sm_ch08

# Water Resources Engineering

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Unformatted text preview: 8 - 1 Chapter 8 8.3.1. A = 200 km 2 , base flow = 20 m 3 /s. Four hour period: 1 2 3 4 5 6 7 8 9 10 11 Rainfall (cm) 1.0 2.5 4.0 2.0 Storm flow (m 3 /s) 20 30 60 95 130 170 195 175 70 25 20 Direct runoff (m 3 /s) 0 10 40 75 110 150 175 155 50 5 0 From the computed direct runoff hydrograph, the volume of direct runoff, V d , can be computed as ¦ ¦ ' ' 10 1 n n 10 1 n n d Q t t Q V = 4(3600)(10 + 40 + 75 + 110 + 150 + 175 + 155 + 50 + 5) = 11.088 x 10 6 m 3 Excess rainfall depth, 0554 . 10 x 200 10 x 088 . 11 A V r 6 6 d d m = 5.54 cm This excess rainfall might have happened from the largest rainfall pulse, from the two largest rainfall pulses, from the three largest rainfall pulses, or from all the given rainfall pulses. First assume that it occurred from the largest rainfall pulse, R m , of 4.0 cm. If I is the infiltration rate, then, r d = R m – I' t 5.54 = 4 – I (4), which gives I = -0.385 cm/hr = -1.54 cm in 4 hours Since this value is unrealistic, the assumption that the rainfall excess occurred form the 4.0 in rainfall pulse must be wrong. Next assume that it occurred from the 2.5 and 4.0 cm rainfall pulses. Then, 5.54 = (2.5 + 4.0 – 2 I (4)), that is, I = 0.12 cm/hr = 0.48 cm in 4 hours. This value also doesn’t make sense since it is less than all the rainfall pulses. Now assume the rainfall excess occurred from three largest rainfall pulses. 5.54 = (2.5 + 4.0 + 2.0 – 3 I (4)), that is, I = 0.247 cm/hr = 0.99 cm ( | 1.0 cm) in 4 hours. Since this infiltration is less than the three largest rainfall pulses assumed to have caused the excess, and is almost equal to the fourth rainfall pulse, the assumption may be acceptable. Hence the rainfall excess hyetograph is 1.5 cm, 3.0 cm and 1.0 cm during the second, the third and the fourth 4 hour intervals, respectively. The 4 hour unit hydrograph can be easily determined using the deconvolution process. Q 1 = P 1 U 1 Q 2 = P 2 U 1 + P 1 U 2 Q 3 = P 3 U 1 + P 2 U 2 + P 1 U 3 Q 4 = P 4 U 1 + P 3 U 2 + P 2 U 3 Q 5 = P 4 U 2 + P 3 U 3 + P 2 U 4 . . . Therefore, U 1 can be determined from the first equation, U 2 from the second equation, and so on. The final result is given in the following table. 8 - 2 Four hour period 1 2 3 4 5 6 7 8 9 10 11 Rainfall (cm) 1.0 2.5 4.0 2.0 ERH (cm) 0.0 1.5 3.0 1.0 Storm flow (m 3 /s) 20 30 65 95 160 225 210 105 35 25 20 Direct runoff (m 3 /s) 0 10 40 75 110 150 175 155 50 5 0 Derived UH (m 3 /s) 0 6.7 13.3 18.9 26.7 34.1 30.7 19.1 10.3 5.0 0 8.3.2. The direct runoff computation is given in the following table. Direct runoff hydrographs corresponding to excess rainfall (m 3 /s) Time (hr) Unit hydrograph ordinate (m 3 /s) 2 3 0 1.5 Direct runoff (m 3 /s) 0 0 0 0 0 0 0 4 6.7 13.3 0.0 0 0 13.3 8 13.3 26.7 20.0 0 0 46.7 12 18.9 37.8 40.0 0 0 77.8 16 26.7 53.3 56.7 0 10.0 120.0 20 34.1 68.1 80.0 0 20.0 168.1 24 30.7 61.5 102.2 0 28.3 192.0 28 19.1 38.3 92.2 0 40.0 170.5 32 10.3 20.6 57.4 0 51.1 129.1 36 5.0 10.0 30.9 0 46.1 87.0 40 0 0 15.0 0 28.7 43.7 44 0 0 0 0 15.4 15....
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## This document was uploaded on 05/19/2011.

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sm_ch08 - 8 1 Chapter 8 8.3.1 A = 200 km 2 base flow = 20 m...

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