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./hr 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./hr during the first and second 2-hr periods, respectively, as shown in column 3 of the table. Thus, between hours 0 and 2, the rate 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. Likewise, 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
Rainfall Data Table

Suppose the 2-hour unit hydrograph, UH2, for this watershed is given. The ordinates of the unit hydrograph are tabulated in column 2 of Table 2 and are plotted in Figure 1. By definition of a unit hydrograph, if the rainfall excess over the watershed had 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 1.2 in. of rainfall excess, alone, would produce direct runoff that can be expressed as 1.2 UH2, tabulated in column 3 of Table 2 and plotted in Figure 1. Note that the ordinates of 1.2 UH2 in column 3 are obtained by multiplying the ordinates of UH2 in column 2 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 UH2. However, because this rainfall excess starts at t = 2 hours, the resulting direct runoff will be delayed by 2 hours with respect to 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. For 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 UH2 at 1 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 the 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 of 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 are obtained by adding the entries in columns 5 and 6 at respective times. The results are tabulated in column 7. Also, the base flow rates, DRH, and TRH are plotted in Figure 2.

Table 2 Calculation of DRH
Calculation of DRH Table

Figure 1: Components of DRH
Components of DRH Graph

Figure 2: DRH and TRH
DRH and TRH Graph

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