CIV E 445 - APPLIED HYDROLOGY
SPRING 2010
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 42 53 68 78 80 84 90 95 100
Snow-water equivalent (mm) 3 3 4 4 4 4 5 5 5 6 7
Incremental area (%) - 24 18 11 15 10 2 4 5 5 5
Average water equivalent per
elevation increment (mm) 		- 3.0 3.5 4.0 4.0 4.0 4.5 5.0 5.0 5.5 6.5

The catchment's overall snow-water equivalent W is the average water equivalent weighted in terms of the incremental area:

W = [(3.0 • 24)+ (3.5 • 18) + (4.0 • 11) + (4.0 • 15) + (4.0 • 10) + (4.5 • 2) + (5.0 • 4) + (5.0 • 5) + (5.5 • 5) + (6.5 • 5) / 100 = 3.93 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) 0.1 0.6 0.9 1.5 1.8 2.6 3.1 2.8 1.7 0.8 0.1
Average velocity (m/s) 0.00 0.40 0.60 0.80 1.05 1.25 1.55 1.20 0.80 0.60 0.00
Unit discharge (m3/s) 0.00 1.20 2.70 6.00 9.45 16.25 24.03 16.80 6.80 2.40 0.00

The discharge is the sum of the unit discharges: 85.63 m3/s.  ANSWER.

Problem 4-3

Using Eq. 3-6:

Q = [(4,500 / 12) - 1] • 120 L/s = 44,880 L/s or 44.88 m3/s.  ANSWER.

Problem 4-4

The result of slopearea.sdsu.edu is attached.

The flood discharge is: Q = 1,500.7 m3/s   ANSWER.