28972761 irriagtion engineering amp hydraulic structures santosh kumar garg 19 edition

7/28/2019 28972761 Irriagtion Engineering Amp Hydraulic Structures Santosh Kumar Garg 19 Edition

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'I V T

RESERVOIRS AND PLANNING FOR DAM RESERVOIRS

943

(xi) Outflow ordinates are plotted so as to obtain the outflow hydrograph. The point at which it crosses the inflow hydrograph gives the peak outflow rate. Fromthis time, the rate of outflow begins to fall due to decrease in the inflow rate. (xii) The time lag between the two peaks is evaluated as to give the time lag.

An example has been solved to make the procedure very clear. Example 18.13. Theinflow flood discharges for a possible worst flood are tabulated in Table 18.22at suitable intervals starting from 0.00 hours on august 20, 1975.Table 18.22T~me from stari in hr

0

6

12

18

24

30

36

42

48

51

60

66

78

90

102

114

Dischargein cumecs

0

50

280

610 1290 1900 2130 1900 1600 1440 1060 780 500 370 220



130

Thisflood approaches a reservoir with an uncontrolled spillway, the crest of which is kept at RL 140.0 m. Determine the maximum reservoir level and the hydrograph of the routed flood. Values of reservoir capacity'(above spillway crest) andoutflow discharge at various elevations are tabulated in Tables 18.23 and 18.24respectively.Table 18.23Elev~~tion metres in Reservoir storage with spillway crest as datum inmillion crtblc metres (m.c.rn.) 140.0 0.0 141.0 15.0 142.0 35.0 143.0 -60.0144.0 145.0 146.0 240.0

.

95.0----140.0-

Table 18.24Elevation 140.0

IOutfow discharge in cumecs0

Solution. The elevation storage curve and the elevation outflow curve are plotte

d with the help of Tables 18.23 and 18.24, as shown in Fig. 18.20 and 18.21 respectively. The hydrograph of the given flood is plotted in Fig. 18.22. Flood routing is carried out by hit and trial method as shown in Table 18.25 and as explained earlier. This table is otherwise self-explanatory.

28972761 irriagtion engineering amp hydraulic structures santosh kumar garg 19 edition

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