Problem 7-1

(1) For shallow concentrated flow, paved, use Fig. 5-18 with slope S = 0.01 to find the average velocity (along the hydraulic length) is: V = 2.05 ft/s = 0.625 m/s. Therefore, the time of concentration is: t_c = L / V = 3350 / 0.625 = 5360 = 1.49 h.

(2) For 48% of the watershed area: for urban 1/3-ac lots, with lawns with 85% grass cover (i.e., in good hydrologic condition), 34% total impervious area, soil group C, use Fig 5-2 (a) to find the pervious area (open space) CN: CN = 74. With pervious area CN = 74 and 34% total impervious area find the composite CN from Fig. 5-16: composite CN = 83.

(3) For 52% of the watershed area: for urban 1/3-ac lots, with lawns with 95% grass cover (i.e., in good hydrologic condition), 24% total impervious, 25% of it unconnected, soil group C, use Fig. 5-2 (a) to find the pervious area (open space) CN: CN = 74. With pervious area CN = 74, 24% total impervious area, 25% of it unconnected, find the composite CN from Fig. 5-17: composite CN = 79.

(4) The runoff curve number for the entire watershed is obtained by areal weighing: CN = [(83 • 48) + ( 79 • 52)] / 100 = 80.92. Use CN = 81.

(5) With CN = 81, use Eq. 5-49 to calculate the initial abstraction I_a: I_a = 1.2 cm

With P = 12 cm, the ratio I_a / P is: I_a / P = 0.1

With P = 12 cm, CN = 81, and R = 2.54, use Eq. 5-9 to find Q: Q = 6.97 cm.

Use Fig. 5-19 (a), with rainfall type IA (Pacific Northwest region), time of concentration t_c = 1.49 h, and ratio I_a / P = 0.1, to find the unit peak discharge q_u: q_u = 95 ft³ / (s-mi² / in)

Converting to SI units: q_u = 95 ft³ / (s-mi² / in) • 0.0046 = 0.409 m³ / (s-km²-cm).

For 0.2% pond and swamp areas, use Table 5-12 to find F: F = 0.97.

Using Eq. 5-47: Q_p = 0.409 m³ / (s-km²-cm) • 8.5 km² • 6.97 cm • 0.97 = 23.5 m³/s.

The 50-y peak discharge is: Q_p = 23.5 m³/s.  ANSWER.

Problem 7-2

Using Online TR-55 (results attached), the 50-yr peak discharge is: 25.57 cubic meters per second.   ANSWER.