Problem 2-1

The mean annual flow is: Q = [ 19.6 m³/s • 86,400 s/d • 365 d/y • 1000 mm/m ] / [ 6937 km² • (1000 m/km)² ] = 89.1 mm/y

The precipitation depth abstracted by the catchment is equal to: (485 - 89.1) = 395.9 mm/y.   ANSWER.

Problem 2-2

Since i = a/t^m, it follows that log i = log a - m log t. Therefore:

log (45) = log a - m log (1)

log (20) = log a - m log (2)

Solving for a and m: a = 45; m = 1.17.  ANSWER.

Problem 2-3

Since at t = ∞, the final infiltration rate is 0.6 mm/h, then: f_c = 0.6 mm/h. Therefore, from Eq. 2-13:

3.5 = 0.6 + (f_o - 0.6) e ^-k; and

1.0 = 0.6 + (f_o - 0.6) e ^-3k

Then: f_c = 0.6 mm/h; f_o = 8.4 mm/h; and k = 0.99 h^-1.   ANSWER.

Problem 2-4

Try several likely values for φ. For instance, assume φ between 1 and 2 cm/h. Therefore: 2 • (2 - φ) + 2 • (3 - φ) + 2 • (4 - φ) + 2 • (2 - φ)= 12.   Solving for φ: φ = 1.25 cm/h. Therefore the assumption of φ being between 1 and 2 cm/h was correct.   ANSWER.