Days inundated above 200 mm at 3 l/s/ha

A drainage rate of 3 l/s/ha is a fixed removal rate. The field is inundated above 200 mm for as long as the water level exceeds 200 mm, so the level history is traced for the design storm.

Data (from the standard problem this question follows)

Initial depth $h_0$ = 40 mm; maximum yearly 3-day rainfall 300 mm with a growth factor of 1.5 for the 10-year return period, so the design 3-day rainfall $R$ = 1.5 × 300 = 450 mm (150 mm/day); ET and deep percolation neglected.

$$D = 3\ \text{l/s/ha} = 3\times\frac{86.4}{10} = 25.92\ \text{mm/day}$$

(1 l/s/ha = 86.4 m³/ha/day = 8.64 mm/day.)

Rise during rain

Net rise = 150 − 25.92 = 124.08 mm/day. Time to reach 200 mm:

$$t_1 = \frac{200-40}{124.08} = 1.29\ \text{days}$$

Peak at the end of day 3:

$$h_{peak} = 40 + 450 - 3\times25.92 = 412.2\ \text{mm}$$

Fall after rain

Time to fall from 412.2 mm to 200 mm:

$$t_2 = \frac{412.2-200}{25.92} = 8.19\ \text{days}$$

Duration above 200 mm = (3 − 1.29) + 8.19 = 9.90 days (permitted: 3 days).

Will the depth exceed 300 mm?

Yes. The maximum depth is 412 mm, which is more than 300 mm, and the level stays above 300 mm for (3 − 2.10) + 4.33 = 5.23 days, against the permitted 1 day. So a drainage rate of 3 l/s/ha is inadequate; about 7.0 l/s/ha (60.6 mm/day) is needed to meet both limits.

If the 300 mm rainfall were used directly, without the growth factor, the peak would be 262 mm and the field would stay above 200 mm for 3.24 days.

Answer: about 9.9 days above 200 mm; yes, the depth rises to about 412 mm, exceeding 300 mm for about 5.2 days.