Chapter 15 Varied Flow in Open Channels Problem 15.1 Water ows in a circular concrete pipe (Manning s = 0 012) with a depth that is half of the pipe diameter (0.8 m). If the slope is 0.004, nd the ow rate. Solution The ow rate is obtained from the Chezy equation. 1 0 2 3 1 2 =
The ow area is 1 = × × 0 82 = 0 251 m2 2 4 The wetted perimeter is
= × 0 8 = 1 26 m 2 131 132 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS The hydraulic radius is 0 251 = = = 0 2 m 1 26 The ow rate is 1 = × 0 251 × 0 22 3 × 0 0041 2 0 012 = 0 45 m3 s 133 Problem 15.2 A troweled concrete ( = 0 012) open channel has a cross-section as shown. The discharge is 400 cfs. The drop of the channel is 10 ft in each horizontal mile (5280 ft). Find the depth of the ow, Solution The ow rate in traditional units is 1 49 2 3 1 2 =
The slope is 10 = = 0 00189 5280 Thus 0 012 × 400 2 3 = = = 74 1 ft8 3 1 2 1 49 1 49 × 0 001891 2 The ow area in terms of depth is = (10 + ) The wetted perimeter is
= 10 + 2 2 so the hydraulic radius is (10 + ) = =
10 + 2 2 Thus (10 + )5 3 5 3 2 3 = = 74 1 (10 + 2 2 )2 3 or (10 + ) = 13 24 (10 + 2 2 )2 5 Solving this equation by iteration gives = 3 26 ft 134 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS Problem 15.3 Find the ow rate in the channel and overbank area that is shown in the following gure. The slope of the channel is 0.001, and the depth in the overbank area is 2 m. The Manning s is 0.04 in the overbank area and 0.03 in the main channel. All the channel sides have a 1:1 slope. Solution The discharge is given by the Chezy equation. 1 0 2 3 1 2 =
For the overbank area = 2 × (50 + 2) × = (100 + ) where is the depth in the overbank area. So when = 2 m, the area is = 204 m2 The wetted perimeter is
= 2 × (50 + 2 )
= 100 + 2 2 So when = 2 m, the wetted perimeter is = 105 6 m. The hydraulic radius for the overbank area is 204 = = = 1 93 m 105 7 For the main channel = 10 + 2 × 5( 5) + 2 × 5 × 5 2 = 20 25 135 With the main channel depth being 7 m, the ow area is 115 m2 The wetted perimeter is
= 10 + 2 × 2 × 5 = 24 1 m so the hydraulic radius of the main channel is 115 = = = 4 77 m 24 1 The ow rate is the sum in each area. 1 1 = × 204 × 1 932 3 × 0 0011 2 + × 115 × 4 772 3 × 0 0011 2 0 04 0 03 = 250 0 + 343 5 = 593 5 m3 s Problem 15.4 Water with a depth of 15 cm and a speed of 6 m/s ows through a rectangular channel. Determine if the ow is critical, subcritical, or supercritical. If appropri- ate, determine the alternative depth. Solution The nature of the ow is determined by the Froude number.
=
6 m/s = rÅ‚ ´ 9 8 m/s2 (0 15 m) = 4 95 Since 1, the ow is supercritical. To nd the alternative depth, note that the speci c energy of subcritical and supercritical ow are the same. µ Å› µ Å› 2 2
+ = + (1) 2 2 1 2 µ Å› 62 = 0 15 + 2 × 9 8 = 1 99 m where subscripts 1 and 2 denote sub- and supercritical, respectively. To solve Eq. (1) for subcritical depth ( 1), speed is needed. The continuity principle gives ( )1 = ( )2 ( )1 = ( )2 136 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS so 2 1 = 2 (2) 1 0 15 = 6 1 0 9 = 1 Combining Eqs. (1) and (2) gives µ Å› 2
+ = 1 99 m 2 1 2 0 92 1 1 + = 1 99 2 × 9 8 or 3 2 1 1 99 1 + 0 04133 = 0 We solved for the roots of this cubic equation using a computer program (MathCad). The solution has three roots: 1 = ( 0 139 0 15 1 979 m). Thus the alternate depth is 1 = 1 979 m 137 Problem 15.5 Water ows at a uniform rate of 400 cfs through a rectangular channel that has a slope of 0.007 and a width of 25 ft. The channel sides are concrete with a rough- ness factor of = 0 015 Determine depth of ow, and whether the ow is critical, subcritical, or supercritical. Solution The nature of the ow is determined by the Froude number.
=
To nd the depth , we can use Manning s equation. 1 49 3 2 1 2 =
µ Å›3 2 1 49 25 400 = (25 × ) (0 007)1 2 0 015 25 + 2 To solve for in Manning s equation, we used a computer program (MathCad) to nd a root for an equation of the form ( ) = 0 The result is = 1 38 ft Discharge is = = Ä„ ó 400 ft3 s = (1 38 × 25 ft2) so = 11 59 ft/s. The Froude number is
=
11 59 ft/s = p (32 2 ft s2) (1 38 ft) = 1 74 Thus ow is supercritical 138 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS Problem 15.6 Water ows in a rectangular channel that ends in a free outfall. The channel has a slope of 0.005, a width of 20 ft, and a depth at the brink of 2 ft. Find the discharge in the channel. Solution A sketch of the situation is At the brink, the depth is 71% of critical depth. 2 1 = 0 071 2 ft = 0 71 = 2 82 ft At section 1, the ow is critical, so the Froude number is 1.0.
1 0 = (1)
From continuity = (2) Combining Eqs. (1) and (2) gives p = 3 rÅ‚ ´ Ä„ ó = 32 2 ft/s2 2 823 ft3 = 26 9 ft2 s 139 Thus = Ä„ ó = 26 9 ft2 s (20 ft) = 538 cfs Problem 15.7 Water ows with an upstream velocity of 6 ft/s and a depth of 12 ft in a rec- tangular open channel. The water passes over a gradual 18-in. upstep. Determine the depth of the water and the change in surface level downstream of the upstep. Solution Assuming no energy losses, the speci c energy is constant across the upstep. 12 22 1 + = 2 + + (1) 2 2 The continuity principle is 1 1 = 2 2 So 1 2 = 1 (2) 2 Combining Eqs. (1) and (2) µ Å›2 2 2 1 1 1 1 + = 2 + + 2 2 2 µ Å› µ Å› µ Å›2 62 ft2 s2 62 ft2 s2 12 ft 12 ft + = 2 + + 18 12 ft 2 × 32 2 ft s2 2 × 32 2 ft s2 2 So 80 50 11 06 = 2 + 2 2 or 3 2 2 11 06 2 + 80 5 = 0 140 CHAPTER 15. VARIED FLOW IN OPEN CHANNELS Solving this cubic equation using a computer program (MathCad) gives three roots 2 = (-2.442, 3.2, 10.30). The negative root is not possible, and the small root (supercritial ow) is unlikely. Thus, the depth of water at section 2 is 2 = 10.3 ft The elevation of the water surface at section 2 is the sum of the depth of the water and the height of the upstep. 2 = 2 + = 10 3 ft + 1 5 ft = 11.8 ft