Problem 1- Tabulated below are total rainfall intensities during each hour of a frontal storm over a drainage basin.

Hour Rainfall intensity 1 0.41 2 0.49 3 0.22 4 0.31 5 6 0.08 7 0.07 8 0.09 9 10 0.06 11 0.11 12 0.12 13 0.15 14 0.23 15 0.28 16 0.26 17 0.21 18 19 20 21 0.03 22 0.02 23 0.01 24

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).

b- The total storm precipitation amount in inches is the summation of total rainfall = ∑ i.t Where : i rainfall intensity t incremental time = 3.68 inch

Hour Rainfall intensity Rainfall depth 1 0.41 2 0.49 3 0.22 4 0.31 5 6 0.08 7 0.07 8 0.09 9 10 0.06 11 0.11 12 0.12 13 0.15 14 0.23 15 0.28 16 0.26 17 0.21 18 19 20 21 0.03 22 0.02 23 0.01 24 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.

Hour Rainfall intensity f index =0.07 f index =0.070 f index =0.080 f index =0.085 1 0.41 0.340 0.330 0.325 2 0.49 0.420 0.410 0.405 3 0.22 0.150 0.140 0.135 4 0.31 0.240 0.230 0.225 5 6 0.08 0.010 0.000 -0.005 7 0.07 -0.010 -0.015 8 0.09 0.020 0.005 9 10 0.06 -0.020 -0.025 11 0.11 0.040 0.030 0.025 12 0.12 0.050 0.035 13 0.15 0.080 0.070 0.065 14 0.23 0.160 0.145 15 0.28 0.210 0.200 0.195 16 0.26 0.190 0.180 0.175 17 0.21 0.130 0.125 18 19 20 21 0.03 -0.040 -0.050 -0.055 22 0.02 -0.060 -0.065 23 0.01 -0.070 -0.075 24 25 3.68 2.000 2.230 1.760 2.070 1.640

Volume of the direct surface runoff = 2015 X 60 X 60 = 7254000 cf 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

The phi index was calculated and found it to be 0.085inch 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 0.215 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 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 11.66   10 8.93 1.49 6.21 4.85 0.81 2.30 20 3.50 2.83 0.47 2.77 30 2.16 1.82 0.30 3.07 40 1.32 0.22 3.29 50 1.16 1.07 0.18 3.47 60 0.99 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.

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