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# solution_set_assign04 - CIVE 486 Assignment 4 Solutions 1...

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CIVE 486 Assignment 4 Solutions 1 Problem 4.10 from text (p. 307): Use lag routing and hydrograph convolution to find the storm hydrograph at point 2. K = 20 min. The procedure to solve this problem is very similar to a question from the last assignment. The rainfall intensity is given (cm/hr), the 10-min unit hydrographs, and the total losses. Therefore convert the intensity into a UH multiplier and perform hydrograph convolution. Then add up the hydrographs from the three areas for a total hydrograph, lag by 20 min to get the output storm hydrograph. Subbasin A Time (min) Q (m 3 /s) 0.417 UH 1.5 UH 0.667 UH Total A 0 0 0 0 10 5 2.1 0 2.1 20 10 4.2 7.5 0 11.7 30 15 6.3 15.0 3.3 24.6 40 20 8.3 22.5 6.7 37.5 50 25 10.4 30.0 10.0 50.4 60 20 8.3 37.5 13.3 59.2 70 15 6.3 30.0 16.7 52.9 80 10 4.2 22.5 13.3 40.0 90 5 2.1 15.0 10.0 27.1 100 0 0 7.5 6.7 14.2 110 0 3.3 3.3 120 0 0 Subbasin B Time (min) Q (m 3 /s) 0.417 UH 1.5 UH 0.667 UH Total B 0 0 0 0 10 5 2.1 0 2.1 20 10 4.2 7.5 0 11.7 30 15 6.3 15.0 3.3 24.6 40 20 8.3 22.5 6.7 37.5 50 25 10.4 30.0 10.0 50.4 60 20 8.3 37.5 13.3 59.2 70 15 6.3 30.0 16.7 52.9 80 10 4.2 22.5 13.3 40.0 90 5 2.1 15.0 10.0 27.1 100 0 0 7.5 6.7 14.2 110 0 3.3 3.3 120 0 0 Subbasin C Time (min) Q (m 3 /s) 0.667 UH 1.667 UH 0.833 UH Total C 0 0 0 0 CIVE 486 Assignment 4 Solution Page 1 of 14

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10 16.7 11.1 0 11.1 20 33.4 22.3 27.8 0.0 50.1 30 50.0 33.4 55.7 13.9 102.9 40 33.4 22.3 83.4 27.8 133.5 50 16.7 11.1 55.7 41.7 108.5 60 0 0 27.8 27.8 55.7 70 0 13.9 13.9 80 0 0 Watershed Time (min) Outflow (m 3 /s) (20 min lag for A and B, normal C) 0 0.0 10 11.1 20 50.1 30 107.1 40 156.8 50 157.6 60 130.7 70 114.8 80 118.4 90 105.9 100 80.0 110 54.2 120 28.3 130 6.7 140 0.0 Storm Hydrograph at Point 2 0.0 20.0 40.0 60.0 80.0 100.0 120.0 140.0 160.0 180.0 0 50 100 150 Time (min) Flow (m 3 /s) out of 2, 1 for getting hydrographs right, 1 for proper lagging CIVE 486 Assignment 4 Solution Page 2 of 14
2 Problem 4.5 from text (p. 306) a) Develop a recursive relation using the continuity equation and S-Q relationship for a linear reservoir See attached hand solutions b) Storage rout the hydrograph through the reservoir using t = 1 hr. Here use the equation developed in part a, substituting in values of t, k and I ave to get the value “Q 2 ”. Initially, it is assumed that at t = 0, Q = 0. Therefore for the time step t = 1 hr, Q 1 (the previous time step’s outflow = 0), I ave is the average inflow between time = 0 and time = 1 (50 m 3 /s) and the constants are given. To calculate the storage, either substitute the Q value into S = KQ, or use S = (Iave-Oave)* t. k = 1.21 Time (hr) Inflow (m 3 /s) Ave Inflow (m 3 /s) Q (m 3 /s) S (m 3 ) 0 0 0 0 1 100 50 29 127000 2 200 150 100 435000 3 400 300 217 945000 4 300 350 295 1284000 5 200 250 269 1170000 6 100 150 199 868000 7 50 75 127 551000 8 0 25 67 293000 9 0 0 28 121000 10 0 0 12 50000 11 0 0 5 21000 12 0 0 2 9000 13 0 0 1 4000 14 0 0 0 1000 15 0 0 0 1000 16 0 0 0 0 CIVE 486 Assignment 4 Solution Page 3 of 14

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Inflow and Outflow From Reservoir (Q4.5)
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