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 |
| 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
| 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
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
| 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.
Discussion