Rainfall excess hydrograph

Ever wondered what a ‘rainfall excess’ hydrograph is and how they are calculated?  Then read on.

‘Rainfall excess’ is the rainfall left over after the initial and continuing loss are removed.  Rainfall excess hydrographs are used in the runoff-routing program RORB.  The RORB manual (Section 3.3.4) describes them as follows:

In catchment studies, the program calculates hyetographs for all sub-areas.  After deducting losses, it converts the hyetograph ordinates to ‘hydrographs’ of rainfall-excess on the sub-areas, in m3/s, and interprets the average ‘discharge’ during a time increments as an instantaneous discharge at the end of the time increment.

Lets look at an example.  I’m using the methods from the 1987 version of Australian Rainfall and Runoff so I can compare results with calculations in RORB.

1. Choose a design rainfall depth

I’m working on a catchment in Gippsland where the 1% AEP 6 hour rainfall is of interest.  Rainfall IFD data is available from the Bureau of Meteorology via this  link.

For the site of interest, the 1% (100-year), 6 hour rainfall depth is 90.9 mm.

2. Select a temporal pattern

Temporal patterns are available in Australian Rainfall and Runoff Volume 2, Table 3.2.  Gippsland is in zone 1 and ARI is > 30 years so we need the bottom row from the table below.   This shows the percentage of the rainfall depth in each 30 min time period


Applying the temporal pattern to the design rainfall depth results in the following hyetograph.


Figure 1: Design rainfall hyetograph

3. Remove the losses

Calculate the rainfall excess hyetograph by removing the initial loss and continuing loss.  For this example,

  • IL = 10 mm and
  • CL = 2 mm/h.

Note that the continuing loss is 2 mm/h and the time step of the hyetograph is 0.5 h so 1 mm is lost per time step.

The rainfall excess hyetograph is shown in Figure 2.


Figure 2: Rainfall excess hyetograph

4. Convert to a hydrograph

The procedure to convert a rainfall excess hyetograph to a rainfall excess hydrograph is explained in the quote at the start of the blog.  We need to:

  • Multiply the rainfall excess by the catchment area (converts rainfall to a volume)
  • Divide by the time step (to calculate volume per unit time)
  • Ensure flow is allocated to the correct time step – the rainfall during a time step produces the instantaneous flow at the end of the time step
  • Ensure the units are correct – calculated flow is is m3/s, rainfall is in mm and catchment area is in km2.

There is also a discussion of this in ARR2016 Book 5, Chapter

Example calculation: in this case, the sub-catchment area is 78.7 km2.  The rainfall in the 3rd time step,  between 1 hour and 1.5 hour, is 8.9 mm so the flow at the end of this time step will be:

Q = \frac{1}{3600} \times  10^{-3} \times 10^6 \times \frac{8.999}{0.5} \times 78.7 = 393.46 \; \mathrm{m^3s^{-1}}

The rainfall excess hydrograph is shown in Figure 3.


Figure 3: Rainfall excess hydrograph

4. Comparison with RORB

Figure 4 shows the rainfall excess hydrograph as calculated by RORB.  The answers look close and I’ve confirmed this by looking at the calculated values.


Figure 4: Rainfall excess hydrograph as calculated by RORB

Calculations are available as a gist.

One thought on “Rainfall excess hydrograph

  1. Pingback: Modelling impervious surfaces in RORB | tonyladson

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