CIV E 445 - APPLIED HYDROLOGY
SPRING 2009
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) 		- 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 1.5 2.6 3.1 2.3 1.3 0.7 0.0
Average velocity (m/s) 0.0 0.40 0.60 0.80 1.05 1.25 1.55 1.20 0.85 0.60 0.00
Unit discharge (m3/s)0.00 1.20 2.70 5.20 7.88 16.25 24.03 13.80 5.20 2.10 0.00

The discharge is the sum of the unit discharges: 78.36 m3/s.  ANSWER.

Problem 4-3

Using Eq. 3-6:

Q = [(10,000 / 16) - 1] • 100 L/s = 62400 L/s or 62.4 m3/s.  ANSWER.

Problem 4-4

The result of slopearea.sdsu.edu is attached.

The flood discharge is: Q = 5712.8 m3/s   ANSWER.