Problem 7-1

(1) For shallow concentrated flow, paved, use Fig. 5-18 with slope S = 0.02 to find the average velocity (along the hydraulic length) is: V = 2.9 ft/s = 0.884 m/s. Therefore, the time of concentration is: t_c = L / V = 2,800 / 0.884 = 3,167 = 0.88 h.

(2) For 40% of the watershed area: for urban 1/3-ac lots, with lawns with 65% grass cover (i.e., in fair hydrologic condition), 34% total impervious area, soil group C, use Table 5-2 (a) to find the pervious area (open space) CN: CN = 79. With pervious area CN = 79 and 40% total impervious area find the composite CN from Fig. 5-16: composite CN = 86.

(3) For 60% of the watershed area: for urban 1/3-ac lots, with lawns with 65% grass cover (i.e., in fair hydrologic condition), 20% total impervious, 25% of it unconnected, soil group C, use Table 5-2 (a) to find the pervious area (open space) CN: CN = 79. With pervious area CN = 79, 20% total impervious area, 25% of it unconnected, find the composite CN from Fig. 5-17: composite CN = 82.

(4) The runoff curve number for the entire watershed is obtained by areal weighing: CN = [(86 • 40) + ( 82 • 60)] / 100 = 83.6. Use CN = 84.

(5) With CN = 84, use Eq. 5-49 to calculate the initial abstraction I_a: I_a = 0.97 cm

With P = 12.5 cm, the ratio I_a / P is: I_a / P = 0.078

With P = 12.5 cm, CN = 84, and R = 2.54, use Eq. 5-9 to find Q: Q = 8.12 cm.

Use Fig. 5-19 (a), with rainfall type I (Southern California region), time of concentration t_c = 0.88 h, and ratio I_a / P = 0.1 (minimum value), to find the unit peak discharge q_u: q_u = 217 ft³ / (s-mi² / in)

Converting to SI units: q_u = 217 ft³ / (s-mi² / in) • 0.0043 = 0.9331 m³ / (s-km²-cm).

For 0% pond and swamp areas, use Table 5-12 to find F: F = 1.00

Using Eq. 5-47: Q_p = 0.9331 m³ / (s-km²-cm) • 10.5 km² • 8.12 cm • 1.00 = 79.56 m³/s.

The 50-y peak discharge is: Q_p = 79.56 m³/s.  ANSWER.

Problem 7-2

Using Online TR-55, the 50-yr peak discharge is: 79.53 cubic meters per second.   ANSWER.