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			0.60			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			8.40			16.25			24.03			12.60			5.20			2.10			0.00

The discharge is the sum of the unit discharges: 77.68 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.

The result of slopearea.sdsu.edu is attached.

The flood discharge is: Q = 5275.1 m³/s   ANSWER.