problem 1- tabulated below are total rainfall intensities during each hour of a frontal storm over a...
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Problem 1-
Tabulated below are total rainfall intensities during each hour of a frontal storm over a drainage basin.
HourRainfall
intensity1 0.412 0.493 0.224 0.315 0.226 0.087 0.078 0.099 0.0810 0.0611 0.1112 0.1213 0.1514 0.2315 0.2816 0.2617 0.2118 0.0919 0.0720 0.0621 0.0322 0.0223 0.0124 0.01
Problem 1-
a. Plot the rainfall hyetograph (intensity versus time).
b. Determine the total storm precipitation amount in inches.
Problem 1- continued
c. If the net storm rain is 2.00 in., determine the exact index (in/hr) for the drainage basin. (Note that by definition the area under the hyetograph above the index line must be 2.00 in.)
d. Determine the area of the drainage basin (acres ) if the net rain is 2.00 in. and the measured volume of direct surface runoff is 2,015 cfs-hr.
Problem 1- continued
e. Using the index calculated in part c, determine the volume of direct surface runoff (acre-ft) that would result from the following storm:
Time in Hour 1 2 3 4
Rainfall intensity (in/hr)
0.40 0.05 0.30 0.20
Solution:
The rainfall hyetograph (intensity versus time).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 240.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
Chart Title
Time (hr)
Ra
infa
ll i
nte
nsit
y i
n/h
r
b- The total storm precipitation amount in inches is the summation of total rainfall = ∑ i.t
Where :i rainfall intensityt incremental time= 3.68 inch
HourRainfall
intensityRainfall depth
1 0.41 0.412 0.49 0.493 0.22 0.224 0.31 0.315 0.22 0.226 0.08 0.087 0.07 0.078 0.09 0.099 0.08 0.0810 0.06 0.0611 0.11 0.1112 0.12 0.1213 0.15 0.1514 0.23 0.2315 0.28 0.2816 0.26 0.2617 0.21 0.2118 0.09 0.0919 0.07 0.0720 0.06 0.0621 0.03 0.0322 0.02 0.0223 0.01 0.0124 0.01 0.01
3.68
If the net storm rain is 2.00 inch, determine the exact index (in/hr) for the drainage basin. (Note that by definition the area under the hyetograph above the index line must be 2.00 in.)
Answer is illustrated in the following table.
HourRainfall
intensityf index =0.07
f index =0.070
f index =0.080
f index =0.080
f index =0.085
f index =0.085
1 0.41 0.340 0.340 0.330 0.330 0.325 0.3252 0.49 0.420 0.420 0.410 0.410 0.405 0.4053 0.22 0.150 0.150 0.140 0.140 0.135 0.1354 0.31 0.240 0.240 0.230 0.230 0.225 0.2255 0.22 0.150 0.150 0.140 0.140 0.135 0.1356 0.08 0.010 0.010 0.000 0.000 -0.0057 0.07 0.000 0.000 -0.010 -0.0158 0.09 0.020 0.020 0.010 0.010 0.005 0.0059 0.08 0.010 0.010 0.000 0.000 -0.00510 0.06 -0.010 -0.020 -0.02511 0.11 0.040 0.040 0.030 0.030 0.025 0.02512 0.12 0.050 0.050 0.040 0.040 0.035 0.03513 0.15 0.080 0.080 0.070 0.070 0.065 0.06514 0.23 0.160 0.160 0.150 0.150 0.145 0.14515 0.28 0.210 0.210 0.200 0.200 0.195 0.19516 0.26 0.190 0.190 0.180 0.180 0.175 0.17517 0.21 0.140 0.140 0.130 0.130 0.125 0.12518 0.09 0.020 0.020 0.010 0.010 0.005 0.00519 0.07 0.000 0.000 -0.010 -0.01520 0.06 -0.010 -0.020 -0.02521 0.03 -0.040 -0.050 -0.05522 0.02 -0.050 -0.060 -0.06523 0.01 -0.060 -0.070 -0.07524 0.01 -0.060 -0.070 -0.07525 3.68 2.000 2.230 1.760 2.070 1.640 2.000
Determine the area of the drainage basin (acres ) if the net rain is 2.00 in. and the measured volume of direct surface runoff is 2,015 cfs-hr.
Volume of the direct surface runoff = 2015 X 60 X 60 = 7254000 cf
Volume of the direct surface runoff = area of the drainage basin X net rain
Area of the drainage basin = 7254000 /(2/12)=43524000 ft2
The acre is equivalent to 43,560 square feet (approximately 4,047 m2)
Area of the drainage basin in acre =43524000/43560= 999.17 acre
Using the index calculated in part c, determine the volume of direct surface runoff (acre-ft) that would result from the following storm:
The phi index was calculated and found it to be 0.085inch
The summation net rain fall depth = (0.315 x 1) + (0.0 x 1) + (0.215 x 1)
+ (0.115 x 1) =0.645 inch
Time in Hour 1 2 3 4
Rainfall intensity (in/hr) 0.40 0.05 0.30 0.20
Net rain fall intensity (in/hr) =rain fall –phi
0.315 0 0.215 0.115
Using the index calculated in part c, determine the volume of direct surface runoff (acre-ft) that would result from the following storm:
The direct surface runoff volume= area x net
rainfall depth= 999.17 x
0.645/12= 53.7 acre-ft
Problem 2-
The infiltration capacity fp is computed in col. 2 of The next table at various times by the given formula.
t fp ∆ t f average ∆ F F cumulation
0 11.66
10 8.93 1.49 1.49
10 6.21
10 4.85 0.81 2.30
20 3.50
10 2.83 0.47 2.77
30 2.16
10 1.82 0.30 3.07
40 1.49
10 1.32 0.22 3.29
50 1.16
10 1.07 0.18 3.47
60 0.99
10 0.95 0.16 3.63
70 0.91
To prepare the revised infiltration curve, the infiltration is cumulated in col. 6 of attached Table for various time intervals.
The infiltration capacity at time zero (col. 2) is plotted against the zero cumulated infiltration, and the infiltration capacity of each subsequent period is plotted against the cumulated infiltration (col. 6) of the corresponding period in Figure , i.e., fp of 11.66 plotted against F of 0.0, fp of 6.21 plotted
against F of 1.49, and so on.
0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.000.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Cumulatedinfiltration curve
F, Cumulated infiltration, in'
fp,
infi
ltera
tion c
apacit
y,
inch/h
From Figure above, corresponding to F= l.08 in.
fp is 7.5 in/hr
Revised Infiltration Capacity for the Storm
Computations of Rainfall Excess by the Horton Method
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