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Current: Rainfall Runoff Process > Runoff Hydrographs > Total Runoff Hydrograph > Example

Total Runoff Hydrograph: Example Problem

Suppose a 4-hour storm occurs over a watershed. Let the rain intensity be 1.0 in./hr during the first 2-hr period, and 0.9 in.during the second 2-hr period as shown in column 2 of Table 1 below. Let the rate of rainfall losses be 0.4 in./hr and 0.2 in./

during the first and second 2-hr periods, respectively, as shown in column 3 of the table. Thus, between hours 0 and 2, the

of rainfall excess is 1.0-0.4= 0.6 in./hr producing an excess of (0.6 in./hr)(2 hrs)= 1.2 in. as shown in columns 4 and 5. Like

between hours 2 and 4, the rate of rainfall excess is 0.9-0.2= 0.7 in./hr producing an excess of (0.7in./hr)(2 hrs)= 1.4 in./hr

Table 1 Rainfall Data

Suppose the 2-hour unit hydrograph, UH2, for this watershed is given. The ordinates of the unit hydrograph are tabulated i

column 2 of Table 2 and are plotted in Figure 1. By definition of a unit hydrograph, if the rainfall excess over the watershed

duration of 2 hrs and a depth of 1.0 in., then the DRH produced would be the same as UH2.

In this example a rainfall excess of 1.2 in. occurs during the first two hours. Then, based on the linearity assumption, this 1

of rainfall excess, alone, would produce direct runoff that can be expressed as 1.2 UH2, tabulated in column 3 of Table 2 a

plotted in Figure 1. Note that the ordinates of 1.2 UH2

in column 3 are obtained by multiplying the ordinates of UH2

in colu

at respective times.

Similarly, the rainfall excess of 1.4 in. occurring between hour 2 and 4, alone, will result in direct runoff expressed as 1.4 U

However, because this rainfall excess starts at t = 2 hours, the resulting direct runoff will be delayed by 2 hours with respe

time zero as shown in column 4 of Table 4 and in Figure 1. Note that the ordinates of the 2-hr lagged 1.4 UH2

tabulated in

column 4 of the table are obtained by multiplying the ordinates of UH2

in column 2 and by lagging the entries by 2 hours. F

example, to calculate the ordinate of 2-hr lagged 1.4 UH2

at 3 hrs (22.4cfs) in column 4, we multiply the ordinate of the UH

hr (16 cfs/in.) in column 2 by 1.4 in.

Based on the linearity assumption we can use the method of superposition to determine the composite DRH resulting from

composite rainfall excess that includes the excess from both 2-hr periods (0 to 2 and 2 to 4). Thus

DRH = 1.2UH2

+ 2-hr lagged 1.4UH2

The ordinates of the DRH tabulated in column 5 of Table 2 are calculated by adding those in columns 3 and 4 at respective

times. The DRH is also plotted in Figure 1.

To calculate the total runoff hydrograph (TRH) resulting from this rainfall, we would add the base flow rate to the ordinates

DRH. For a constant base flow (BF) rate of 50 cfs as tabulated in column 6 of Table 1, for example, the ordinates of TRH a

obtained by adding the entries in columns 5 and 6 at respective times. The results are tabulated in column 7. Also, the bas

rates, DRH, and TRH are plotted in Figure 2.

Table 2 Calculation of DRH

Figure 1: Components of DRH

Runoff Hydrograph: Example Problem http://www.mem.odu.edu/rain/html/runOffHydrographs_totalRun

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Figure 2: DRH and TRH

Runoff Hydrograph: Example Problem http://www.mem.odu.edu/rain/html/runOffHydrographs_totalRun

12/28/2012