Drag and viscosity

Drag at high Reynolds numbers is given by the formula: 0.5*$$\rho$$*V2*Cd*A
Why does at the molecular (or at the micro) level the drag is independent of the viscosity?
 The density is related with the viscosity of the fluid: density=viscosity/kinematic viscosity

 Quote by uros The density is related with the viscosity of the fluid: density=viscosity/kinematic 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 and viscosity

 Quote by GT1 So still no matter how high is the viscosity (concrete..)...
Drag is a force exerted by the motion through fluids, doesn't make sense to relate it with solids (or very viscous fluids).

 Quote by GT1 ... as long as we are at the high turbulent zone, the drag is independent of the viscosity. It sounds counterintuitive to me.
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
 Recognitions: Science Advisor 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 $\rho V^2 C_d A / 2$ formula is relevant or not. For some flow situations (e.g. in the transition between laminar and turbulent flow) $C_d$ is a function of Re, not a constant value.
 Recognitions: Homework Help 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/Externa...r-measurements

 Quote by AlephZero 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 $\rho V^2 C_d A / 2$ formula is relevant or not. For some flow situations (e.g. in the transition between laminar and turbulent flow) $C_d$ is a function of Re, not a constant value.
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.

Recognitions: