Drag at high Reynolds numbers is given by the formula: 0.5*[tex]\rho[/tex]*V^{2}*C_{d}*A Why does at the molecular (or at the micro) level the drag is independent of the viscosity?
Kinematic viscosity is Viscosity/Density Viscosity/Kinematic viscosity = Viscosity/Viscosity/Density = Density. So still no matter how high is the viscosity (concrete..), as long as we are at the high turbulent zone, the drag is independent of the viscosity. It sounds counterintuitive to me.
Drag is a force exerted by the motion through fluids, doesn't make sense to relate it with solids (or very viscous fluids). The density (ρ) is proportional to the viscosity (μ), the kinematic viscosity (η) expresses this ratio, with the kinematic viscosity you can turn viscosity into density and vice versa. η=μ/ρ Since the drag depends of the density it also is indirectly related to the viscosity. If you're more comfortable expressing the drag through the viscosity instead of the density change the density by the ratio of viscosity (μ) and kinematic viscosity (η). F=0.5*ρ*v²*A*Cd F=0.5*μ/η*v²*A*Cd
Reynolds number represents the the ratio of the inertia forces to the viscous forces in the flow. So for high Reynolds numbers (e.g. Re >= 10^6) the viscous forces can be neglected. For "creeping" or Stokes flow at very low Reynolds number (Re << 1) the inertia forces can be neglected. Of course the Reynolds number depends on the viscosity (and other physical parameters), so in that sense the viscosity DOES influence the drag calculation - but only to the extent of deciding whether your [itex]\rho V^2 C_d A / 2[/itex] formula is relevant or not. For some flow situations (e.g. in the transition between laminar and turbulent flow) [itex]C_d[/itex] is a function of Re, not a constant value.
In the real world, Cd is also a function of velocity once you get past about 4/10 the speed of sound. For external ballistics, tables are used for Cd versus speed versus projectiles: http://en.wikipedia.org/wiki/External_ballistics#Doppler_radar-measurements
Why at high Reynolds numbers the viscosity can be neglected? Eventually all the energy is dissipated through viscosity, the energy has to pass from high scale turbulence (where viscosity is can be neglected) to small scale turbulence (where viscosity can't be neglected) so at some point the viscosity is important. Also you can see that at high Reynolds numbers the drag coefficient (Cd) vs. Reynolds number is almost flat line.
"The viscous forces acting on the body can be neglected relative to the inertia forces on the body" is exactly what "high Reynolds number" means. "The viscous forces on the body can be neglected compared with the inertia forces" does NOT mean the same thing as "the effects of viscosity on the fluid flow can be neglected". For inviscid flow, the lift and drag forces would be indentically zero for any shape of body. But if you are considering the wing of a large aircraft for example, the aerodynamic forces in the tip vortex a few kilometers behind the plane, where the energy in the vortex is being slowly dissipated by the air viscosity, do not contribute anything to the drag force on the wing. The calculation of Re uses the velocity, density, and a length parameter which is often related to the cross section area. The velocity, density and area are explicitly included in the drag formula. So in a sense, the "effect of Re on the drag" is mainly represented in the formula by the velocity, density and area, not by Re itself. The second-order effects are represented empirically, for a particular shape of body, by the variation of Cd with Re.