河流泥沙动力学习题

1、Problems 1.Consider steady and uniform open channel flow. The turbulent eddy viscosity is Derive the time-averaged velocity profile (let =0.2)Assume the turbulent diffusion coefficient of suspended sediment is equal to the turbulent eddy viscosity, calculate the profile of relative suspended sedimen

2、t 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 usedSolution :A reduction yields the relationAveraging over turbulence in the same way as before yields the resultWhere For fully tur

3、bulent flow, the Reynolds stress is usually far in excess of the viscous stress , which can be dropped.Integrating the equation under the condition of vanishing shearstress at the water surface z = H yields the resultAnd Reynolds flux of streamwisemomentum in the z direction: Then we can integrate t

4、his equation: AlsoWhere the turbulent eddy viscosity (let =0.2) So Then The time-average velocity profile of the flow depth The balance equation of suspended sediment is This equation can be integrated under the condition of vanishingnet sediment flux in the z direction at the water surface to yield

5、 the resultThe Reynolds flux F can be related to the gradient of the mean concentration asAssuming the turbulent diffusion coefficient of suspended sediment is equal to the turbulent eddy viscosity The balance equation thus reduces to:SoIntegrating both side of this equation yields under the conditi

6、on that assuming the boundary condition is at the height of b=0.05h, when the sediment concentration is Cb.Then we can caculate the integration by the way of numerical integration.Where y=0.1h,we can caculate the integration approximately as: . So we get the relative suspended sediment concentration

7、 along the flowdepth as following:While the Rouse formulaWe compare it with the Rouse formula in a figure for suspension index z=0.05, 0.2, 0.6, 1.5 as followsThe tablec/cbt=z/hf(t)AiCalRouseCalRouseCalRouseCalRouse0.10 00111111110.15 0.3353 0.0138 0.9407 0.9413 0.7829 0.7851 0.4799 0.4839 0.1595 0.

8、1629 0.20 0.4612 0.0337 0.9235 0.9251 0.7273 0.7323 0.3848 0.3926 0.0918 0.0966 0.25 0.5918 0.0600 0.9091 0.9118 0.6831 0.6913 0.3187 0.3304 0.0573 0.0627 0.30 0.7256 0.0929 0.8963 0.9005 0.6453 0.6574 0.2687 0.2841 0.0374 0.0430 0.35 0.8606 0.1326 0.8844 0.8902 0.6116 0.6281 0.2288 0.2478 0.0250 0.

9、0306 0.40 0.9952 0.1790 0.8729 0.8808 0.5807 0.6018 0.1958 0.2180 0.0170 0.0222 0.45 1.1276 0.2321 0.8617 0.8718 0.5515 0.5777 0.1677 0.1928 0.0115 0.0163 0.50 1.2560 0.2917 0.8506 0.8631 0.5235 0.5549 0.1435 0.1709 0.0078 0.0121 0.55 1.3786 0.3575 0.8393 0.8545 0.4963 0.5331 0.1223 0.1515 0.0052 0.

10、0089 0.60 1.4936 0.4293 0.8278 0.8458 0.4696 0.5117 0.1036 0.1340 0.0035 0.0066 0.65 1.5996 0.5067 0.8159 0.8368 0.4431 0.4903 0.0870 0.1179 0.0022 0.0048 0.70 1.6949 0.5890 0.8033 0.8273 0.4164 0.4684 0.0722 0.1028 0.0014 0.0034 0.75 1.7784 0.6758 0.7898 0.8170 0.3892 0.4455 0.0589 0.0884 0.0008 0.

11、0023 0.80 1.8489 0.7665 0.7750 0.8053 0.3608 0.4206 0.0470 0.0744 0.0006 0.0015 0.85 1.9058 0.8604 0.7581 0.7914 0.3303 0.3923 0.0360 0.0604 0.0004 0.0009 0.90 1.9492 0.9568 0.7372 0.7733 0.2953 0.3576 0.0258 0.0457 0.0002 0.0004 0.95 1.9836 1.0551 0.7067 0.7449 0.2494 0.3080 0.0155 0.0292 0.0001 2

12、The graph (by Grapher 4.0)Conclusion :Firstly,we can get the idea that suspension index z plays an important role in calculating the relative suspended sediment concentration,for the smaller z is,the more evenly the relative concentration distribute in the z direction.Secondly,we can know from the f

13、igure that the relative concentrations we calculate is marginal bigger than those according to Rouse formulation with the same suspension index,especially adjacent to the surface of bed.But both they present the tendency that the sediment concentration rises from the flow surface to the bed and fina

14、lly tends to infinity. Consider Run 20 of the flume experiments of Coleman (1986), find the best value of in the modified suspension index for satisfactory agreement between profile of suspended sedimentconcentration calculated from the Rouse formula and measured data.Reference:Coleman, N. L. (1986)

15、. “Effects of suspended sediment on the open-channel velocity distribution.” Water Resources Research, AGU, 22(10), 1377-1384.SolutionsMeasured dataCsuspension indexz=s/u*Zhang RJ formulation 10-6m2CSo s=7.35810-4m/sThen z=s/u*=7.35810-4/(0.4Rouse formulaFirst,we need to discuss whether 1 or not.Let

16、 b=zmin=6 mmThe tablemeasured datac/cb（Rouse formula）Z(mm)z/hcc/cb=160.0353 1111120.0706 0.5217 0.7206 0.6948 0.7424 180.1059 0.3565 0.5904 0.5568 0.6194 240.1412 0.2652 0.5096 0.4729 0.5418 300.1765 0.2087 0.4525 0.4143 0.4863 460.2706 0.1435 0.3537 0.3152 0.3888 690.4059 0.1130 0.2690 0.2325 0.303

17、1 910.5353 0.0696 0.2128 0.1792 0.2449 1220.7176 0.0330 0.1492 0.1208 0.1774 1370.8059 0.0174 0.1197 0.0946 0.1452 1520.8941 0.0087 0.0871 0.0664 0.1087 1620.9529 0.0048 0.0588 0.0429 0.0761 The graph (by Excel)So 1Then let =0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9 and compare it with the Rouse formula i

18、n a figureThe table z/hc/cb0.035 1 1 1 1 1 1 1 1 1 1 0.071 0.038 0.194 0.335 0.441 0.579 0.626 0.664 0.695 0.522 0.106 0.005 0.072 0.173 0.268 0.349 0.416 0.471 0.518 0.557 0.357 0.141 0.001 0.034 0.106 0.185 0.260 0.325 0.382 0.431 0.473 0.265 0.176 0.000 0.019 0.071 0.138 0.205 0.267 0.322 0.371 0

19、.414 0.209 0.271 0.000 0.006 0.031 0.074 0.125 0.177 0.227 0.273 0.315 0.143 0.406 0.000 0.001 0.013 0.038 0.072 0.112 0.153 0.194 0.232 0.113 0.535 0.000 0.000 0.006 0.021 0.045 0.076 0.110 0.145 0.179 0.070 0.718 0.000 0.000 0.002 0.009 0.022 0.042 0.066 0.093 0.121 0.033 0.806 0.000 0.000 0.001 0

20、.005 0.014 0.029 0.048 0.070 0.095 0.017 0.894 0.000 0.000 0.000 0.002 0.008 0.017 0.031 0.047 0.066 0.009 0.953 0.000 0.000 0.000 0.001 0.003 0.009 0.017 0.029 0.043 0.005 The graph (by Excel)Finally we need to turn to the program to determine the value of , and I find that I neednt have done the f

21、ormer work if I had used the program in the beginning.The program:#include #include void main()double x=0.5,r; double s112=0.023,0.012,0.0082,0.0061,0.0048,0.0033,0.0026,0.0016, 0.00076,0.0004,0.0002,0.00011; double s212=6,12,18,24,30,46,69,91,122,137,152,162; double a12=0,b12=0,c10000=0; int i,j,k,

22、m,n; for(k=0;k12;k+) ak=s1k/0.023; for(i=0;i10000;i+) for(j=0;j12;j+) cj=pow(170/s2j-1)/(170/6-1),0.44866/x); ci+=fabs(aj-bj); x+=0.000001; r=c0; for(m=1;mcm) r=cm;for(n=0;n10000;n+)if(cn=r) x=0.5+0.000001*n; printf(x=%f,sum=%f,x,r);At last,we can know=0.511011 When Where is the result that we calcu

23、late from the modified Rouse formulationis the rusult of Run 20.3.Consider suspended sediment transport in steady and uniform flows.The settling velocity of a single sediment particle in still water is 0,the turbulent diffusion coefficient of suspended sediment s=u*h/6,where is von Karmen constant,u* is bed shear velocity, h is flow depth,Also,the volumetric concentration of suspended sediment at the height y=a above the bed is equal to Ca.Derive the distribution of suspended sediment co