pro00060

pro00060 - More Practice Problems Related to Open Channels...

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Unformatted text preview: More Practice Problems Related to Open Channels CIVL 252, Spring 2002 1. Water flows in a rectangular, concrete, open channel that is 12.0 m wide at a depth of 2.5m. The channel slope is 0.0028. Find the water velocity and the flow rate. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29( 29 ( 29 ( 29 ( 29 s m Av Q s m v m p A R n s R n v w / 178 945 . 5 5 . 2 . 12 / 945 . 5 0028 . 765 . 1 ) 013 . / . 1 ( 765 . 1 5 . 2 . 12 5 . 2 / 5 . 2 . 12 / 013 . ) / . 1 ( 3 2 / 1 3 / 2 2 / 1 3 / 2 = = = = = = + + = = = = Water flows in the symmetrical trapezoidal channel lined with asphalt shown in Fig. 1 The channel bottom drops 0.1 ft vertically for every 100ft of length. What are the water velocity and flow rate? 2. Water is to flow at a rate of 30 m 3 /s in the concrete channel shown in Fig 2. Find the required vertical drop of the channel bottom per kilometer of length. ( 29 ( 29 ( 29 ( 29 [ ] ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29( 29 000746 . 236 . 1 ) 013 . / . 1 ( 419 . 2 236 . 1 3 . 10 / 40 . 12 / 03 . 10 6 . 1 6 . 1 6 . 3 . 2 . 4 6 . 2 6 . 3 / 419 . 2 40 . 12 / 30 / ) / . 1 ( 40 . 12 2 / 6 . 3 6 . 1 . 2 . 4 . 2 6 . 3 2 / 1 3 / 2 2 2 2 / 1 3 / 2 2 = = = = = = +- +- + + = = = = = = +- + = s s m p A R m p s m A Q s R n v m A w w This slope represents a drop of the channel bottom of 0.000746 m per meter of length, or 0.746 m per kilometer of length. 16ft 3 1 4.5ft 2 .0 m 1 .6 m 3 .6 m 4 .0 m 3. A 500-mm-diameter concrete pipe on a 1/500 slope is to carry water at a flow rate of 0.040 m 3 /s. Find the depth of flow. See Fig as below. ( 29 ( 29 ( 29 ( 29 ( 29 1 01163 . / , / sin 01163 . 500 1 013 . / . 1 / 040 . / 040 . / ) / . 1 ( 3 / 2 3 / 5 3 / 2 2 / . 1 3 / 2 2 / 1 3 / 2 = = = = = = = w w p A p A ceR AR R A A A Q s R n v Equation (1) contains two unknowns, A and p w ; however, both unknowns can be expressed in terms of the unknown depth of flow, d. The applicable area in this problem is the shaded area (AECDA) in Fig as below. AB=BC=0.25m (both area radii), BE = 0.25-d. Therefore, AE=CE= ( 29 ( 29 , 25 . 25 ....
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pro00060 - More Practice Problems Related to Open Channels...

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