CIV E 445 - APPLIED HYDROLOGY
SPRING 2010
SOLUTIONS TO HOMEWORK 3 , CHAPTER 2

Problem 3-1

E = 29.43 cm  ANSWER.

Problem 3-2

E = 27.05 cm  ANSWER.

Problem 3-3

E = 11.6 cm  ANSWER.

Problem 3-4

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

At x = L = 320,000 m: elevation y = 14.66 m.

From Fig. 2-19, the area comprised between elevation y = 14.66 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 between the limits of 0 and 320,000:

At = ∫0320,000 y dx = ∫0320,000 100e-0.000006x dx = 14,223,217

The area comprised between elevation 0 and elevation 14.66 m is:

Ab = 14.66 • 320,000 = 4,691,200 m2.

Therefore, the area Ap is:

Ap = At - Ab = 9,532,017 m2.

S2 = 2Ap/L2 = 0.000186   ANSWER.