Runoff and effective rainfall, 4-hr storm, 80 km2

Effective (excess) rainfall in each hour of a sub-basin is $P_t - \phi$ (if positive; otherwise zero). The sub-basin values are then area-weighted for the whole catchment.

Step 1: Effective rainfall (mm) in each sub-basin

Example (15 km², $\phi = 10$): 16-10 = 6, 48-10 = 38, 22-10 = 12, 10-10 = 0.

Step 2: Area-weighted effective rainfall for 80 km²

$$\bar P_{e,t} = \frac{\sum A_i P_{e,i,t}}{80}$$

Check: $15\times6 + 25\times1 + 0 + 0 = 115$ (1st hour); $15\times38 + 25\times27 + 35\times19 + 5\times26 = 2040$ (2nd hour); $15\times12 + 25\times5 + 5\times2 = 315$ (3rd hour).

Step 3: Runoff from the catchment

Total effective rainfall $= 1.4375 + 25.5 + 3.9375 + 0 = 30.875$ mm.

$$V = 30.875\times10^{-3}\times 80\times10^6 = 2,470,000\ \text{m}^3 = 2.47\ \text{million m}^3$$

Average runoff rate over the 4 h $= 2.47\times10^6/(4\times3600) = 171.5$ m³/s.

Answer: Runoff depth = 30.875 mm, volume = 2.47 × 10⁶ m³; hourly effective rainfall = 1.44, 25.50, 3.94 and 0 mm.