The total rainfall in the 6-h period is: P = 6 in.
With runoff curve number CN = 80 and total rainfall P = 6 in., use Eq. 5-8 to calculate the direct runoff Q: Q = 3.78 in.
Assume φ-index between 0 and 0.5 in./h.
Therefore: [ (1.0 - φ) × 2 h + (1.5 - φ) × 2 h + (0.5 - φ) × 2 h ] =
3.78 in.
Solving for φ: φ = 0.37 in/h.
Rainfall intensities and depths are as follows.
| Time | Total rainfall
(in/hr) | Abstracted rainfall
(in/hr)
| Effective rainfall
(in/hr) | Effective rainfall depth
(in) |
| [1] | [2] | [3]
| [4] | [5] |
| 0-2 | 1.0 | 0.37
| 0.63 | 1.26 |
| 2-4 | 1.5 | 0.37
| 1.13 | 2.26 |
| 4-6 | 0.5 | 0.37
| 0.13 | 0.26 |
For each time interval, the effective rainfall depth (in) is shown in Col. 5.
The unit hydrograph is convoluted with the effective rainfall depth pattern as shown in the following table.
| Time | UH ordinates
(cfs) | 1.26 × UH ordinates
| 2.26 × UH ordinates | 0.26 × UH ordinates | Composite hydrograph
(cfs)
|
| [1] | [2] | [3]
| [4] | [5] | [6] |
| 0 | 0 | 0
| 0 | 0 | 0 |
| 2 | 100 | 126
| 0 | 0 | 126 |
| 4 | 200 | 252
| 226 | 0 | 478 |
| 6 | 150 | 189
| 452 | 26 | 667 |
| 8 | 100 | 126
| 339 | 52 | 517 |
| 10 | 50 | 63
| 226 | 39 | 328 |
| 12 | 0 | 0
| 113 | 26 | 139 |
| 14 | 0 | 0
| 0 | 13 | 13 |
| 16 | 0 | 0
| 0 | 0 | 0 |
| Sum | 600 | 0
| 0 | 0 | 2268 |
The composite hydrograph for the effective storm pattern is shown in Col. 6.
ANSWER.
To verify that the composite-hydrograph ordinates are correct, the radio of sums is: 2268 / 600 = 3.78.
The sum of the effective rainfall depth is: 1.26 + 2.26 + 0.26 = 3.78. Therefore, the volumes under hyetograph and hydrograph
are the same.
(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: tc = L / V = 3850 / 0.625 = 6150 = 1.71 h.
(2) For 42% 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 58% 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 × 42) + ( 79 × 58)] / 100 = 80.7. Use CN = 81.
(5) With CN = 81, use Eq. 5-49 to calculate the initial abstraction Ia:
Ia = 1.2 cm
With P = 10 cm, the ratio Ia / P is: Ia / P = 0.12
With P = 10 cm, CN = 81, and R = 2.54, use Eq. 5-9 to find Q:
Q = 5.25 cm.
Use Fig. 5-19(a), with rainfall type IA (Pacific Northwest region), time of concentration
tc = 1.71 h, and ratio Ia / P = 0.12, to find the unit peak discharge
qu:
qu = 85 ft3 / (s-mi2 / in)
Converting to SI units:
qu = 85 ft3 / (s-mi2 / in) × 0.0043 =
0.366 m3 / (s-km2-cm).
For 1% pond and swamp areas, use Table 5-12 to find F: F = 0.87.
Using Eq. 5-47:
Qp =
0.366 m3 / (s-km2-cm) × 9.5 km2 × 5.25 cm × 0.87 = 15.9 m3/s.
The 25-y peak discharge is: Qp = 15.9 m3/s.
ANSWER.