sm_ch09

Water Resources Engineering

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9 - 1 Chapter 9 9.1.1. The computation of the storage outflow function is given in table below. The computation of the first Q t S 2 ± ' value is shown as an example. For S = 70,000,000 Q = 0 and ' t = 1 hr, Q t S 2 ± ' = 0 1(3600) ) 70000000 ( 2 ± = 38889 m 3 /s. Storage, S (10 6 m 3 ) 70 80 85 100 115 Outflow, Q (m 3 /s) 0 50 150 350 700 Q t S 2 ± ' (m 3 /s) 38889 44494 47372 55906 64589 9.2.1. The storage outflow from problem 9.1.1 was used to route the inflow hydrograph, as given in the following table. The computed result is given in the following table. Note that although the routing can be continued beyond the last time step shown in the table, the result for only one time step after the peak outflow was observed is given here. S o = 70 x 10 6 m 3 Time (hr) Inflow, I (m 3 /s) I j + I j+1 (m 3 /s) 2S j / ' t – Q j (m 3 /s) 2S j+1 / ' t + Q j+1 (m 3 /s) Q j (m 3 /s) 0 0 0 1 40 40 38889 38929 0.4 2 60 100 38928 39028 1.24 3 150 210 39026 39236 3.10 4 200 350 39229 39579 6.16 5 300 500 39567 40067 10.51 6 250 550 40046 40596 15.23 7 200 450 40566 41016 18.97 8 180 380 40978 41356 22.02 9 220 400 41314 41714 25.20 10 320 540 41663 42203 29.56 11 400 720 42144 42864 35.46 12 280 680 42793 43473 40.89 13 190 470 43391 43861 44.35 14 150 340 43773 44113 46.60 15 50 200 44020 44220 47.56 16 0 50 44124 44174 47.15 9.2.2. The computation for this problem is similar to that for problem 9.2.1. Hence, the 2S/ ' t + Q versus Q relationship given in problem 9.2.1 has been used. The following storage-outflow relation was obtained for initial storage of 80 x 10 6
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9 - 2 m 3 , which is used to route the given inflow hydrograph. Note that at an initial storage of 80 x 10 6 m 3 , there was some initial outflow which was on a recession limb of a previous hydrograph. Therefore, initially the outflow decreases from a 50 m 3 /s to 47.9 m 3 /s and then increases to a peak of 156.0 m 3 /s due to the new inflow hydrograph given in the problem. S (10 6 ) Q 2S/ ' t + Q 80 50 44494 85 150 47372 100 350 55906 115 700 64589 S o = 80 x 10 6 m 3 Time (hr) Inflow, I, (m 3 /s) I j +I j+1 2S j /&t-Q j 2S j+1 /&t+Q j+1 Q j (m 3 /s) 0 0 0 44394 44494 50.0 1 40 40 44339 44434 47.9 2 60 100 44342 44439 48.1 3 150 210 44448 44552 52.0 4 200 350 44677 44798 60.6 5 300 500 45030 45177 73.7 6 250 550 45404 45580 87.7 7 200 450 45660 45854 97.3 8 180 380 45833 46040 103.7 9 220 400 46012 46233 110.4 10 320 540 46309 46552 121.5 11 400 720 46753 47029 138.1 12 280 680 47130 47433 151.4 13 190 470 47289 47600 155.3 14 150 340 47317 47629 156.0 15 50 200 47210 47517 153.4 16 0 50 46968 47260 146.1 17 0 0 46696 46968 136.0 18 0 0 46443 46696 126.5 19 0 0 46208 46443 117.7 20 0 0 45989 46208 109.5 21 0 0 45785 45989 101.9 9.2.3. For this problem, there may be no specifc answer as different programmers may use different styles and programming languages. A sample computer program written in FORTRAN is given below. PROGRAM SAMPLE5 (INPUT, OUTPUT, TAPE5=INPUT, TAPE6=OUTPUT) DIMENSION S(30), QS(30), Q(50), SFUNC(30) C ******************************************************************************************* C C THIS PROGRAM IS FOR RESERVOIR LEVEL POOL ROUTING FOR PROBLEM 9.2.3. C C ********************************************************************************************
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9 - 3 READ (5, 101) SJ0, DT, DTT, JSS, JNN, TIM 101 FORMAT(3F10.0, 2I10, F10.0) READ (5, 102) (S(JS), JS = 1, JSS) READ (5, 102) (QS(JS), JS = 1, JSS) READ (5, 102) (Q(JN), JN = 1, JNN) 102 FORMAT(8F10.2) C ******************************************************************************************** C C INPUT DATA C CONV = CONVERSION FACTOR = 1 C SJ0 = INITIAL RESERVOIR STORAGE C DT = ROUTING INTERVAL, MIN C DTT = TIME INTERVAL FOR HYDROGRAPH, MIN C JSS = NUMBER OF VALUES DESCRIBING DISCHARGE-STORAGE RELATION C JNN = NUMBER OF VALUES DESCRIBING INFLOW HYDROGRAPH C TIM = TOTAL TIME FOR ROUTING COMPUTATIONS, MIN C S(JS) = RESERVOIR STORAGE C QS(JS) = SPILLWAY DISCHARGE C Q(JN) = RESERVOIR INFLOW C C ********************************************************************************************* DT = DT * 60.0 DTT = DTT * 60.0 TIM = TIM * 60.0 NTIM = TIM/DT + 1 WRITE(6, 200) 200 FORMAT(5X, ‘LEVEL POOL ROUTING’,// +5X, ‘DISCHARGE STORAGE RELATION’,//) WRITE(6, 202) 202 FORMAT(8X,’STORAGE’,2X,’DISCHARGE’,2X’,STORAGE FUNCTION’,) DO 50 JS = 1, JSS SFUNC(JS) = 2.0*S(JS)/DT+QS(JS)
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sm_ch09 - 9-1 Chapter 9 9.1.1. The computation of the...

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