 10[latex]^{ 6}[/latex]

Using three parts: Part #3 y′ = .6580 to y′ = .9926

w′ = -0.5316203+ 9.2891735y′- 14.5119655y′² +12.0485209y′³− 3.9610566y′ [latex]^{4}[/latex]

Average absolute difference = 0.000086512 = 8.651189 × 10[latex]^{−5}[/latex]

Question 6.13: Previously, it has been shown that, if a point outflow takes...

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