Phi-index and rainfall excess duration, 0.340 Mm3 runoff

Runoff depth

$$R = \frac{V}{A} = \frac{0.340\times10^{6}\ \text{m}^3}{500\times10^{4}\ \text{m}^2} = 0.068\ \text{m} = 6.8\ \text{cm}$$

Rainfall in each 2-h interval

Total rain $P = 12.6$ cm, so total loss $= 12.6 - 6.8 = 5.8$ cm.

Trial for $\phi$

Let the loss per 2-h interval be $\phi \times 2$. Try intervals with rain above the loss: 2-4, 6-8, 8-10, 10-12, 12-14 (5 intervals). Their rain $= 1.8 + 1.3 + 3.2 + 3.5 + 1.0 = 10.8$ cm.

$$10.8 - 5\times(2\phi) = 6.8 \;\Rightarrow\; \phi = 0.40\ \text{cm/h}$$

Check: loss per interval $= 0.8$ cm. The five intervals have rain $> 0.8$ cm (the smallest is 1.0), and the other intervals (0.8, 0.2, 0.6, 0.2) have rain $\le 0.8$ cm, so they give no excess. The assumption is correct.

Duration of rainfall excess

Five intervals of 2 h each produce excess: $5\times2 = 10$ h (periods 2-4 h and 6-14 h).

Answer: $\phi$-index = 0.40 cm/h; duration of rainfall excess = 10 h.