CIV E 445 - APPLIED HYDROLOGY
SPRING 2014
HOMEWORK No. 6 SOLUTION

  1. From Table 5-2 (a):

    (1) for a industrial district, with 72% impervious area, soil group B: CN = 88;

    (2) for a residential district, with 1/4-ac average lot size, 38% impervious area, soil group C: CN = 83.

    The area-weighted runoff curve number is: CN = [ (88 × 0.25) + (83 × 0.75) ] = 84.25 Use CNII = 84.

    The total rainfall is: P = 2.5 cm/h × 2 h = 5 cm.

    Using Eq. 5-9, with P = 5 cm, R = 2.54, and CN = 84: Q = 1.8 cm.

    The total runoff volume is:

    V = 1.8 cm × 15 ha × 10,000 m2/ha × 0.01 m/cm = 2700 m3.  ANSWER.

  2. The total rainfall in the 24-h period is: P = 201 mm.

    Therefore: Q = [ (5 - 3) × 3 + (8 - 3) × 3 + (10 - 3) × 3 + (15 - 3) × 3 + (12 - 3) × 3 + (10 - 3) × 3 + (5 - 3) × 3] = 132 mm.

    Using Eq. 5-9, with P = 201 mm, R = 25.4 mm/in and Q = 132 mm, by trial and error: CN = 77.  ANSWER.

  3. The data are: A = 520 km2, L = 25 km, LC = 14 km, Ct = 1.5, Cp = 0.62.

    Using Eq. 5-19:   t1 = 8.70 h.   ANSWER.

    Using Eq. 5-21:   Tbt = 28.05 h.  ANSWER.

    Using Eq. 5-22:   Qp = 103.07 m3/s.  ANSWER.

    Using Eq. 5-24:   tr = 1.58 h.  ANSWER.

    Using Eq. 5-26:   tp = 9.49 h.  ANSWER.

    Using Eq. 5-27:   Tb = 98.09 h.  ANSWER.

    Using Eq. 5-28:   W50 = 33.71 h.  ANSWER.

    Using Eq. 5-29:   W75 = 19.23 h.  ANSWER.

  4. The data are: A = 11.2 k m2, CN = 82, L = 4.2 km, Y = 0.02.

    Using Eq. 5-30: t1 = 1.72 h.

    Using 5-36: tr = 0.38 h.

    Using Eq. 5-34: tp = 1.91 h. ANSWER.

    Using Eq. 5-39: Qp = 12.20 m3/s. ANSWER.

    The time base is: Tb = 5 tp = 9.55 h. ANSWER.

    The SCS dimensionless unit hydrograph ordinates (Table 5-6) are shown below.

    The SCS unit hydrograph ordinates, calculated by using Table 5-6, are shown in the following table.