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Unformatted text preview: 16 – 1 Chapter 16 16.1.1. L = 100 m, n = 0.4, S = 0.02 m/m, T = 10 years, in SI units, K = 26.285 0.4 0.3 0.4 0.6 0.6 0.3 0.4 0.6 0.6 c i 38 . 777 (0.02) i (0.4) ) 26.285(100 S i n KL t in seconds. 0.4 c i 96 . 12 t in minutes Assume i, then compute t c and compare t c to t D (since t c = t D is assumed) from the IDF curve at T = 10 years. If both are not equal (close), take t D = t c and determine i. Using this value of i, determine a new t c . Repeat this process until t D = t c . The result of such a computation is shown in tabular form below. Assumed i (m/hr) Assumed i (in/hr) t D from IDF curve (min) Computed t c (min) 0.025 0.043 0.050 1.0 1.7 1.97 120 54 43 56.66 45.61 42.94 Thus, i = 1.97 in/hr is OK. t c = 43 minutes. 16.1.2. L = 200 m, n = 0.01, S = 0.003 m/m, design frequency = 10 years, K = 26.285 0.4 0.3 0.4 0.6 0.6 0.3 0.4 0.6 0.6 c i 61 . 227 (0.003) i (0.01) ) 26.285(200 S i n KL t in seconds. 0.4 c i 79 . 3 t in minutes, i in m/hr. This problem is solved in a similar way as problem 16.1.1. The result is given in the following table. Assumed i (m/hr) Assumed i (in/hr) t D from IDF curve (min) Computed t c (min) 0.050 0.107 0.117 0.119 2.0 4.2 4.6 4.7 43.0 12.6 9.3 8.9 12.6 9.3 8.9 8.9 Thus, t c = 43 minutes, for which i = 0.119 m/hr = 119 mm/hr. 16.1.3. L = 200 ft, n = 0.10, S = 0.005 ft/ft, design frequency = 25 years, K = 56 Durationintensity data: Duration (min) 5 10 25 30 60 Rainfall intensity (in/hr) 7.3 5.7 4.8 3.3 2.1 0.4 0.3 0.4 0.6 0.6 0.3 0.4 0.6 0.6 c i 21 . 1656 (0.005) i (0.10) 56(200) S i n KL t in seconds. 16 – 2 0.4 c i 60 . 27 t in minutes, i in in/hr. This problem is solved in a similar way as problems 16.1.1 and 16.1.2. The durationrainfall intensity relationship is first plotted on a loglog graph paper as shown. The computation result is given in the following table. Intensityduration curve for Colorado Springs (T = 25 years) 1 10 1 10 100 Duration (min) Intensity (in/hr) Assumed i (in/hr) t D from IDF curve (min) Computed t c (min) 3.0 4.5 4.6 35.0 17.8 15.1 17.8 15.1 15.0 Thus, t c = 15.0 minutes, for which i = 0.119 m/hr = 4.6 in/hr. 16.1.4. L = 400 ft, n = 0.45, ¡ ¢ 77 . D 12 t 57 i £ , S = 0.01 ft/ft Thus, ¡ ¢ 3 . 4 . 77 . D 6 . 6 . c ) 01 . ( 12 t 57 ) 45 . ( ) 400 ( 56 t » ¼ º « ¬ ª £ = 997.70(t D + 12) 0.31 (in seconds) t c = 16.63(t D + 12) 0.31 (in minutes) This equation can be solved iteratively, in essence, in a similar way to the solutions to problems 16.1.1 – 16.1.3. Assume t D value and compute t c , then let the new t D value equal to the computed t c value. Doing so results in t c = 63.6 min after a few iterations. 16.1.5. L = 500 ft, B = 32 ft, S = 0.005 ft/ft, n = 0.016, C = 0.9 0.4 0.3 0.4 0.6 0.6 0.3 0.4 0.6 0.6 c i 76 . 955 (0.005) i (0.016) 56(500) S i n KL t in seconds....
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