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.