河流动力学大作业.doc

Problem 1Calculate the variation in H and b= CfU2 in x upstream of x1 (here set equal to 0) until H is within 1 percent of Hn.A WORKED EXAMPLE (constant Cz):S = 0.00025 Cz= 22 qw= 5.7 m2/s D = 0.6 mmR = 1.65 H1= 30m Hn= 3.01m Hc= 1.49mSolution:where program mainfs=0.00025fCz=22fqw=5.7fg=9.81fdx=3000fH1=30 fdH=3.01fH=fH150if(fH-fdH)0.01*fdH) thenfsH1=fCz*2.0*fqw*2.0/(fg*fH1*3.0)fno=1-fqW*2.0/(fg*fH1*3.0)ffH1=(fs-fsH1)/fnofH=fH1+fdx*ffH1print*, fHpausefdh=fHgoto 50endifendHn+ =2.928926543，Hn- =3.739186042，Cf+ =0.00143480257120,Cf- =0.0032283057851Sheet 1xSfFr2H+Ub01.7599854848E-070.00012266394835300.190.05179637281940001.9482861183E-070.000135787749329.0005814020.196547783680.05542790299480002.1644210827E-070.0001508514917828.0012250160.20356252260.059454919373120002.4137258667E-070.00016822703827.001940040.211095943160.063936950784160002.7028107528E-070.0001883750982326.0027374380.219207689710.068945149262200003.0399304022E-070.0002118709893125.0036303550.227966895970.074565119457240003.435480976E-070.000239439282124.0046346690.237454144940.080900583892280003.9026764405E-070.0002720009371923.0057696940.247763933820.088078177948320004.4584799723E-070.0003107382201522.0070591150.259007801560.096253789698360005.1249037876E-070.0003571852943821.0085322260.271318335740.10562103896400005.9308493791E-070.0004133564783320.0102256070.284854359560.11642275907440006.9147539012E-070.000481930687919.01218540.299807722250.12896675609480008.1284614047E-070.0005665212460618.0144704730.316412298030.14364775933520009.6429928358E-070.0006720780286817.0171568580.334956070960.16097849155560001.1557325066E-060.0008054993277816.0203441190.35579760070.18163443048600001.4012055507E-060.0009765842206115.0241646270.379388814050.20651955956640001.7211214664E-060.001199552817214.0287973890.406307101160.23686504312680002.1458077305E-060.001495542155913.034489150.437301372870.27438086937720002.7215832142E-060.00189683463712.0415874550.473359515220.32149510826760003.5214049984E-060.002454278427711.0505940370.515809374680.38174258352800004.6643058772E-060.003250834624210.0622539950.56647347630.46041695269840006.3524878956E-060.00442742996379.0777106310.62791162130.56570392012880008.9456727168E-060.00623477605678.09878646440.703809147840.71072560344920001.3118215981E-050.00914287180987.12851975960.799604994050.91736706054960002.0208887973E-050.0140847865616923488257791.22364348351000003.3028621118E-050.0230196277745.23995393471.08779582251.69780152571040005.7666045677E-050.0401909271954.35161928621.30985723362.46172803671080000.000106209334980.0740236581063.55006839541.60560286883.69886445741120000.000189124333610.131812095562.92892654321.94610548135.4340658635Graph 1Sheet 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237Problem 2Calculate sediment fall velocity using formulations of Dietrich EW (1982) and Zhang RJ (textbook, page 48-50)g= 9.81 m/s2R= 1.65 (quartz)= 1.00x10-6m2/s(water at 20 degree at Celsius)= 1000 kg/m3(water)Solution：We know Vs (Zhang RJ)：Then use the formula：where and b1=2.891394 b2=0.95296 b3=0.056835 b4=0.002892 b5=0.000245D (mm)D（m）RepRfVs (Dietrich)Vs (Zhang RJ)0.0313 0.00003 0.7025 0.0394 0.0009 0.0006 0.0625 0.00006 1.9869 0.1039 0.0033 0.0025 0.1250.00013 5.6198 0.2371 0.0107 0.0095 0.25 0.00025 15.8952 0.4294 0.0273 0.0309 0.5 0.00050 44.9583 0.5256 0.0473 0.0700 10.00100 127.1613 0.3439 0.0437 0.1195 2 0.00200 359.6665 0.0883 0.0159 0.1809 we can get Rf.Finally we can get the Vs (Dietrich) using the formula Conclusion：The figure told us that there is no difference between Vs (Dietrich) and Vs(Zhang RJ) when D less than 0.5mm,but the sediment fall velocity calculated by Zhang RJ is smaller than that calculated by Dietrich when D0.5mm.Both of the two formulations use the condition of equilibrium Fg=Fd in the formulation of Dietrich.while in the formulation of Zhang RJ, He take the particle irregularity in shape into condition. As we all know, the effect of sediment fall velocity is small to fine sediment particle, and the effect is increasing as the increasing of the size of particle. The result and analysis consist.Problem 3Consider the incipient motion of sediment in open channel flow. Let sediment particle diameter d varies from 0.01 mm to 10 mm.(1) Calculate the critical bed shear velocity u* for incipient motion of the sediment specified above, using the formulation by Brownlie(1981).(2) Calculate the critical mean velocity Uc for incipient motion of the sediment specified above, based on the critical bed shear velocity obtained above and the integration along flow depth of the log velocity distribution. Here let flow depth h=0.15 m, and roughness height Ks=2d.(3) Plot Uc (cm/s) against d(mm) in a log-log coordinate setting.refer to Figure 4-6, page 70 on textbook.We know that:g= 9.81 m/s2 R= 1.65 = 1.0106m2/s= 1000kg/m3 = 0.4 Solution：(1) We can get the critical Shields number：Where Then we can easily get .(2) If 100,B=8.5,the near-wall flow is in hydraulically smooth regimes.If 3 ,B =5.5+ln,the near-wall flow is in hydraulically rough regimes.If 3100, B should be calculated by the diagram B against as follows.we can get flow velocity averaged over turbulence Finally we integrate the equation and get the UcThe tableDRep*u*u*ks/v0.000010.1272261770.7580132470.0110768140.2215362850.000020.3598499690.4062092410.0114674370.4586974910.000041.0178094120.2176821440.0118718360.9497468490.000082.878799750.116657930.0122907561.9665209810.000168.1424752990.0629020080.0127634694.0843100110.0003223.0303980.0375330580.0139430718.923565470.0006465.139802390.0320718770.01822756523.331282860.00128184.2431840.037254130.02778233171.122766550.00256521.11841920.0447569960.043065245220.49405520.005121473.9454720.0507865230.064876244664.33273840.014023.2449590.0546272920.0940332211880.66442BuCLOGDLOGuC1.7320777830.23858044-21.3776348353.5515891320.247987539-1.698971.3944298585.3711004890.257761418-1.397940011.4112179127.1906649550.26792259-1.096910011.4280093338.4410914920.272069489-0.795880021.434679849.3073182190.285130563-0.494850021.4550437729.3518746350.341973041-0.193820031.5339918718.8208581490.4583370070.107209971.6611849248.50.6220215730.4082399651.7938054488.50.824631180.7092699611.9162597528.51.03786887912.016142489(3) We Plot Uc (cm/s) against d(mm) in a log-log coordinate setting as follow.Then compare it with Figure 4-6page 70 on textbook,we can find it work good when D is greater than 0.17 mm, but the trend is ascending while Figure 4-6 is opposite when D is smaller.We surmise that this phenomenon is behind that Shields number is required to initiate motion of the grains of a bed composed of non-cohesive particles and we ignore the impact of cohesive force using the formula Plot Uc (cm/s) against d(mm)Problem 4Consider steady and uniform open channel flow.The turbulent eddy viscosity is (1) derive the time-average velocity profile (let =0.2)(2)assume the turbulent diffusion coefficient of suspended sediment is equal to the turbulent eddy viscosity,calculate the profile of relative suspended sediment concentration along the flow depth,and compare it with the Rouse formula in a figure for suspension index z=/u* =0.05, 0.2, 0.6, 1.5.(numerical integration may be used)Solution :(1) For full turbulent flow,the Reynolds stress is usually far in excess of the viscous stress,which can be dropped.We can get this equation:Remove at the two sides ,so we can integrate this equation:So the time-average velocity profile of the flow depth is :(1) Assume the turbulent diffusion coefficient of suspended sediment is equal to the turbulent eddy viscosity, so we can get the balance of the suspension equation as: According to , Then we get:Then we calculate the integration by the way of numerical integration. Where z=0.1h, we can calculate the integration approximately as:.So we get the relative suspended sediment concentration along the flow depth as follows:While the Rouse formulaSupposing ,we can get the data in Table 1 and plot z/h againstin the Table 1. And if we let z=/u* vary as 0.05 , 0.2 , 0.6 , 1.5 and compare these two formulation in a graph, we can get eight curves as Graph 1. Table c/cbs=0.05s=0.2s=0.6s=1.5t=z/hf(t)AiCalRouseCalRouseCalRouseCalRouse0.100111111110.120.262840.002630.989680.98981 0.95936 0.95987 0.88298 0.88437 0.73262 0.140.311080.008370.980670.98108 0.92491 0.92646 0.79122 0.79520 0.55685 0.56389 0.160.360350.015080.972680.97341 0.89510 0.89781 0.71717 0.72369 0.43556 0.44553 0.180.410570.022790.965430.96653 0.86872 0.87269 0.65560 0.66463 0.34802 0.36012 0.20.461650.031510.958750.96026 0.84494 0.85028 0.60322 0.61474 0.28262 0.29630 0.220.513470.041260.952520.95449 0.82319 0.83001 0.55783 0.57182 0.23241 0.24725 0.240.565940.052060.946650.94911 0.80306 0.81147 0.51791 0.53434 0.19303 0.20871 0.260.618950.063910.941050.94406 0.78425 0.79433 0.48234 0.50120 0.16158 0.17784 0.280.672400.076820.935680.93928 0.76650 0.77837 0.45034 0.47158 0.13610 0.15272 0.30.726170.090810.930500.93473 0.74965 0.76339 0.42129 0.44488 0.11520 0.13201 0.320.780160.105870.925460.93037 0.73355 0.74924 0.39471 0.42060 0.09788 0.11473 0.340.834240.122010.920540.92617 0.71807 0.73581 0.37025 0.39838 0.08342 0.10017 0.360.888310.139240.915710.92211 0.70313 0.72298 0.34761 0.37790 0.07124 0.08779 0.380.942250.157550.910960.91816 0.68863 0.71068 0.32656 0.35894 0.06094 0.07719 0.40.995940.176930.906260.91431 0.67453 0.69883 0.30691 0.34128 0.05218 0.06804 0.421.049270.197380.901590.91054 0.66076 0.68736 0.28849 0.32476 0.04470 0.06010 0.441.102120.218890.896960.90683 0.64727 0.67624 0.27118 0.30924 0.03830 0.05318 0.461.154380.241460.892330.90317 0.63402 0.66539 0.25487 0.29460 0.03279 0.04711 0.481.205920.265060.887710.89955 0.62098 0.65479 0.23947 0.28074 0.02806 0.04176 0.51.256640.289690.883070.89596 0.60812 0.64439 0.22489 0.26758 0.02398 0.03704 0.521.306410.315320.878420.89238 0.59540 0.63416 0.21107 0.25503 0.02047 0.03285 0.541.355140.341930.873740.88880 0.58281 0.62406 0.19796 0.24304 0.01744 0.02912 0.561.402700.369510.869020.88522 0.57031 0.61405 0.18549 0.23153 0.01482 0.02579 0.581.449000.398030.864240.88161 0.55788 0.60411 0.17363 0.22047 0.01256 0.02282 0.61.493920.427460.859410.87798 0.54551 0.59420 0.16234 0.20980 0.01062 0.02016 0.621.537360.457770.854510.87429 0.53317 0.58429 0.15157 0.19948 0.00894 0.01777 0.641.579220.488940.849530.87055 0.52084 0.57435 0.14129 0.18946 0.00750 0.01563 0.661.619410.520920.844450.86673 0.50850 0.56434 0.13148 0.17973 0.00627 0.01369 0.681.657830.553690.839260.86282 0.49612 0.55422 0.12211 0.17023 0.00521 0.01196 0.71.694400.587220.833950.85879 0.48367 0.54395 0.11315 0.16094 0.00431 0.01039 0.721.729030.621450.828480.85463 0.47113 0.53348 0.10457 0.15183 0.00354 0.00898 0.741.761630.656360.822850.85031 0.45845 0.52276 0.09635 0.14286 0.00288 0.00771 0.761.792140.691900.817020.84578 0.44559 0.51172 0.08847 0.13400 0.00233 0.00657 0.781.820480.728020.810950.84102 0.43250 0.50028 0.08090 0.12521 0.00186 0.00555 0.81.846580.764690.804600.83596 0.41910 0.48836 0.07361 0.11647 0.00147 0.00463 0.821.870390.801860.797900.83054 0.40532 0.47582 0.06659 0.10773 0.00114 0.00381 0.841.891840.839480.790770.82467 0.39102 0.46251 0.05979 0.09894 0.00087 0.00308 0.861.910890.877510.783090.81822 0.37606 0.44821 0.05318 0.09004 0.00065 0.00243 0.881.927500.915900.774700.81100 0.36019 0.43260 0.04673 0.08096 0.00047 0.00187 0.91.941610.954590.765330.80274 0.34307 0.41524 0.04038 0.07160 0.00033 0.00137 0.921.953210.993530.754530.79296 0.32413 0.39538 0.03405 0.06181 0.00021 0.00095 0.941.962251.032690.741510.78080 0.30231 0.37167 0.02763 0.05134 0.00013 0.00060 0.961.968731.072000.724430.76432 0.27542 0.34128 0.02089 0.03975 0.00006 0.00032 0.981.972621.111410.697660.73753 0.23691 0.29588 0.01330 0.02590 0.00002 0.00011 Conclusion:(1)We can learn about it that suspension index make difference in calculating the relative suspended sediment concentration along the flow depth. Its slightly change will affect the value of significantly.(2)It can be known that with the same suspension index, the value of derived from Rouse formula is bigger than that from the calculated formula. But both of them present the tendency that relative suspended sediment concentration varies along the flow depth.Problem 5Consider Run 20 of the flume experiments of Coleman (1986), find the best value of in the modified suspended index: for satisfactory agreement between profile of suspended sediment concentration calculated from the Rouse formula and measured data.Reference: Coleman.N.L(1986).Effects of suspended sediment on the open-channel velocity distribution.Water Resources Research,AGU,22(10),1377-1384.Solution:Table 2 : Measured dataZ(mm)6121824304669911221371521620.610.480.330.260.160.0760.040.020.011Get the following data from the textbook: D=0.015mm h=170mm Q=0.064m3/s u*=0.041m/s , T=23.9suspension indexFirst , we calculate by Zhang RJ formulation But we should figure out when T=23.9 before it.By linear Interpolation as follows :t05101520253040V(10-6m2/s)1.7921.5191.3081.1411.0070.8940.8010.656From the curve we see that v=0.96010-6m2/s at the temperature of 23.9.And we can calculate the settling velocity:=7