SPRING 2003
SOLUTIONS TO HOMEWORK 4 , CHAPTER 3
Problem 4-1
| Elevation (%) | 0 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | 100 | ||||||||||||||||||||||
| Cumulative Area (%) | 0 | 24 | 40 | 52 | 66 | 77 | 82 | 86 | 95 | 98 | 100 | ||||||||||||||||||||||
| Snow-water equivalent (mm) | 3 | 3 | 3 | 4 | 4 | 5 | 5 | 7 | 7 | 8 | 8 | ||||||||||||||||||||||
| Incremental area (%) | - | 24 | 16 | 12 | 14 | 11 | 5 | 4 | 9 | 3 | 2 | ||||||||||||||||||||||
| Average water equivalent per elevation increment (mm) | 0 | 3.0 | 3.0 | 3.5 | 4.0 | 4.5 | 5.0 | 6.0 | 7.0 | 7.5 | 8.0 |
The catchment's overall snow-water equivalent W is the average water equivalent weighted in terms of the incremental area: W = [(3.0 • 24)+ (3.0 • 16) + (3.5 • 12) + (4.0 • 14) + (4.5 • 11) + (5.0 • 5) + (6.0 • 4) + (7.0 • 9) + (7.5 • 3) + (8.0 • 2) / 100 = 4.18 mm . ANSWER.
Problem 4-2
| Vertical no | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | ||||||||||||||||||||||
| Width (m) | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | ||||||||||||||||||||||
| Depth (m) | 0.0 | 0.6 | 0.9 | 1.3 | 1.6 | 2.6 | 3.1 | 2.1 | 1.3 | 0.7 | 0.0 | ||||||||||||||||||||||
| Average velocity (m/s) | 0.0 | 0.40 | 0.70 | 0.85 | 1.05 | 1.30 | 1.55 | 1.20 | 0.85 | 0.70 | 0.00 | ||||||||||||||||||||||
| Unit discharge (m3/s) | 0.00 | 1.20 | 3.15 | 5.53 | 8.40 | 16.90 | 24.03 | 12.60 | 5.53 | 2.45 | 0.00 |
The discharge is the sum of the unit discharges: 79.78 m3/s. ANSWER.
Problem 4-3
Using Eq. 3-6:
Q = [(12,000 / 15) - 1] • 100 L/s = 79900 L/s or 79.9 m3/s. ANSWER.
Problem 4-4
The upstream conveyance Ku (Eq. 3-7a) is 378,378; the downstream conveyance Kd (Eq. 3-7b) is 293,837; the reach conveyance K (Eq. 3-8) is 333,439. The first approximation to the energy slope (Eq. 3-9) is 0.0002816.
The first approximation to the peak discharge (Eq. 3-10) is 5595 m3/s. Since the reach is contracting, the loss coefficient k=1. The result of the iteration is summarized below.
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The flood discharge is: Q = 5275 m3/s ANSWER.