QUESTIONS

  1. What is reservoir routing?

    Reservoir routing uses mathematical relations to calculate outflow from a reservoir once inflow, initial conditions, reservoir characteristics and operational rules are known.

  2. What is a linear reservoir?

    A linear reservoir is a reservoir in which outflow and storage are linearly related.

  3. What is the proportional or Sutro® spillway?

    In the proportional or Sutro® weir, the cross-sectional flow area, above the rectangular section, grows in proportion to the half-power of the hydraulic head. Therefore, outflow is linearly related to hydraulic head and a spillway rating based on Eq. 8-7 is applicable.

  4. What is the differential equation of storage? What principle is it based on?

    The differential equation of storage is:

                   dS
    I - O  =  _____
                   dt

    in which I = inflow, 0 = outflow, S = storage, and t = time.

    The differential equation of storage is based on the principle of conservation of mass.

  5. What is the xt plane?

    The xt plane is a cartesian coordinate system in which space (x) is plotted in abscissas and time (t) in the ordinates.

  6. Above what value of the storage constant K will one of the routing coefficients be negative?

    When Δt /K exceeds 2 (i.e., K less than Δt/2), one of the routing coefficients in reservoir routing (C2) will be negative.

  7. Explain why in reservoir routing with uncontrolled outflow, the peak outflow occurs when inflow and outflow coincide.

    Since storage and outflow are directly related, maximum storage (S) is associated with maximum or peak outflow (O). Maximum storage occurs when dS/dt = 0 (maximization principle), which implies that inflow is equal to outflow (I = O). Therefore, when inflow equals outflow, both outflow and storage are at their maximum values.

  8. Explain why there is no lag between inflow and outflow in reservoir routing.

    From a mathematical standpoint, the lack of initial lag is attributed to the infinite propagation velocity of short surface waves in an ideal reservoir.

  9. What is the storage indication quantity?

    The storage indication quantity is [(2S/Δt) + O]. The chosen value of time interval Δt should be such that the resulting linearization of the inflow hydrograph is a close approximation of the actual shape of the hydrograph. For smoothly rising hydrographs, a minimum value of tp/Δt = 5 is generally recommended, in which tp is the time-to-peak of the inflow hydrograph. In actual practice, a computer-aided calculation would normally use a much greater ratio, say greater than 10.

  10. What is an appropriate value of time interval to choose in reservoir routing?

    Values of Δt /K > 2 are not used in reservoir routing. Therefore, Δt has to be less than 2K.

  11. What is surcharge storage?

    Surcharge storage is the reservoir storage measured above a certain elevation, usually the elevation of the crest of the emergency spillway.

  12. Name three applications of reservoir routing.

    Reservoir routing applications include: (1) principal and emergency spillway design, (2) reservoir operation studies, (3) optimal use of water resources.

  13. What is a detention basin? When is it used?

    A detention basin is a small reservoir, built typically in an urban setting, designed to hold and diffuse storm runoff to mitigate and reduce regional downstream floods and channel erosion.

    The detention basin is a widely used method for controlling peak discharge in urban areas. It is generally the least expensive and most reliable of the measures that are usually considered for controlling storm runoff.

PROBLEMS

  1. Route the following inflow hydrograph through a linear reservoir:

    Time (h) 0 1 2 3 4 5 6 7 8 9 10
    Inflow (m3/s) 0 10 20 30 40 50 40 30 20 10 0

    Assume baseflow = 0 m3/s, K = 3 h, Δt = 1 h.

    With K = 3 h, and Δt = 1 h, the routing coefficients (Eqs. 8-16 to 8-18) are:

    C0 = 1/7 = 0.143; C1 = 1/7 = 0.143; and C2 = 5/7 = 0.714

    The routing computations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 C0I2
    		 C1I1
    		 C2O1
    		 Outflow
    (m2/s)
    0 0 - - - 0.00
    1 10 1.43 0.00 0.00 1.43
    2 20 2.86 1.43 1.02 5.31
    3 30 4.29 2.86 3.79 10.94
    4 40 5.72 4.29 7.81 17.82
    5 50 7.15 5.72 12.72 25.59
    6 40 5.72 7.15 18.27 31.14
    7 30 4.29 5.72 22.23 32.24
    8 20 2.86 4.29 23.02 30.17
    9 10 1.43 2.86 21.54 25.83
    10 0 0 1.43 18.44 19.87
    11 0 0 0 14.19 14.19
    12 0 0 0 10.13 10.13
    13 0 0 0 7.23 7.23
    14 0 0 0 5.16 5.16
    15 0 0 0 3.68 3.68
    16 0 0 0 2.63 2.63
    17 0 0 0 1.88 1.88
    18 0 0 0 1.34 1.34
    19 0 0 0 0.96 0.96
    20 0 0 0 0.68 0.68

    The calculated peak outflow is 32.24 m3/s, and it occurs at 7 h.  ANSWER.

  2. Route the following triangular inflow hydrograph through a linear reservoir: peak inflow = 120 m3/s, baseflow = 0 m3/s, time-to-peak = 6 h, time base = 16 h, storage constant K = 2 h, and time interval Δt = 1 h.

    With K = 2 h, and Δt = 1 h, the routing coefficients (Eqs. 8-16 to 8-18) are:

    C0 = 1/5 = 0.2; C1 = 1/5 = 0.2; and C2 = 3/5 = 0.6

    The routing computations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 C0I2
    		 C1I1
    		 C2O1
    		 Outflow
    (m2/s)
    0 0 - - - 0.00
    1 20 4.00 0.0 0.0 4.00
    2 40 8.0 4.0 2.4 14.40
    3 60 12.0 8.0 8.64 28.64
    4 80 16.0 12.0 17.18 45.18
    5 100 20.0 16.0 27.11 63.11
    6 120 24.0 20.0 37.87 81.87
    7 108 21.6 24.0 49.12 94.72
    8 96 19.2 21.6 56.83 97.63
    9 84 16.8 19.2 58.58 94.58
    10 72 14.4 16.8 56.75 87.95
    11 60 12.0 14.4 52.77 79.17
    12 48 9.6 12.0 47.50 69.10
    13 36 7.2 9.6 41.46 58.26
    14 24 4.8 7.2 34.96 46.96
    15 12 2.4 4.8 28.18 35.38
    16 0 0.0 2.4 21.23 23.63
    17 0 0.0 0.0 14.18 14.18
    18 0 0.0 0.0 8.51 8.51
    19 0 0.0 0.0 5.11 5.11
    20 0 0.0 0.0 3.07 3.07

    The calculated peak outflow is 97.63 m3/s, and it occurs at 8 h.  ANSWER.

  3. Route the following inflow hydrograph through a linear reservoir.

    Time (h) 0 1 2 3 4 5 6 7 8 9 10 11 12
    Inflow (m3/s) 10 20 50 80 90 100 90 60 50 40 30 20 10

    Assume baseflow = 10 m3/s, K = 4 h, Δt = 1 h.

    With K = 4 h, and Δt = 1 h, the routing coefficients (Eqs. 8-16 to 8-18) are:

    C0 = 1/9 = 0.111; C1 = 1/9 = 0.111; and C2 = 7/9 = 0.778

    The routing computations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 C0I2
    		 C1I1
    		 C2O1
    		 Outflow
    (m2/s)
    0 10 - - - 10.00
    1 20 2.22 1.11 7.78 11.11
    2 50 5.55 2.22 8.64 16.41
    3 80 8.88 5.55 12.77 27.20
    4 90 9.99 8.88 21.16 40.03
    5 100 11.10 9.99 31.14 52.23
    6 90 9.99 11.10 40.63 61.72
    7 60 6.66 9.99 48.02 64.67
    8 50 5.55 6.66 50.31 62.52
    9 40 4.44 5.55 48.64 58.63
    10 30 3.33 4.44 45.61 53.38
    11 20 2.22 3.33 41.53 47.08
    12 10 1.11 2.22 36.63 39.96
    13 10 1.11 1.11 31.09 33.31
    14 10 1.11 1.11 25.92 28.14
    15 10 1.11 1.11 21.89 24.11
    16 10 1.11 1.11 16.32 18.54
    17 10 1.11 1.11 16.32 18.54
    18 10 1.11 1.11 14.42 16.64
    19 10 1.11 1.11 12.94 15.16
    20 10 1.11 1.11 11.79 14.01
    21 10 1.11 1.11 10.90 13.12
    22 10 1.11 1.11 10.21 12.43
    23 10 1.11 1.11 9.67 11.89
    24 10 1.11 1.11 9.25 11.47

    The calculated peak outflow is 64.67 m3/s, and it occurs at 7 h.  ANSWER.

  4. Develop a spreadsheet to route a triangular inflow hydrograph through a linear reservoir. Inputs to the program are the following: peak inflow, baseflow, time-to-peak, time base, storage constant, and time interval. Test your work using Problem 8- 2.

    TBD

  5. Use the spreadsheet developed in Problem 8-4 to route the following inflow hydrograph: peak inflow = 750 m3/s, baseflow = 50 m3/s, time-to-peak = 3 h, time base = 8 h, storage constant K = 1.5 h, time interval Δt = 0.5 h.

    The spreadsheet developed in Problem 8-4 is used to route the following triangular inflow hydrograph:

    peak inflow = 750 m3/s; base flow = 50 m3/s; time-to-peak = 3 h; time base = 8 h; K = 1.5 h; and Δt = 0.5 h.

    The result are as follows: Peak outflow = 562.859 m3/s, and it occurs after 4.5 h.  ANSWER.

  6. Develop a spreadsheet to route an inflow hydrograph of arbitrary shape through a linear reservoir. Inputs are the following: inflow hydrograph ordinates, baseflow, reservoir storage constant, and time interval. Test your work using Problem 8-3.

    TBD

  7. Use the spreadsheet developed in Problem 8-6 to study the sensitivity of the outflow hydrograph to the chosen value of storage constant K. Use the following inflow hydrograph:

    Time (h) 0 1 2 3 4 5 6 7 8 9 10 11 12
    Inflow (m3/s) 0 10 30 50 80 100 90 60 40 30 20 10 0

    Assume baseflow = 0 m3/s, and Δt = 1 h. Report calculated peak outflow and time-to-peak for (a) K = 1 h, (b) K = 2 h, (c) K = 3 h, and (d) K = 4 h. Verify your results using the online calculator ONLINE ROUTING01.

    The spreadsheet developed in Problem 8-6 is used to study the sensitivity of the outflow hydrograph to the chosen value of storage constant K. The time interval is: Δt = 1 h. The results are summarized in the following table.

    Case Storage
    constant K
    (h)    		 Peak outflow
    (m3/s)    		 Time-to-peak
    (h)
    a 1 89.314 6
    b 2 74.165 7
    c 3 64.931 7
    d 4 56.745 7

    It is seen that an increase in storage constant K causes a decrease in peak outflow and a small increase in time-to-peak. However, it should be noted that the rather coarse temporal resolution used for this problem (i.e., a time interval Δt = 1 h) precludes a more precise determination of time-to-peak, and therefore, of peak outflow. The calculated values are only approximations to the more precise values which can be obtained using a higher temporal resolution (i.e., a smaller time interval).  ANSWER.

  8. Using a spreadsheet, solve Problem 8-1 by the storage indication method. Verify with ONLINE ROUTING02.

    With K = 3 h: S = 3 (O )

    With Δt = 1 h, the storage-indication-outflow relation O ∝ [(2S / Δt) + O ] is:

    O = (1/7) [(2S / Δt) + O ]

    The calculations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 [ (2S / Δt) - O ]
    (m3/s)    		 [ (2S / Δt) + O ]
    (m3/s)    		 Outflow
    (m2/s)
    0 0 0 0 0
    1 10 7.14 10.00 1.43
    2 20 26.52 37.14 5.31
    3 30 54.66 76.52 10.93
    4 40 89.04 124.66 17.81
    5 50 127.88 179.04 25.58
    6 40 155.62 217.88 31.13
    7 30 161.16 225.62 32.23
    8 20 150.82 211.16 30.17
    9 10 129.16 180.82 25.83
    10 0 99.40 139.16 19.88
    11 0 71.00 99.40 14.20
    12 0 50.72 71.00 10.14
    13 0 36.24 50.72 7.24
    14 0 25.88 36.24 5.18
    15 0 18.48 25.88 3.70
    16 0 13.20 18.48 2.64
    17 0 9.42 13.20 1.89
    18 0 6.72 9.42 1.35
    19 0 4.80 6.72 0.96
    20 0 3.42 4.80 0.69

    The calculated peak outflow is 32.23 m3/s, and it occurs at 7 h.  ANSWER.

  9. Using a spreadsheet, solve Problem 8-2 by the storage indication method. Verify with ONLINE ROUTING02.

    With K = 2 h: S = 2 (O )

    With Δt = 1 h, the storage-indication-outflow relation O ∝ [(2S / Δt) + O ] is:

    O = (1/5) [(2S / Δt) + O ]

    The calculations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 [ (2S / Δt) - O ]
    (m3/s)    		 [ (2S / Δt) + O ]
    (m3/s)    		 Outflow
    (m2/s)
    0 0 0.00 0.00 0.00
    1 20 12.00 20.00 4.00
    2 40 43.20 72.00 14.40
    3 60 85.92 143.20 28.64
    4 80 135.56 225.92 45.18
    5 100 189.34 315.56 63.11
    6 120 245.60 409.34 81.87
    7 108 284.16 473.60 94.72
    8 96 292.90 488.16 97.63
    9 84 283.74 472.90 94.58
    10 72 263.84 439.74 87.95
    11 60 237.50 395.84 79.17
    12 48 207.30 345.50 69.10
    13 36 174.78 291.30 58.26
    14 24 140.86 234.78 46.96
    15 12 106.12 176.86 35.37
    16 0 70.86 118.12 23.62
    17 0 42.52 70.86 14.17
    18 0 25.52 42.52 8.50
    19 0 15.34 25.52 5.10
    20 0 9.20 15.34 3.07

    The calculated peak outflow is 97.63 m3/s, and it occurs at 7 h.  ANSWER.

  10. Using a spreadsheet, solve Problem 8-3 by the storage indication method. Verify with ONLINE ROUTING02.

    With K = 4 h: S = 4 (O )

    With Δt = 1 h, the storage-indication-outflow relation O ∝ [(2S / Δt) + O ] is:

    O = (1/9) [(2S / Δt) + O ]

    The calculations are shown in the following table.

    Time
    (h)    		 Inflow
    (m3/s)    		 [ (2S / Δt) - O ]
    (m3/s)    		 [ (2S / Δt) + O ]
    (m3/s)    		 Outflow
    (m2/s)
    0 10 70.00 90.00 10.00
    1 20 77.78 100.00 11.11
    2 50 114.94 147.78 16.42
    3 80 190.50 244.94 27.22
    4 90 280.38 360.50 40.06
    5 100 365.86 470.38 52.26
    6 90 432.34 555.86 61.76
    7 60 452.94 582.34 64.70
    8 50 437.84 562.94 62.55
    9 40 410.54 527.84 58.65
    10 30 373.76 480.54 53.39
    11 20 329.60 423.76 47.08
    12 10 279.68 359.60 39.96
    13 10 233.08 299.68 33.30
    14 10 196.84 253.08 28.12
    15 10 168.66 216.84 24.09
    16 10 146.74 188.66 20.96
    17 10 129.68 166.74 18.53
    18 10 116.42 149.68 16.63
    19 10 106.10 136.42 15.16
    20 10 98.08 126.10 14.01
    21 10 91.84 118.08 13.12
    22 10 86.98 111.84 12.43
    23 10 83.20 106.98 11.89
    24 10 80.26 103.20 11.47

    The calculated peak outflow is 64.70 m3/s, and it occurs at 7 h.  ANSWER.

  11. Use ONLINE ROUTING03 to solve the following reservoir routing problem: emergency spillway width L = 15 m, rating coefficient Cd = 1.886, rating exponent y = 1.5, emergency spillway crest elevation = 730 m, dam crest elevation = 735 m, initial pool elevation = 730.5 m, baseflow = 10 m3/s. At spillway crest elevation, the reservoir storage is 3,000,000 m3, increasing linearly to 4,000,000 m3 at dam crest elevation. The inflow hydrograph is the following:

    Time (h) 0 1 2 3 4 5 6 7 8 9 10 11 12
    Inflow (m3/s) 10 30 70 150 210 250 170 110 70 50 30 20 10

    Set Δt = 1 h. Report peak outflow, time-to-peak, maximum pool elevation, and effective freeboard.

    The discharge rating is: Q = Cd L H y = 1.886 × 15 × H 1.5 = 28.29 H 1.5

    The following table shows the elevation, hydraulic head, outflow, and storage.

    Elevation
    (m)    		 Hydraulic
    head
    (m)    		 Outflow
    (m3/s)    		 Storage
    (m3
    730 0 0 3,000,000
    731 1 28.290 3,200,000
    732 2 80.016 3,400,000
    733 3 146.999 3,600,000
    734 4 226.320 3,800,000
    735 5 316.292 4,000,000

    The program ONLINE ROUTING03 is run with the given data, and the results are summarized as follows:

    peak outflow Qp = 217.686 m. ANSWER.

    time-to-peak tp = 5 h.  ANSWER.

    maximum pool elevation = 733.891 m. ANSWER.

    effective freeboard = 1.109 m. ANSWER.

  12. Using the data of Problem 8-11, modify the volumetric characteristics of the reservoir to the following: storage at spillway crest elevation, 6,000,000 m3; storage at dam crest elevation, 8,000,000 m3. Run ONLINE ROUTING03 using Δt = 1 h. Report peak outflow, time-to-peak, maximum pool elevation, and effective freeboard. Compare with the results of Problem 8-11, explaining the differences.

    The program ONLINE ROUTING03 is run with the given data, and the results are summarized as follows:

    peak outflow Qp = 188.12 m. ANSWER.

    time-to-peak tp = 6 h.  ANSWER.

    maximum pool elevation = 733.518 m. ANSWER.

    effective freeboard = 1.482 m. ANSWER.

    Since this reservoir is twice as large, the peak outflow is reduced, the time-to-peak has increased, the maximum pool elevation has decreased and the effective freeboard has increased. The larger reservoir has resulted in a safer design. ANSWER.

  13. Determine the actual freeboard for the following dam, reservoir, and flood conditions: dam crest elevation = 125 m; emergency spillway crest elevation = 120 m; coefficient of spillway rating Cd = 1.7; exponent of spillway rating y = 1.5; width of emergency spillway (rectangular cross section) L = 18 m.

    Elevation-storage relation:

    Elevation (m) 120 121 122 123 124 125
    Storage (m3) 3,000,000 3,050,000 3,150,000 3,350,000 3,750,000 4,250,000

    Inflow hydrograph to reservoir:

    Time (h) 0 1 2 3 4 5 6 7 8 9 10
    Inflow (m3/s) 0 10 15 30 55 85 105 125 150 135 110

    Time (h) 11 12 13 14 15 16 17 18 19 20 21
    Inflow (m3/s) 95 72 55 38 29 14 9 7 2 1 0

    Assume the initial reservoir pool level at spillway crest. Use ONLINE ROUTING03.

    Using the program ONLINE ROUTING03 with the given reservoir data and design flood conditions, the peak outflow is:

    Qp = 140.246 m3/s.

    The maximum water surface elevation corresponding to this peak outflow is: 122.741 m. Since the dam crest elevation is at 125 m, the actual freeboard is:

    FB = 125.0 - 122.741 = 2.259 m. ANSWER.

  14. Design the emergency spillway width L (to a 0.1 m accuracy) for the following dam, reservoir, and flood conditions: dam crest elevation = 483 m; emergency spillway crest elevation = 475 m; coefficient of spillway rating = 1.7; exponent of spillway rating = 1.5.

    Elevation-storage relation:

    Elevation (m) 475 477 479 481 483
    Storage (m3) 5100,000 5300,000 5600,000 6400,000 7600,000

    Inflow hydrograph to reservoir:

    Time (h) 0 1 2 3 4 5 6 7 8 9 10 11 12
    Inflow (m3/s) 0 10 30 50 90 150 250 350 280 210 190 170 130

    Time (h) 13 14 15 16 17 18 19 20 21 22 23 24
    Inflow (m3/s) 100 90 75 50 40 30 15 10 5 2 1 0

    Assume minimum design freeboard = 3 m, and initial reservoir pool level at spillway crest. Use ONLINE ROUTING03.

    With the given reservoir and design flood conditions, the objective is to determine the emergency spillway width (rectangular cross section) that produces a maximum water surface elevation of 480.0 m, i.e., the diference between dam crest elevation (483.0 m) and design freeboard (3.0 m). The calculation proceeds by trial and error, using program ONLINE ROUTING03. The trials are summarized in the following table.

    Emergency
    spillway width
    (m)    		 Maximum
    water surface elevation
    (m)
    10.0 481.088
    12.0 480.617
    14.0 480.199
    15.0 480.029
    15.1 480.013
    15.2 479.961

    The emergency spillway width that assures a minimum freeboard (within 0.1 m accuracy) equal to 3.0 m is: B = 15.2 m. ANSWER.

  15. A development is planned for a 85-acre watershed that outlets into an existing channel designed for present conditions. If the channel capacity is exceeded, damages will be substantial. The watershed is in the Type I storm distribution region. The present channel capacity is 150 cfs. The planned development will produce a storm runoff depth of 3 in. and a peak discharge of 320 cfs at the watershed outlet. Size a detention basin to reduce the post-development peak flow to pre-development conditions.

    The ratio of peak outflow to peak inflow, i.e., pre-development peak flow to unattenuated post development peak flow is: Qo / Qi = 150 / 320 = 0.469

    From Fig. 8-13, for a Type I storm : Vs / Vr = 0.193

    The storm runoff volume is: Vr = 85 ac. × (3 in / 12 in) = 21.25 ac-ft.

    Therefore, the detention basin storage volume is: Vs = 0.193 × 21.25 = 4.09 ac-ft.  ANSWER.

    The same results are obtained by using ONLINE_TR55_DETENTION.

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