Water flowing at the normal depth in a rectangular concrete channel that is 12.0 m wide encounters an obstruction, as shown in Fig. 10-12, causing the water level to rise above the normal depth at the obstruction and for some distance upstream. The water discharge is 126 m^3/s, and the channel

Question

Water flowing at the normal depth in a rectangular concrete channel that is 12.0 m wide encounters an obstruction, as shown in Fig. 10-12, causing the water level to rise above the normal depth at the obstruction and for some distance upstream. The water discharge is 126 m³/s, and the channel bottom slope is 0.00086. If the depth of water just upstream from the obstruction [latex]\left(y_{0}\right)[/latex] is 4.55 m, find the distance upstream to the point where the water surface is at the normal depth.

Accepted Answer

[latex]y_{c}=\sqrt[3]{\left(Q / b^{\prime} ight)^{2} / g}=\sqrt[3]{(126 / 12.0)^{2} / 9.81}=2.24 m[/latex].

[latex]V=Q / A=(1.0 / n) R^{2 / 3} S^{1 / 2}[/latex].

[latex]126 /(12.0 y)=(1.0 / 0.013)[12.0 y /(12.0+2 y)]^{2 / 3}(0.00086)^{1 / 2}[/latex].

[latex](2.256)[12.0 y /(12.0+2 y)]^{2 / 3}-10.5 / y=0[/latex].

A trial-and-error solution of this equation gives a value of y, the normal depth [latex]\left(y_{N} ight),[/latex], of 2.95 m.
Since [latex]y_{N}>y_{c}[/latex], the flow is subcritical, and computations should proceed upstream. The problem now is to determine the distance from the point where the depth is 4.55 m to the point upstream where the depth is 2.95 m. This will be done (arbitrarily) in 10 equal-depth increments of 0.16 m. The computations are given in the table below; they apply expression (A) of Problem 10.46. The summation of values in column 8 of the table (4568 m) gives the answer to the problem (i.e., the distance upstream from the obstruction to the point where the water surface is at the normal depth).

(8) L(m) [latex]\frac{[(4)+(1)]_{2}-[(4)+(1)]_{1}}{0.00086-(7)}[/latex]	(7) S[latex]\left[\frac{0.013 imes(3)}{(6)^{2 / 3}} ight]^{2}[/latex]	(6) [latex]R_{m}( m )[/latex]	(5) R(m)[latex]\frac{12.0 imes(1)}{12.0+2 imes(1)}[/latex]	(4) V²/2g(m) [latex]\frac{(2)^{2}}{2 imes 9.807}[/latex]	(3) [latex]V_{m}( m / s )[/latex]	(2) V(m/s)[latex]\frac{126}{12.0 imes(1)}[/latex]	(1) y(m)
			2.588	0.2716		2.308	4.55
-236	0.0002662	2.562			2.350
			2.535	0.2917		2.392	4.39
-243	0.0002946	2.508			2.437
			2.481	0.3141		2.482	4.23
-253	0.0003272	2.453			2.531
			2.425	0.3394		2.580	4.07
-266	0.0003654	2.396			2.633
			2.367	0.3676		2.685	3.91
-284	0.0004098	2.338			2.743
			2.308	0.3997		2.800	3.75
-311	0.0004626	2.277			2.863
			2.246	0.4362		2.925	3.59
-353	0.0005246	2.214			2.993
			2.182	0.4777		3.061	3.43
-429	0.0005989	2.150			3.136
			2.117	0.5257		3.211	3.27
-613	0.0006893	2.083			3.294
			2.048	0.5811		3.376	3.11
-1580	0.0007997	2.013			3.468
			1.978	0.6458		3.559	2.95
-4568 m

Question 10.50: Water flowing at the normal depth in a rectangular concrete ...

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