CIV E 445 - APPLIED HYDROLOGY
SPRING 2014
HOMEWORK No. 3 SOLUTION

  1. At x = 0 km: elevation Emax = 100 m.

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

    From Fig. 2-19 of the texbook (see below), the area comprised between elevation y = Emin = 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.

  2. The average precipitation intensity is: 50 mm / 2 hr = 25 mm/hr.

    Since the time of concentration is 1 hr, and the rainfall duration is 2 hr, the catchment is superconcentrated. The maximum possible flow rate is obtained by assuming zero hydrologic abstractions, i.e., Qe = IA.

    Qe = (25 mm/hr × 35 ha × 10,000 m2/ha) / (3,600 s/hr × 1,000 mm/m)

    Qe = 2.43 m3/s.

  3. n = R2/3 S1/2 / V = (3.2)2/3 (0.0005)1/2 / 1.8 = 0.02698 ≅ 0.027.   ANSWER.

    C = V / R1/2 S1/2 = 1.8 / (3.2)1/2 (0.0005)1/2 = 45 m1/2/s   ANSWER.

    f = S / F2 = S / [V2 / (gR)] = 0.0005 / [1.82 / (9.81 × 3.2)] = 0.00484   ANSWER.

    fD = 8 f = 0.03876   ANSWER.

  4. The drainage area in square miles is A = 980 × (0.621371)2 = 350. 58 mi2

    Applying the Creager formula, with C = 100: Q [C = 100] = 239,900 cfs.

    Applying the Creager formula, with C = 30: Q [C = 30] = 71,970 cfs.

    Converting to SI units: Q [C = 100] = 6,793 m3/s   ANSWER.

    Applying the Creager formula, with C = 30: Q [C = 30] = 2,038 m3/s   ANSWER.

    These results are verified with CREAGER.