Phi-index and cumulative infiltration curve

Direct runoff depth: Total direct runoff volume $= \sum Q\,\Delta t = (30+60+45+30+15)\times3600 = 648\,000$ m³.

Note on area: with $A = 0.2$ km² the runoff depth ($648000/2\times10^5 = 3.24$ m) is far greater than the rainfall (5.18 cm), which is impossible. The area must be 200 km² (0.2 × 10³ km²); this is assumed.

$$R = \frac{648\,000}{200\times10^6} = 0.00324\ \text{m} = 0.324\ \text{cm}$$

Total rainfall $= 1.05+1.28+0.80+0.75+0.70+0.60 = 5.18$ cm, so the total loss $= 5.18 - 0.324 = 4.856$ cm.

Finding $\phi$

Runoff occurs only when rainfall rate exceeds $\phi$. Try the hours with rain greater than $\phi$: 1.05 and 1.28 cm/h (the others are below $\phi$ and infiltrate fully).

$$(1.05 - \phi) + (1.28 - \phi) = 0.324 \Rightarrow \phi = \frac{2.33 - 0.324}{2} = 1.003\ \text{cm/h}$$

Check: $1.05, 1.28 > 1.003$ and $0.80, 0.75, 0.70, 0.60 < 1.003$. Valid.

Cumulative infiltration (loss) curve

Infiltration in each hour $= \min(\text{rain rate}, \phi)$:

The last ordinate (4.856 cm) equals the total loss, confirming the result.

Answer: $\phi$-index = 1.003 cm/h; cumulative infiltration ordinates = 1.003, 2.006, 2.806, 3.556, 4.256, 4.856 and 4.856 cm at 1 to 7 h.