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

  1. The accumulated effective rainfall depth is:  (0.5 + 1 + 2 + 3 + 1 + 0.5) × 2 = 16 cm.

    The total runoff volume is: 45 km2 • 16 cm = 720 km2-cm = 7.2 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)
    0.5
    cm/hr    		 1.0
    cm/hr    		 2.0
    cm/hr    		 3.0
    cm/hr    		 1.0
    cm/hr    		 0.5
    cm/hr
    0 - 0 - - - - - 0 0
    2 9 4.5 0 - - - - 4.5 12.5
    4 21 10.5 9.0 0 - - - 19.5 54.167
    6 15 7.5 21.0 18.0 0 - - 46.5 129.167
    8 - 0. 15.0 42.0 27.0 0 - 84.0 233.333
    10 - 0 0 30.0 63.0 9.0 0.0 102.0 283.333
    12 - - 0 0 45.0 21.0 4.5 70.5 195.833
    14 - - - 0 0 15.0 10.5 25.5 70.833
    16 - - - - 0 0 7.5 7.5 20.833
    18 - - - - - 0 0 0 0
    Sum 45 - - - - - - 360.0 -

    The sum of outflow hydrograph ordinates is 360 km2-cm/h.
    The integration of the outflow hydrograph results in: 360 km2-cm/h • 2 h = 720 km2-cm = 7.2 hm3, which is the same as the total runoff volume. The same results obtained with ONLINE ROUTING 06 (shown below). ANSWER.


  2. The total amount of rainfall is: 1 + 2 + 5 + 4 + 3 + 0.5 = 15.5 cm/hr • 1 hr = 15.5 cm.

    The total volume of runoff is: 1000 km2 • 15.5 cm = 15,500 km2-cm = 155 hm3.

    The Courant number is: C = Δt/K = 1 / 2 = 0.5, which results in the following coefficients: 2C1 = 0.4; and C2 = 0.6.

    The calculations are shown in the following table.

    Time
    (hr)    		 N = 1 N = 2 Outflow
    (m3/s)
    Inflow
    (km2-cm/hr)    		 Outflow
    (km2-cm/hr)    		 Inflow
    (km2-cm/hr)    		 Outflow
    (km2-cm/hr)
    0 - 0 - 0 -
    - 1000 - 200 - -
    1 - 400 - 80 222.22
    - 2000 - 720 - -
    2 - 1040 - 336 933.333
    - 5000 - 1832 - -
    3 - 2624 - 934.40 2595.556
    - 4000 - 2899.2 - -
    4 - 3174.4 - 1720.32 4778.667
    - 3000 - 3139.52 - -
    5 - 3104.64 - 2288 6355.556
    - 500 - 2853.71 - -
    6 - 2062.78 - 2406.28 6684.125
    - 0 - 1650.23 - -
    7 - 1237.67 - 2103.86 5844.061
    - 0 - 990.13 - -
    8 - 742.60 - 1658.37 4606.588
    - 0 - 594.08 - -
    9 - 445.56 - 1232.65 3424.044
    - 0 - 356.45 - -
    10 - 267.34 - 882.17 2450.481
    - 0 - 213.87 - -
    11 - 160.40 - 614.85 1707.921
    - 0 - 128.32 - -
    12 - 96.24 - 420.24 1167.332
    - 0 - 76.99 - -
    13 - 57.74 - 282.94 785.947
    - 0 - 46.19 - -
    14 - 34.65 - 188.24 522.897
    - 0 - 27.72 - -
    15 - 20.79 - 124.03 344.535
    - 0 - 16.63 - -
    16 - 12.47 - 81.07 225.2
    - 0 - 9.97 - -
    17 - 7.48 - 52.63 146.207
    - 0 - 5.99 - -
    18 - 4.49 - 33.97 94.376
    - 0 - 3.59 - -
    19 - 2.69 - 21.82 60.617
    - 0 - 2.15 - -
    20 - 1.62 - 13.95 38.765
    - 0 - 1.29 - -
    21 - 0.97 - 8.89 24.696
    - 0 - 0.78 - -
    22 - 0.58 - 5.64 15.68
    - 0 - 0.46 - -
    23 - 0.35 - 3.57 9.925
    - 0 - 0.28 - -
    24 - 0.21 - 2.25 6.265
    - 0 - 0.17 - -
    25 - 0.13 - 1.42 3.945
    - 0 - 0.10 - -
    26 - 0.08 - 0.89 2.479
    - 0 - 0.06 - -
    27 - 0.04 - 0.56 1.554
    Sum - - - 15499.01
    ≅ 15500 OK    		 -


    The results are the same as calculated by ONLINEROUTING08, shown below.

  3. For the combination 10, 30, 20, 40, the ratio of peak flows is: 21.312 / 21.05 = 1.012.

    The combination 40, 10, 30, 20 will produce the largest ratio of peak flows: 20.128 / 18.504 = 1.087.