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# Hw assignments on flood routing update

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### Hw assignments on flood routing update

1. 1. Assignmentson FloodRoutingSolution HW 4©S.Nimtim
2. 2. Assignments on Flood Routing September 21, 20121. Storage vs. outflow characteristics for a proposed reservoir is given below. Calculate the storage- outflow function 2S/∆t + Q vs. Q for each of the tabulated values if ∆t = 2 h. Plot a graph of the storage-outflow function. Discharge Storage Q S 𝟐𝑺 ∆𝒕 + 𝑸 (cfs.) (ft2) 57 75 20890 227 81 22727 519 87.5 24824 1330 100 29108 2270 110.2 32881 ©S.Nimtim
3. 3. Assignments on Flood Routing September 21, 20122. Use the level pool routing method to route the hydrograph given below through the reservoir whosestorage-outflow characteristics are given in Prob. 1 .What is the maximum reservoir discharge andstorage? Assume that the reservoir initially contains 75 x 10^6 m3 of storage. Time Index Time Inflow 𝟐𝑺 𝟐𝑺 𝒋+𝟏 Outflow 𝑰 𝒋 + 𝑰 𝒋+𝟏 − 𝑸𝒋 + 𝑸 𝒋+𝟏 (j) (Minuit) (cms) ∆𝒕 ∆𝒕 (cms) 1 0 60 - 20776 - 57 2 2 100 160 20776 20936 61.26 3 4 232 332 20985 21146 80.65 4 6 300 532 21287 21517 115.02 5 8 520 820 21776 22117 170.22 6 10 1310 1830 22907 23606 349.69 7 12 1930 3240 24609 26147 769.00 8 14 1460 3390 25759 27999 1120.00 9 16 930 2390 25855 28149 1147.05 10 18 650 1580 25408 27435 1013.55 11 650 24552 26058 752.96 12 23590 24552 480.72 13 22896 23590 347.15 14 22395 22896 250.50 15 22002 22395 196.37 16 21682 22002 160.1 17 21421 21682 130.26 18 21209 21421 160.12 19 21036 21209 86.5 20 20963 21036 72.84 21 20899 20963 63.76 22 20840 20899 58.11 ©S.Nimtim
4. 4. Assignments on Flood Routing September 21, 20123. A watershed · divided into two subareas A and B. The surface runoff from subarea A enters a channel isat point 1 and flows to point 2 where the runoff from subarea B is added to the hydrograph and thecombined flow routed through a reservoir. Determine the discharge hydrograph from the reservoir,assuming that the reservoir is initil1 empty. What the areas of subareas A. and B in square miles?The reservoir has the following storage-outflow characteristics:Time (hr) 0 0.5 1 1.5 2.0Asculum RD 0 1 3 4 4.5Absolute RD 0 1 2 1 0.5Rainfall excess 0 0.6 1.6 0.6 0.1 ∅ = 0.8 𝑖𝑛/ℎ𝑟Rainfall excess 0 0.9 1.9 0.9 0.4 ∅ = 0.2𝑖𝑛/ℎ𝑟The areas of subareas A Time UH A Q1 Q2 Q3 Q4 𝑸 (𝒇𝒕 𝟑 /𝒔) 0 0 0 0 0.5 100 60 0 60 1 200 120 160 0 280 1.5 300 180 320 60 0 560 2 400 240 480 120 10 850 2.5 350 210 640 180 20 1050 3 300 180 560 240 30 1010 3.5 250 150 480 210 40 880 4 200 120 400 180 35 735 4.5 150 90 320 150 30 590 5 100 60 240 120 25 445 5.5 50 30 160 90 20 300 6 0 0 80 60 15 155 6.5 0 30 10 40 7 0 5 5 7.5 0 0 2400 × 0.5 × 3600 × 12 𝐴𝑟𝑒𝑎 𝐴 = = 1.86 𝑚𝑖𝑙𝑒 2 3 × 1760 2 ©S.Nimtim
5. 5. Assignments on Flood Routing September 21, 2012 Time UH A Q1 Q2 Q3 Q4 𝑸 (𝒇𝒕 𝟑 /𝒔) 0 0 0 0 0.5 200 180 0 180 1 500 450 380 0 830 1.5 800 720 950 180 0 1850 2 700 630 1520 450 80 2680 2.5 600 540 1330 720 200 2790 3 500 450 1140 630 320 2540 3.5 400 360 950 540 280 2130 4 300 270 760 450 240 1720 4.5 200 180 570 360 200 1310 5 100 90 380 270 160 900 5.5 50 45 190 180 120 535 6 0 0 95 90 80 265 6.5 0 45 40 85 7 0 20 20 7.5 0 0 4350 × 0.5 × 3600 × 12 𝐴𝑟𝑒𝑎 𝐵 = = 3.37 𝑚𝑖𝑙𝑒 2 3 × 1760 2Routing Equation by Muskingum method.K = 0.5 hourX = 0.25 ∆𝑡 − 2𝑘𝑥 𝐶1 = = 0.2 2𝑘 1 − 𝑥 + ∆𝑡 ∆𝑡 + 2𝑘𝑥 𝐶2 = = 0.6 2𝑘 1 − 𝑥 + ∆𝑡 2𝑘 1 − 𝑥 − ∆𝑡 𝐶3 = = 0.2 2𝑘 1 − 𝑥 + ∆𝑡 𝐶1 + 𝐶2 + 𝐶3 = 1 0.2 + 0.6 + 0.2 = 1 ©S.Nimtim
6. 6. Assignments on Flood Routing September 21, 2012Routing Inflow 𝑪 𝟏 + 𝑰 𝒋+𝟏 𝑪 𝟐 𝑰𝒋 𝑪 𝟑 𝑸𝒋 OutflowPeriod j 𝑪 𝟏 = 𝟎. 𝟐 𝑪 𝟐 = 𝟎. 𝟔 𝑪 𝟑 = 𝟎. 𝟐 Q (hr) (cfs) (cfs) 0 0 - - - 0 0.5 180 12 0 0 12.00 1 830 56 36 2.40 94.40 1.5 1850 112 168 18.88 298.88 2 2680 170 336 59.78 565.78 2.5 2790 210 510 113.16 833.16 3 2540 202 630 166.63 998.63 3.5 2130 176 606 199.73 981.73 4 1720 147 528 196.35 871.35 4.5 1310 118 441 174.27 733.27 5 900 89 354 146.65 589.65 5.5 535 60 267 117.93 444.93 6 265 31 180 88.99 299.99 6.5 85 8 93 66.00 161.00 7 20 1 24 32.20 57.20 7.5 0 0 3 11.44 14.44 8 0 0 0 2.888 2.888 Storage Q Outflow 𝟐𝒔 + 𝑸 ∆𝒕 (ac.ft.) (cu.ft.) (cfs) (cfs) 0 0 0 0 220 9583200 2 10650 300 13068000 20 14540 400 17424000 200 19560 500 21780000 300 24500 600 26136000 350 29390 700 30492000 450 34330 1100 47916000 1200 54440 ©S.Nimtim
7. 7. Assignments on Flood Routing September 21, 2012Time Time Inflow 𝑰 𝒋 + 𝑰 𝒋+𝟏 (𝟐𝒔 ∆𝒕) − 𝑸 𝒋 𝟐𝑺 𝒋 ∆𝒕 + 𝑸 𝒋+𝟏 OutflowIndex (minute) (cfs) (cfs) (cfs) (cfs) (cfs) j 1 0 0.00 - 0.00 - 0.00 2 0.5 192.00 192 191.93 192.00 0.04 3 1.0 924.40 1116 1307.84 1308.33 0.25 4 1.5 2148.88 3073 4379.47 4381.12 0.82 5 2.0 3245.78 5395 9770.46 9774.13 1.84 6 2.5 3623.16 6869 16448.85 16639.40 95.28 7 3.0 3538.63 7162 23046.64 23610.64 282.00 8 3.5 3111.73 6650 28984.57 29697.00 356.21 9 4.0 2591.35 5703 33760.98 34687.65 463.34 10 4.5 2043.27 4635 37192.34 38395.60 601.63 11 5.0 1489.65 3533 39348.24 40725.26 688.51 12 5.5 979.93 2470 40359.31 41817.82 729.26 13 6.0 564.99 1545 40439.27 41904.23 732.48 14 6.5 245.99 811 39834.07 41250.25 708.09 15 7.0 77.20 323 38822.60 40157.26 667.33 16 7.5 14.44 92 37672.31 38914.24 620.97 17 8.0 0.00 14 36536.37 37686.75 575.19 18 8.5 0.00 0 35471.80 36536.37 532.29 4000.00 3500.00 3000.00 2500.00 Q (cfs) 2000.00 Inflow 1500.00 Outflow 1000.00 500.00 0.00 0 2 4 6 8 10 Time (hr) ©S.Nimtim
8. 8. Assignments on Flood Routing September 21, 20124. using the inflow and outflow hydrograph given below for a channel, determine K and X. Time (min) Channel inflow Channel outflow เศษ ส่ วน K (cfs) (cfs) 0 0 0 5400 15 360.00 3 60 0 15030 24.75 607.27 6 120 13 22050 36.75 600.00 9 180 42 26730 44.25 604.07 12 240 81 29880 49.5 603.64 15 300 127 32310 54.25 595.58 18 364 178 36090 60.25 599.00 21 446 231 40680 67.5 602.67 24 530 293 43830 73.25 598.36 27 613 363 45810 76.25 600.79 30 696 437 46890 77.75 603.09 33 776 514 47160 79 596.96 36 855 593 46980 78.5 598.47 39 932 672 40590 67.75 599.11 42 948 757 27090 44.75 605.36 45 932 822 14670 24.75 592.73 48 914 861 7650 12.75 600.00 51 911 879 5850 9.25 632.43 54 921 888 6930 11.75 589.79 57 941 897 8280 14 591.43 60 958 910 8910 14.75 604.07 63 975 924 8370 13.75 608.73 66 982 940 6120 10 612.00 69 980 954 2790 4.75 587.37 72 969 964 -1080 -1.5 720.00 75 951 968 -5130 -8.75 586.29 78 925 965 -9540 -15.5 615.48 81 890 956 -13680 -23 594.78 ©S.Nimtim
9. 9. Assignments on Flood Routing September 21, 2012 84 852 938 -17550 -24.75 709.09 87 810 919 -20340 -37 549.73 90 767 884 -22590 -37.25 606.44 93 717 851 -25020 -41.5 602.89 96 668 812 -26550 -44.75 593.30 99 618 769 -27900 -46 606.52102 566 725 -28980 -49 591.43105 514 677 -29700 -49 606.12108 462 629 -30240 -50.5 598.81111 410 579 -30420 -51 596.47114 359 528 -30420 -50 608.40117 309 478 -30150 -50.25 600.00120 261 427 -26190 -43.75 598.63123 248 373 -19710 -33.25 592.78126 238 332 -15030 -24.75 607.27129 229 302 -11610 -19.75 587.85132 222 278 -9000 -15 600.00135 216 260 -7200 -12 600.00138 210 246 -5940 -9.5 625.26141 205 235 -5040 -9 560.00144 199 225 -4410 -7.25 608.28147 194 217 5400 15 K= 599.16 K = 599.157921 min 9.98596536 hr X= 0.25 ©S.Nimtim
10. 10. Assignments on Flood Routing September 21, 2012 ความสัมพันธ์ระหว่างเศษ ส่วน ของค่า K 60000 50000 40000 30000 0.22 20000 0.15 0.23 เศษ 10000 0.24 0 -0.00006 -0.00004 -0.00002 0 0.00002 0.00004 0.00006 0.00008 0.0001 Millions -10000 -20000 -30000 -40000 ส่ วน ความสั มพันธ์ ระหว่างเศษและส่ วนของค่า K ที่x = 0.25 60000 50000 40000 30000 20000 10000เศษ 0 -60 -40 -20 0 20 40 60 80 100 -10000 -20000 -30000 -40000 ส่ วน ©S.Nimtim
11. 11. Assignments on Flood Routing September 21, 2012 5. A 4400-foot reach of channel has a Muskingum K = 0.24 h and X =0.25. Route the following inflow hydrograph through this reach. Assume the initial outflow =739 cfs. K = 0.24 hr C1 = 0.44 X =0.25 C2 = 0.72 ∆t = 0.5 hr C3 = -0.16 C1 + C2 + C3 = 1.00Routing Time Inflow C1Ij+1 C2 Ij C3 Q j Outflowperiod , j (hr) (cfs) 0.441860465 0.720930233 -0.162790698 (cfs) (min) 1 0.0 819 - - - 739 2 0.5 1012 447.1627907 590.4418605 -120.3023256 917.3023256 3 1.0 1244 549.6744186 729.5813953 -149.3282856 1129.927528 4 1.5 1537 679.1395349 896.8372093 -183.9416907 1392.035054 5 2.0 1948 860.744186 1108.069767 -226.6103575 1742.203596 6 2.5 2600 1148.837209 1404.372093 -283.6145389 2269.594763 7 3.0 5769 2549.093023 1874.418605 -369.468915 4054.042713 8 3.5 12866 5684.976744 4159.046512 -659.9604416 9184.062814 9 4.0 17929 7922.116279 9275.488372 -1495.079993 15702.52466 10 4.5 20841 9208.813953 12925.55814 -2556.224944 19578.14715 11 5.0 21035 9294.534884 15024.90698 -3187.140234 21132.30163 12 5.5 20557 9083.325581 15164.76744 -3440.142125 20807.9509 13 6.0 19485 8609.651163 14820.16279 -3387.340844 20042.47311 14 6.5 14577 6441 14047.32558 -3262.728181 17225.5974 15 7.0 9810 4334.651163 10509 -2804.167019 12039.48414 16 7.5 6448 2849.116279 7072.325581 -1959.916023 7961.525837 17 8.0 4558 2014 4648.55814 -1296.062346 5366.495794 18 8.5 0 0 0 -873.6155944 4492.8802 19 9.0 0 0 0 -731.3991023 3761.481097 20 9.5 0 0 0 -612.3341321 3149.146965 21 10.0 0 0 0 -512.6518315 2636.495134 22 10.5 0 0 0 -429.1968822 2207.298251 23 11.0 0 0 0 -359.3276223 1847.970629 24 11.5 0 0 0 -300.832428 1547.138201 25 12.0 0 0 0 -251.8597072 1295.278494 26 12.5 0 0 0 -210.8592897 1084.419204 27 13.0 0 0 0 -176.5333588 907.8858454 ©S.Nimtim
12. 12. Assignments on Flood Routing September 21, 201228 13.5 0 0 0 -147.7953702 760.090475229 14.0 0 0 0 -123.7356588 636.354816530 14.5 0 0 0 -103.5926445 532.762171931 15.0 0 0 0 -86.72872566 446.033446332 15.5 0 0 0 -72.6100959 373.423350433 16.0 0 0 0 -60.78984773 312.633502634 16.5 0 0 0 -50.89382601 261.739676635 17.0 0 0 0 -42.60878457 219.130892136 17.5 0 0 0 -35.6724708 183.458421337 18.0 0 0 0 -29.86532439 153.593096938 18.5 0 0 0 -25.0035274 128.589569539 19.0 0 0 0 -20.93318573 107.656383740 19.5 0 0 0 -17.52545782 90.13092592 ©S.Nimtim