CIV E 445 - APPLIED HYDROLOGY
SPRING 2009
SOLUTIONS TO HOMEWORK 3 , CHAPTER 2

Problem 3-1

E = 25.75 cm  ANSWER.

Problem 3-2

E = 24.15 cm  ANSWER.

Problem 3-3

E = 11.54 cm  ANSWER.

Problem 3-4

At x = 0 km: elevation y = 100 m.

At x = L = 250,000 m: elevation y = 28.65 m.

From Fig. 2-19, the area comprised between elevation y = 28.65 and the longitudinal profile is:   Ap = YL/2.

Therefore, the S2 slope is:   S2 = Y/L = 2Ap/L2.

The total area, At, below the longitudinal profile is obtained by integration:

At = ∫0250,000 y dx = ∫0250,000 100e-0.000005x dx = 14,269,904

The area comprised between elevation 0 and elevation 28.65 m is:

Ab = 28.65 X 250,000 = 7,162,500 m2.

Therefore, the area Ap is:

Ap = At - Ab = 7,107,404 m2.

S2 = 2Ap/L2 = 0.000227. ANSWER.