河流泥沙动力学习题.doc

Problems 1.Consider steady and uniform open channel flow. The turbulent eddy viscosity is Derive the time-averaged velocity profile (let Π=0.2) 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=ω/(κu*)=0.05, 0.2, 0.6, 1.5. [numerical integration may be used] Solution : A reduction yields the relation Averaging over turbulence in the same way as before yields the result Where For fully turbulent 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 result And Reynolds flux of streamwisemomentum in the z direction: Then we can integrate this equation: Also Where the turbulent eddy viscosity (let Π=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 the result The Reynolds flux F can be related to the gradient of the mean concentration as Assuming the turbulent diffusion coefficient of suspended sediment is equal to the turbulent eddy viscosity The balance equation thus reduces to: So Integrating both side of this equation yields under the condition 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 along the flow depth as following: While the Rouse formula We compare