Phi-index and W-index of isolated storm, 500 ha

Given: $A = 500$ ha $= 5\times10^6$ m²; runoff volume $= 0.361\ \text{Mm}^3 = 3.61\times10^5$ m³; baseflow negligible; 18 h of rain.

Step 1: Runoff depth and rainfall

$$R = \frac{3.61\times10^5}{5\times10^6} = 0.0722\ \text{m} = 7.22\ \text{cm}$$

Total rainfall $P = 12.6$ cm. Rainfall in each 2-h block (cm): 0.8, 1.8, 0.2, 1.3, 3.2, 3.5, 1.0, 0.6, 0.2.

Step 2: $\phi$-index by trial

Find $\phi$ (cm per 2 h) so that the sum of $(P_i - \phi)$ over the blocks with $P_i > \phi$ is 7.22 cm. Try $\phi = 0.73$ cm per 2 h; the blocks 0.8, 1.8, 1.3, 3.2, 3.5, 1.0 exceed it (0.2, 0.2 and 0.6 do not):

$$(12.6 - 0.2 - 0.2 - 0.6) - 6\times\phi' = 7.22 \Rightarrow 11.6 - 6\phi' = 7.22 \Rightarrow \phi' = 0.73\ \text{cm per 2 h}$$ $$\phi = \frac{0.73}{2} = 0.365\ \text{cm/h}$$

Step 3: Duration of rainfall excess

Rainfall excess occurs in 6 blocks of 2 h, so the duration of rainfall excess $t_r = 6\times2 = 12$ h.

Step 4: W-index

Other losses (depression storage, interception) $= 0.1\times10^6/5\times10^6 = 0.02$ m $= 2.0$ cm.

$$W = \frac{P - R - S_d}{t_r} = \frac{12.6 - 7.22 - 2.0}{12} = \frac{3.38}{12} = 0.282\ \text{cm/h}$$

Answer: $\phi$-index = 0.365 cm/h; duration of rainfall excess = 12 h; W-index = 0.282 cm/h.