(a) A wide rectangular channel flows 4 ft deep on a slope of 4 ft in 10,000 ft. Using the theoretical formula for velocity in Problem 10.4, calculate values of theoretical velocities at depth increments of 1/10 depth, assuming the channel is smooth.

(b) Compare the average of the 0.2 and 0.8 depth values of velocity with the 0.6 depth velocity.
Use kinematic viscosity as $1.50 \times 10^{-5} ft ^{2} / sec$ .

Question

(a) A wide rectangular channel flows 4 ft deep on a slope of 4 ft in 10,000 ft. Using the theoretical formula for velocity in Problem 10.4, calculate values of theoretical velocities at depth increments of 1/10 depth, assuming the channel is smooth.

(b) Compare the average of the 0.2 and 0.8 depth values of velocity with the 0.6 depth velocity.
Use kinematic viscosity as [latex]1.50 imes 10^{-5} ft ^{2} / sec[/latex].

Accepted Answer

[latex] ext { (a) Since } v_{*}=\sqrt{ au_{o} / ho}=\sqrt{g R S}=\sqrt{g h S} ext { and } y_{o}=v / 9 v_{*} [/latex],

[latex]v=2.5 v_{*} \ln \left(y / y_{o} ight)=(2.5)(2.303) \sqrt{g h S} \log \left(9 v_{*} y / v ight)[/latex].

[latex]=5.76 \sqrt{(32.2)(4)(0.0004)} \log \frac{9 y \sqrt{(32.2)(4)(0.0004)}}{1.5 imes 10^{-5}}[/latex].

[latex]=1.307 \log (136,188 y)[/latex] (A).

Using (A) we obtain the following values of velocity v:

v (ft/sec)	y (ft)	Dist. Down (%)
7.50	4.0	0
7.44	3.6	10.
7.37	3.2	20.
7.29	2.8	30.
7.21	2.4	40.
7.10	2.0	50.
6.98	1.6	60.
6.81	1.2	70.
6.58	0.8	80.
6.19	0.4	90.
6.03	0.3	92.5
5.80	0.2	95.0
5.40	0.1	97.5
4.10	0.01	99.75

[latex] ext { (b) Average of the } 0.2 ext { and } 0.8 ext { depth values is } V=\frac{1}{2}(7.37+6.58)=6.98 ft / sec [/latex].

The 0.6 depth value is 6.98 ft/sec. Seldom does such exact agreement occur.

Question 10.7: (a) A wide rectangular channel flows 4 ft deep on a slope of...

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