CIV E 445 - APPLIED HYDROLOGY
SPRING 2014
HOMEWORK No. 12 SOLUTION

  1. The total rainfall depth is: P = 16 cm. With CN = 81, and R = 2.54 cm/in, the effective rainfall depth (i.e., runoff) (Eq. 5-9) is: Q = 10.56 cm.

    Assuming φ between 0 and 1 cm/h: [(1 - φ) • 1 + (2 - φ) • 1 + (5 - φ) • 1 + (4 - φ) • 1 + (3 - φ) • 1 + (1 - φ) • 1] = 10.56.

    From which: φ = 0.9067 cm/h. The total and effective rainfall pattern is:

    Time (h)Total rainfallφ-index Effective rainfall
    0-1 10.9067 0.0933
    1-2 20.9067 1.0933
    2-3 50.9067 4.0933
    3-4 40.9067 3.0933
    4-5 30.9067 2.0933
    5-6 10.9067 0.0933
    Total 16- 10.5598
    ≅ 10.56

    The accumulated effective rainfall depth is:  10.5598 cm.

    The total runoff volume is: 85 km2 • 10.5598 cm = 897.583 km2-cm = 8.97583 hm3.

    The calculations are shown in the following table:

    Time
    (hr)    		 Time-area histogram subareas
    (km2)    		 Partial flows (km2-cm/hr)
    for indicated rainfall increments    		 Outflow
    (km2-cm/hr)    		 Outflow
    (m3/s)
    0.0933
    cm/hr    		 1.0933
    cm/hr    		 4.0933
    cm/hr    		 3.0933
    cm/hr    		 2.0933
    cm/hr    		 0.0933
    cm/hr
    0 - 0 - - - - - 0 0
    1 17 1.5861 0 - - - - 1.5861 4.40
    2 21 1.9593 18.5861 0 - - - 20.5454 57.07
    3 31 2.8923 22.9593 69.5861 0 - - 95.4377 265.10
    4 16 1.4923 33.8923 85.9593 52.5861 0 - 173.9305 483.14
    5 - 0 17.4923 126.8923 64.9593 35.5861 0 244.9305 680.36
    6 - - 0 65.4928 95.8923 43.9593 1.5861 206.9305 574.81
    7 - - - 0 49.4928 64.8923 1.9593 116.3444 323.18
    8 - - - - 0 33.4928 2.8923 36.3851 101.07
    9 - - - - - 0 1.4928 1.4928 4.15
    10 - - - - - - 0 0 0
    Sum 85 - - - - - - 897.583 -

    The sum of outflow hydrograph ordinates is 897.583 km2-cm/h.
    The integration of the outflow hydrograph results in: 897.583 km2-cm/h • 1 h = 897.583 km2-cm = 8.97583 hm3, which is the same as the total runoff volume. The same results obtained with ONLINE ROUTING 06 (shown below). ANSWER.

  2. Since Δt = 1 h, and K = 3 h: Δt /K = 1/3, and from Eq. 8-16 to 8-18, the linear reservoir-routing coefficients are the following:

    C0 = 1/7; C1 = 1/7; C2 = 5/7.

    The unit runoff volume is:   (20 + 40 + 60 + 40 + 24 + 16) km2 • 1 cm = 200 km2-cm = 2 hm3.

    Since the duration of the SI unit hydrograph (1 cm of runoff) is 2 h, the rainfall intensity is 0.50 cm/h.

    The peak outflow is 83.899 m3/s, and it occurs at time = 5 hr.

    The same results obtained with ONLINE ROUTING 07.   ANSWER.

  3. The results for Example 10-3 are given below, for K = 12 hr, and N = 3. The peak outflow is 308.611 km2-cm/hr, or 857.253 m3/s, and it occurs at time = 36 hr.

    The results for K = 12 hr, and N = 4 are shown below. The peak outflow is 723.115 m3/s, and it occurs at time = 48 hr.

    The results for K = 12 hr, and N = 5 are shown below. The peak outflow is 429.853 m3/s, and it occurs at time = 84 hr.

    As K and/or N increase, the peak outflow is reduced and the time-to-peak increases, as expected.

  4. The time of concentration is: tc = 6 hr.

    The effective rainfall duration for this unit hydrograph is: tr = 1 hr.

    The time base of the translated-only hydrograph (Eq. 10-4) is:

    Tb = tc + tr = 7 hr.

    The evaluation of K can be based on Eq. 10-4, using two consecutive ordinates after 7 hr.

    Using Eq. 10-4 and the ordinates at 7 hr and 8 hr:

    K = - [(47 + 28) / 2] / [(28 - 47) / 1] = 1.97 hr.

    Since this a unit hydrograoph derived from data, it may be appropriate to average several values of K computed at the tail of the hydrograph. Using the ordinates at 8 and 9 hr: K = 2.04 hr. Using the ordinates at 9 and 10 hr: K = 1.92 hr. Using the ordinates at 10 and 11 hr: K = 2.00 hr.

    The average of four values is: K = 1.98 hr, say K = 2 hr.