CE 4603: Water Resources Engineering ====== Homework #3 Due 2/9 ====== //Solve problems 5.2, 5.6, 5.11, 5.140, and 5.15 of Water Resources Engineering by David Chin, 2nd edition, ages 585-587. Hint: Problem 5.6, for Atlanta, using Fig. 5.5-5.7 you should get R101=2.7 in., R1024=5.7 in., R1001=3.75 in.// ====== 5.2 ====== n= 50 TR = (n+1)/m ^Rank (m) ^Return (yr)^ 5 ^ 10 ^ 15 ^ 20 ^ 25 ^ 30 ^ ^1 ^51| 26.2| 45.8| 60.5| 72.4| 81.8| 89.7| ^//(Interpoloated)//^ //40.0//| //25.8//| //45.0//| //59.5//| //71.2//| //80.4//| //88.2//| ^2 ^25.5| 25.3| 44.0| 58.1| 69.6| 78.6| 86.3| ^3 ^17| 24.2| 42.2| 55.8| 66.8| 75.5| 82.8| {{hw_3_5-2.jpg}} ^ Duration (min)^ 5 ^ 10 ^ 15 ^ 20 ^ 25 ^ 30 ^ ^ Intensity (mm/hr)| 309.74| 270.14| 237.86| 213.58| 193.01| 176.47| ====== 5.6 ====== x=R10,1/R10,24= 47% a1= 28.3 b1= 9.1 c1= 0.79 Tp= 9.49 a= 77.32 {{hw_3_5-6.jpg}} ^ Duration (min)^ 5^ 10^ 15^ 20^ 25^ 30^ ^ Intensity (in/hr)| 9.56| 7.52| 6.26| 5.39| 4.76| 4.27| i = a/(t+b1c1) ====== 5.11 ====== {{ hw_3_5-11.jpg?200}} td = 50 min = 50/60 hr i = 818/(td + 8.54)0.76) = 818/(50 + 8.54)0.76) = 37.18 mm/hr P50 = 37.18 mm/hr * 50/60 = 31mm tp/td = 3 tbar / tp -1 = 3 x 0.44 - 1 = 0.32 -> tp = td x 0.32 = 50/60 x 0.32 = 0.267 hr = tp ½iptd = 37.18 -> ip = 37.18/(½td) = 2 x 37.18/ (50/60) = 89 mm/hr = ip ====== 5.14 ====== a=64.1 b1=8.16 c1=0.76 T=10 t=50 min = 50/60 hr Δt=50/7 i=a/(t+b1)c1=64.1/(nΔt+8.16)0.76 {{hw_3_5-14.jpg}} ^Increment ^//T (min)// ^I (mm/hr) ^It (mm) ^Rainfall (mm) ^Intensity (mm/hr) ^//Ordered//^ |1| //7.14//| 8.06| 0.96| 0.96| 8.06| //0.16//| |2| //14.29//| 6.03| 1.43| 0.47| 1.99| //0.31//| |3| //21.43//| 4.88| 1.74| 0.31| 0.87| //0.87//| |4| //28.57//| 4.14| 1.97| 0.23| 0.48| //8.06//| |5| //35.71//| 3.62| 2.16| 0.18| 0.31| //1.99//| |6| //42.86//| 3.23| 2.31| 0.15| 0.21| //0.48//| |7| //50//| 2.92| 2.44| 0.13| 0.16| //0.21//| //Please note the text does not specify whether the values for the empirical formula for i yields mm/hr as assumed here, or in/hr.// ====== 5.15 ====== Storms dumping the same amount of rain in a shorter period of time would be even more intense, and hence less frequent. Assuming the same return period will tend to overestimate the frequency of these shorter-duration storms and is therefore a conservative estimate.