Problem 12-1

The total rainfall depth is: P = 16 cm. With CN = 81, and R = 2.54 cm/in., the effective rainfall depth (i.e., runoff) (Eq. 5-9) is: Q = 10.56 cm. Assuming φ between 0 and 1 cm/h: [(1 - φ) • 1 + (2 - φ) • 1 + (5 - φ) • 1 + (4 - φ) • 1 + (3 - φ) • 1 + (1 - φ) • 1] = 10.56.
From which: φ = 0.907 cm/h. The total and effective rainfall pattern is:

The accumulated effective rainfall depth is:  10.56 cm.

The total runoff volume is: 85 km² • 10.56 cm = 897.6 km²-cm = 8.976 hm³.

The calculations are shown in the following table:

The sum of outflow hydrograph ordinates is 897.6 km²-cm/h.
The integration of the outflow hydrograph results in:

897.6 km²-cm/h • 1 h = 897.6 km²-cm = 8.976 hm³, which is the same as the total runoff volume. The same results obtained with ONLINE ROUTING 06.   ANSWER.

Problem 12-2

Since Δt = 1 h, and K = 3 h: Δt /K = 1/3, and from Eq. 8-16 to 8-18, the linear reservoir-routing coefficients are the following:

C₀ = 1/7; C₁ = 1/7; C₂ = 5/7.

The unit runoff volume is:   (20+40+60+40+24+16) km² • 1 cm = 200 km²-cm = 2 hm³.

Since the duration of the SI unit hydrograph (1 cm of runoff) is 2 h, the rainfall intensity is 0.50 cm/h.

The same results obtained with ONLINE ROUTING 07.   ANSWER.

Problem 12-3

For K = 12 hr, and N = 4, the peak flow is 723.115 m3/s at 48 hours.   ANSWER.

For K = 18 hr, and N = 5, the peak flow is 429.853 m3/s at 84 hours.   ANSWER.