Problem 6-1

From Table 5-2(a), for a industrial district, with 72% impervious area, soil group B: CN = 88; for a residential district, with 1/4-ac average lot size, 38% impervious area, soil group C: CN = 83. The area-weighted runoff curve number is: CN = [ (88 • 0.25) + (83 • 0.75) ] = 84.25 Use CN = 84. The total rainfall is: P = 2.5 cm/h • 2 h = 5 cm. Using Eq. 5-9, with P = 5 cm, R = 2.54, and CN = 84: Q = 1.8 cm. The total runoff volume is:

V = 1.8 cm • 15 ha • 10,000 m²/ha • 0.01 m/cm = 2700 m³.  ANSWER.

Problem 6-2

The total rainfall in the 24-h period is: P = 201 mm. Therefore: Q = [ (5 - 3) • 3 + (8 - 3) • 3 + (10 - 3) • 3 + (15 - 3) • 3 + (12 - 3) • 3 + (10 - 3) • 3 + (5 - 3) • 3] = 132 mm. Using Eq. 5-9, with P = 201 mm, R = 25.4 mm/in and Q = 132 mm, by trial and error: CN = 77.  ANSWER.

Problem 6-3

The data are: A = 520 km², L = 25 km, L_C = 14 km, C_t = 1.5, C_p = 0.62.

Using Eq. 5-19:   t₁ = 8.70 h.   ANSWER.

Using Eq. 5-21:   T_bt = 28.05 h.  ANSWER.

Using Eq. 5-22:   Q_p = 103.07 m³/s.  ANSWER.

Using Eq. 5-24:   t_r = 1.58 h.  ANSWER.

Using Eq. 5-26:   t_p = 9.49 h.  ANSWER.

Using Eq. 5-27:   T_b = 98.09 h.  ANSWER.

Using Eq. 5-28:   W₅₀ = 33.71 h.  ANSWER.

Using Eq. 5-29:   W₇₅ = 19.23 h.  ANSWER.

Problem 6-4

The data are: A = 11.2 k m², CN = 82, L = 4.2 km, Y = 0.02. Using Eq. 5-30: t₁ = 1.72 h. Using 5-36: t_r = 0.38 h. Using Eq. 5-34: t_p = 1.91 h. Using Eq. 5-39: Q_p = 12.20 m³/s.

The time base is: T_b = 5 t_p = 9.55 h.

The SCS unit hydrograph ordinates, calculated by using Table 5-6, are shown in the following table.