sm_ch17

Water Resources Engineering

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Unformatted text preview: 17 – 1 Chapter 17 17.3.1. Q = 45000 cfs, vertical upstream face ogee crest, bridge weir width = 24 in = 2 ft, K p = 0.05, K a = 0.10, bridge span (center to center) d 25 ft, H o = 10 ft H e t 0.75 ft, P = 40 ft Using the weir equation, 3/2 e CLH Q 3/2 e CH Q L This equation shows that L is minimum when H e is at its maximum. Thus, for the least cost, H e = H o = 10 ft. 4 10 40 H P o , 1 H H o e For 4 H P o , C o = 3.95 (from Figure 17.3.3) Hence, 3 . 360 ) 10 ( 95 . 3 45000 CH Q L 2 / 3 3/2 e ft Also, ¡ ¢ ¡ ¢ > @ N 3 . 362 ) 10 ( 10 . 0.05 N 2 3 . 360 H K NK 2 L L e a p ' £ £ £ £ £ Let the number of piers be 14. 14 3 . 362 L ' £ = 376.3 Span length = 1 . 25 1 14 3 . 376 £ ft > 25 ft Hence N = 15 is taken. The shape of the ogee crest is determined using equation (17.3.4): n o o H x K H y ¸ ¸ ¹ · ¨ ¨ © § ¤ 10 H h o a From Figure 17.3.2(b), n = 1.872 for vertical face ogee spillway and H h o a . K = 0.5. Hence, 1.872 872 . 1 x 01671 . 10 x-0.50(10) y ¤ ¸ ¹ · ¨ © § The following table gives the data points of the spillway surface profile downstream of the crest and up to a point at which x = 30 ft. x (ft) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 4.0 6.0 10.0 20.0 30.0 y (ft) 0.0 -0.018 -0.067 -0.143 -0.246 -0.373 -0.525 -0.899 -1.921 -4.997 -18.292 -39.075 17.3.2. Gate openings: i) D = 5 ft, ii) D = 10 ft. 4 ft d H d 18 ft, P = 20 ft, H e = 12 ft, L = 100 ft. 17 – 2 For both openings, both orifice flow and weir flow conditions may exist. i) D = 5 ft: For 4 d H d 5, Q = CLH 3/2 = C(100)H 3/2 = 100CH 3/2 For 5 < H d 18, 2gH C 500 2gH C(5)(100) 2gH CDL Q ii) D = 10 ft For 4 d H d 10, Q = CLH 3/2 = C(100)H 3/2 = 100CH 3/2 For 10 < H d 18, 2gH C 1000 2gH C(10)(100) 2gH CDL Q C = 0.68 ( T = 90 o ) from Figure 17.3.8 For 67 . 1 12 20 H P o , C o = 3.925 from Figure 17.3.3 The remaining computation is performed in the following table. Case Total head (ft) H e (ft) D (ft) C Q (cfs) C o o e H H o C C C Q (cfs) i 4 5 6 8 10 12 14 16 18 4 5 3.5 5.5 7.5 9.5 11.5 13.5 15.5 5 5 5 5 5 5 5 5 5 - - 0.68 0.68 0.68 0.68 0.68 0.68 0.68 5105 6399 7472 8410 9253 10025 10742 3.925 3.925 0.33 0.42 0.88 0.90 3.454 3.533 2763 3950 ii 4 6 8 10 12 14 16 18 4 6 8 10 7 9 11 13 10 10 10 10 10 10 10 10 - - - - 0.68 0.68 0.68 0.68 14438 16371 18099 19675 3.925 3.925 3.925 3.925 0.33 0.50 0.67 0.83 0.88 0.92 0.95 0.98 3.454 3.611 3.729 3.847 2763 5307 8438 12165 17.3.3. L = 100 ft, H o = 8 ft, P = 12 ft. a) 3/2 e CLH Q , 5 . 1 8 12 H P o For 5 . 1 H P o , C o = 3.92 (from Figure 17.3.3) At H e = H o , C = C o . Hence, 8870 8) 3.92(100)( Q 3/2 cfs. b) H e = 12 ft, 5 . 1 8 12 H H o e 17 – 3 For 5 . 1 H H o e , 08 . 1 C C o 234 . 4 ) 08 . 1 ( 92 . 3 C C C C o o ¸ ¸ ¹ · ¨ ¨ © § . Hence, Q = 4.234(100)(12) 3/2 = 17600 cfs....
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sm_ch17 - 17 – 1 Chapter 17 17.3.1. Q = 45000 cfs,...

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