Flood Routing.ppt:flood routing and control
- 1. Flood Routing By Brook A. (PhD)
- 2. Reservoir routing • Flood routing is the process of determining the reservoir stage, storage volume and of the outflow hydrograph corresponding to a known hydrograph of inflow into the reservoir. • For this, the capacity curve of the reservoir, i.e., ‘storage vs pool elevation’ and ‘outflow rate vs pool elevation’ , curves are required. • Storage volumes for different pool elevations are determined by planimetering the contour map of the reservoir site. • For example, the volume of water stored (V) between two successive contours having areas A1 and A2 (planimetered) and the contour interval d, is given by
- 3. • Cone formula, V • Prismoidal formula, Where Am = (A1+A2)/2, i.e., area midway between the two successive contours. The Prismoidal formula is more accurate. The outflow rates are determined by computing the discharge through the sluices and the spillway discharge for different water surface elevations of the reservoir (i.e., pool elevations): • Discharge through sluices, • Discharge over spillway crest, • Outflow from the reservoir O = Qsl +Qsp
- 4. Where: h = height of water surface of reservoir above the centre of sluice H = height of water surface of reservoir above the crest of spillway Cd = coefficient of discharge for the sluice C = coefficient of spillway A = area of sluice opening L = length of spill way • The problem in flood routing is to determine the relation between the inflow, the outflow and the storage as a function of time. • The problem can be solved by applying the hydrologic equation I = O + ∆S (5) • Where I = inflow rate, O = outflow rate, ∆S = incremental storage, at any instant.
- 5. • Taking a small interval of time, t (called the routing period and designating the initial and final conditions by subscripts 1 and 2 between the interval, eq. (5) may be written as (6) • The routing period, t selected should be sufficiently short such that the hydrograph during the interval 1-2 can be assumed as a straight line, i.e., Imean = (I1+I2)/2. • The above equation can be rearranged as (7)
- 6. • After selecting a routing period t, curves of O vs S, and O vs S Ot/2 on either side of O – S curve are drawn. • At the beginning of the routing period all the terms on the left side of the equation are known and the value of the right side terms is found out. • Corresponding to this O2 and (S-Ot/2) are read from the graph, which become the initial values for the next routing period and so on.
- 7. • This method of flood routing is called the ISD (Inflow-storage –discharge) method. • Here it is assumed that the outflow (i.e., discharge) from the reservoir is a function of the pool elevation provided that the spillway and the sluices have no gates (i.e., uncontrolled reservoirs) or with constant gate openings, if provided with control gates for which pool elevation vs. discharge curves are drawn.
- 8. • Equation (6) may be rearranged as: (8) • After selecting a routing period t, a curve of 2S/t + O vs O can be drawn since (2S/t + O) – 2O = 2S/t – O, a curve of 2S/t – O vs O can also be drawn. • At the beginning of the routing period all terms on the left of the equation are known. • This method is called Storage Indication Method.
- 9. Example: • For the reservoir with constant gate openings for the sluices and spill way, pool elevation vs storage and discharge (outflow) curves are shown in fig 1. The inflow hydrograph into the reservoir is given below: Time(hr) 0 6 12 18 24 30 36 42 Inflow(cumec) 50 70 160 300 460 540 510 440 Time(hr) 48 54 60 66 72 78 84 90 Inflow(cumec) 330 250 190 150 120 90 80 70
- 10. Pool elevation vs storage and discharge
- 11. • Pool elevation at the commencement = 110 m • Discharge at the commencement = 124 cumec • Route the flood through the reservoir by a) ISD method, and b) storage indicator method, and C) compute the outflow hydrograph, the maximum pool elevation reached, the reduction in the flood peak and the reservoir lag.
- 12. Solution a) Flood routing by ISD method Take the routing period as 6hr or 1/4day. • It is easier to work the flow rates in cumec and the storage volumes in terms of cumec – ¼ day. • Hence, the storage in Mm3 is converted to cumec-1/4day by multiplying by 46.3, Table 1. • Corresponding to an initial pool elevation of 110 m, O =124 cumec, S = 49.1 Mm3=49.1*46.3 =2270 cumec- 1/4day, • Ot/2 = O/2*t = 124/2 cumec*1/4day = 62 cumec-1/4day, • S + Ot/2 = 2270 + 62 = 2332 cumec-1/4day, and • S - Ot/2 = 2270 – 62 = 2208 cumec – ¼ day.
- 13. Tabl 1: Tabulation for drawing (i) S Ot/2 and ii) 2S/t O curves for routing the flood through the reservoir Pool elevation (m) Outflow O (cumec) Storage S Computation for ISD method (t = 6hr = ¼ day) Computation for storage indicator method (Mm3) (cumec- 1/4 day) Ot/2 (cumec – ¼ day) S + Ot/2 (cumec – ¼ day) S – Ot/2 (cumec – ¼ day) 2S/t (cumec) 2S/t + O (cumec) 2S/t – O (cumec) 100 60 8.7 400 30 430 370 800 860 740 102 70 15.1 700 35 735 665 1400 1470 1330 104 86 23.4 1480 43 1123 1037 2160 2246 2074 106 100 32.0 1480 50 1530 1430 2960 3060 2860 108 110 40.0 1850 55 1905 1795 3700 3810 3590 110 124 49.1 2270 62 2332 2208 4540 4664 4416 112 138 58.3 2700 69 2769 2631 5400 5538 5262 113 310 63.0 2920 155 3075 2765 5840 6150 5530 114 550 68.3 3160 275 3435 2885 6320 6870 5770 115 800 73.5 3400 400 3800 3000 6800 7600 6000 116 1030 78.8 3650 515 4165 3135 7300 8330 6270 117 1280 83.8 3880 640 4520 3240 7760 9040 6480 118 1520 90.0 4160 760 4920 3400 8320 9840 6800 120 - 101.0 4680 - - - - - - •1 cumec- ¼ day = 1 * 6*60 = 21600 m3 , 1 million m3 (Mm3) = 106/21600 = 46.3 cumec – ¼ day.
- 14. Fig 2: Reservoir routing by ISD method
- 15. • First O vs S curve is drawn. • For a particular O on the S curve, Ot/2 abcissa units may be set off on either side of the S curve and this is repeated for other values of O. • The points obtained on either side of S curve plot S + Ot/2 and S – Ot/2 curves as shown in fig 2.
- 16. • For routing the flood by the ISD method, table 2, for the known outflow at the commencement of 124 cumec, S – Ot/2 is read from the curve as 2208 cumec-1/4 day and to this (I1 + I2)*t/2 = (50 +70)/2 cumec * ¼ day = 60 cumec- ¼ day is added to get the right hand side of eq. (7) i.e., S + Ot/2 = 2268 and corresponding to this O = 120 cumec is read from the graph which is the outflow at the beginning of the next routing period.
- 17. • Corresponding to this O = 120 cumec, the pool elevation of 109.2 m is read from the pool elevations vs O curve. • Corresponding to this O = 120 cumec, S – Ot/2 =2040 is read from the graph and (I1 + I2)*t/2 = (70 +160)t/2 = 115 cumec – ¼ day is added to get S + Ot/2 = 2155 for which O is read as 116 cumec and pool elevation as 108.4 m. • Thus the process is repeated till the flood is completely routed through the reservoir and the outflow hydrograph is obtained as shown in fig 3.
- 18. Table 2: reservoir routing – ISD method Time (hr) Inflow I (cumec) (I1 + I2)t/2 (cumec – ¼ day) Outflow O (cumec) S – Ot/2 (cumec – ¼ day) S + Ot/2 (cumec – ¼ day) Pool Elevation (m) 0 50 124 110.0 60 + 2208 2268 6 70 120 109.2 115 + 2040 2155 12 160 116 108.4 230 + 1960 2190 18 300 119 109.1 380 2020 2400 24 460 122 109.6 500 2080 2580 30 540 130 110.8 525 2380 2905 36 510 195 112.5 475 2730 3205 42 440 395 113.4 385 2820 3205 48 330 395 113.4 290 2920 3110 54 250 335 113.1 220 2790 3010 60 190 265 112.8 170 2760 2930 66 150 210 112.6 135 2740 2875 72 120 170 112.4 105 2720 2825 78 90 145 112.3 85 2700 2785 84 80 132 111.2 75 2650 2725 90 70 130 110.8
- 19. Fig 3: Reservoir routing
- 20. • b) flood routing by storage indicator method: corresponding to the initial pool elevation of 110 m, O =124 cumec, S = 2270 cumec – ¼ day, 2S/t = (2*2270 cumec – ¼ day)/(1/4 day) = 4540 cumec, 2S/t + O = 4540 + 124 = 4664 cumec and 2S/t – O = 4540 – 124 = 4416 cumec. • Thus, for other values of O, values of 2S/t + O and 2S/t – O are computed and ‘O vs 2S/t + O and 2S/t – O curves are drawn as shown in fig 4.
- 21. Fig 4: Reservoir routing by modified Puls method
- 22. • For routing the flood by the modified Puls method (storage indicator method), Table 3 , corresponding to the initial pool elevation of 110 m, O = 124 cumec, 2S/t + O = 4664 cumec and 2S/t – O = 4416 cumec are read off. • For this 2S/t – O = 4416 cumec, (I1 +I2) = 50 + 70 = 120 cumec is added to get the right hand side of eq (8), i.e., 2S/t + O = 4416 + 120 = 4536 cumec. • For this value of 2S/t + O, O = 123 cumec, and 2S/t – O = 4290 cumec are read off from the curves. • For O = 123 cumec, the pool elevation of 109.8 m is read off from the O vs pool elevation curve.
- 23. Table 3 Reservoir routing – modified Puls method Time (hr) Inflow I (cumec) 2S/t – O* (cumec) 2S/t + O (cumec) Outflow O (cumec) Pool elevation (m) 0 50 4416 4464 124 110.0 6 70 + 4290 4536 123 109.8 12 160 + 4276 4520 122 109.6 18 300 4482 4736 126 111.8 24 460 4986 5248 131 111.0 30 540 5506 5986 240 112.7 36 510 5696 6556 430 113.5 42 440 5716 6646 465 113.6 48 330 5666 6486 410 113.4 54 250 5586 6246 330 113.0 60 190 5526 6026 250 112.7 66 150 5466 5866 200 112.5 72 120 5436 5736 150 112.3 78 90 5476 5646 135 111.6 84 80 5278 5546 134 111.4 90 70 5428 130 110.8 2S/t – O = (2S/2 + O) - 2O
- 24. • These values become the initial values for the next routing period. Again, for 2S/t – O = 4290 cumec, I1 + I2 = 70 +160 = 230 cumec is added to get the right hand side of equation (8), i.e., 2S/t + O = 4290 + 230 = 4520 cumec for which O and 2S/t – O values are read off and pool elevation obtained, which become the initial values for the next routing period. • Thus the process is repeated till the flood is completely routed through the reservoir and the outflow hydrograph is obtained as shown in fig 3 by dashed line.
- 25. Results : ISD method Modified Puls method i) Maximum pool elevation reached 113.5 m* 113.6 m i) Reduction in flood peak 132 cumec 75 cumec i) Reservoir lag 14.5 hr 12hr * To pass the crest of the out flow hydrograph