CE322 Basic Hydrology

Jorge A. Ramirez

Level Pool Routing - Example

Use the Storage-Indication Method to route the Input hydrograph tabulated below.

 Time (h) Input Hydrograph (m3/s) Time (h) Input Hydrograph (m3/s) 0 0 0.00 90 420.000 6 50.000 96 320.000 12 130.000 102 270.000 18 250.000 108 200.000 24 350.000 114 150.000 30 540.000 120 100.000 36 735.000 126 72.000 42 1215.000 132 45.000 48 1800.000 138 25.000 54 1400.000 144 10.000 60 1050.000 150 0.000 66 900.000 156 0.000 72 740.000 162 0.000 78 620.000 84 510.000

This hydrograph flows into a reservoir whose storage and discharge characteristics are as presented in the following table. The initial storage in the system is 1'000,000 m3, and the initial outflow is 20 m3/s.

 H (m) O (m3/s) S (m3) 130 20 1.00E+06 131 34 1.69E+06 132 57 2.85E+06 133 96 4.80E+06 134 162 8.12E+06 136 463 2.31E+07 137 781 3.91E+07 138 1318 6.59E+07 139 2226 1.11E+08

&#9;Reservoir or level pool routing refers to routing for systems whose storage and outflow are related by a function of the type S(t) = f[O(t)] which is of the invariable type (unique, non-hysteretic). These relationships imply that for a given set of conditions (e.g. stage) the outflow is unique, independent of how that stage is achieved. Reservoirs or systems with horizontal water surfaces have S vs. O relationships of the invariable type. Such systems have a pool that is wide and deep compared to its length in the direction of flow, and low flow velocities in the reservoir. For such systems, the peak outflow occurs when the outflow hydrograph intersects the inflow hydrograph.

&#9;The Storage-Indication method is a level pool routing procedure for calculating the outflow hydrograph of a system with horizontal water surface, given its inflow hydrograph, and storage outflow characteristics. The solution involves integrating the continuity equation as indicated below, and rearranging terms such that all the unknown quantities are on the left-hand side of the equation.

Storage-Indication Routing Equation:

For a level pool reservoir, the storage is a unique function of elevation; and the outflow is a unique function of elevation. Thus, the left-hand side of the equation above is a unique function of elevation in the system, only. Usually, the storage-elevation relationship is available from topographic surveys, and the outflow-elevation relationship is available from hydraulic considerations with respect to the outlet structures (e.g. spillways, etc.)

&#9;The solution involves the development of the function 2S/Dt + O = f(O) and then solving it sequentially for every time step. These steps are illustrated below.

1. Develop the function 2S/Dt + O vs. O. Use a Dt of 6 hours, as suggested by the time interval of the inflow hydrograph.

 1 2 3 4 5 H (m) O (m3/s) S (m3) 2S/Dt (m3/s) 2S/Dt + O (m3/s) 130 20 1.00E+06 92.59259 112.5926 131 34 1.69E+06 156.4815 190.4815 132 57 2.85E+06 263.8889 320.8889 133 96 4.80E+06 444.4444 540.4444 134 162 8.12E+06 751.8519 913.8519 136 463 2.31E+07 2142.593 2605.593 137 781 3.91E+07 3615.741 4396.741 138 1318 6.59E+07 6103.704 7421.704 139 2226 1.11E+08 10303.7 12529.7

&#9;In the table above, Columns 1-3 are given. Columns 2 and 5 correspond to the desired function, 2S/Dt + O vs. O , which has been graphed above.

B - Proceed with the routing of the inflow hydrograph by using the Storage-Indication routing equation sequentially for every time step:

t = 0 -- i = 0. Initial Conditions: So = 1'000,000 m3; Oo = 20 m3/s.

t = 6 -- i = 1

(Io + I1) = (0 + 50) m3/s = 50 m3/s

(2So /Dt - Oo) = (2 x 1'000,000 m3)/(6 x 3600 s) + 20 m3/s = 72.593 m3/s

(2S1 /Dt + O1) = (Io + I1) + (2So /Dt - Oo) = 122.593 m3/s

Using the relationship (2S/Dt + O) vs. O developed in Part A, obtain the outflow O1 corresponding to the value of (2S1 /Dt + O1) obtained above. This is done by entering the graph with the value of (2S1 /Dt + O1) and exiting with the value of O1. Use interpolation as indicated below.

O1 = 20 m3/s + [(34 - 20)/(190.4815 - 112.5925)] (122.593 - 112.5925) m3/s = 21.797 m3/s

t = 12 -- i = 2

(I1 + I2) = (50 + 130) m3/s = 180 m3/s

(2S1 /Dt - O1) = (2S1 /Dt + O1) - 2 x O1 = 122.593 m3/s - 2 x 21.797 m3/s = 78.998 m3/s

(2S2 /Dt + O2) = (I1 + I2) + (2S1 /Dt - O1) = 258.998m3/s

Using the relationship (2S/Dt + O) vs. O developed in Part A, obtain the outflow O2 corresponding to the value of (2S2 /Dt + O2) obtained above. This is done by entering the graph with the value of (2S2 /Dt + O2) and exiting with the value of O2. Use interpolation as indicated below.

O2 = 34 m3/s + [(57 - 34)/(320.8889 - 190.4815)] (258.998 - 190.4815) m3/s = 46.084 m3/s

Proceed as above for every time step. Results are tabulated below.

 Time (h) I (m3/s) Ii + Ii+1 (m3/s) 2Si/Dt - OI (m3/s) 2SI+1/Dt + Oi+1 (m3/s) O (m3/s) 0 0 20 6 50 50 72.593 122.593 21.797 12 130 180 78.998 258.998 46.084 18 250 380 166.829 546.829 97.129 24 350 600 352.572 952.572 168.889 30 540 890 614.794 1504.794 267.142 36 735 1275 970.509 2245.509 398.933 42 1215 1950 1447.644 3397.644 603.621 48 1800 3015 2190.403 5205.403 924.556 54 1400 3200 3356.291 6556.291 1164.37 60 1050 2450 4227.552 6677.552 1185.896 66 900 1950 4305.76 6255.76 1111.018 72 740 1640 4033.723 5673.724 1007.694 78 620 1360 3658.336 5018.336 891.347 84 510 1130 3235.641 4365.641 775.479 90 420 930 2814.684 3744.684 665.234 96 320 740 2414.216 3154.216 560.402 102 270 590 2033.411 2623.411 466.164 108 200 470 1691.084 2161.084 383.912 114 150 350 1393.261 1743.261 309.571 120 100 250 1124.119 1374.119 243.892 126 72 172 886.334 1058.334 187.707 132 45 117 682.921 799.921 141.863 138 25 70 516.195 586.195 104.087 144 10 35 378.022 413.022 73.366 150 0 10 266.291 276.291 49.134 156 0 0 178.022 178.022 31.761 162 0 0 114.501 114.501 20.343