# Tag Info

4

The question you ask is actually the central question of a huge sub-discipline of fluid dynamics. Some have even referred to it as "the last great unsolved problem in classical physics." If you get a complete answer, please let me know! (And don't tell anyone else. Just keep between us, eh?) Generally, there are always small fluctuations in any flow, ...

7

The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by $$\text{Re}=\frac{\rho~u~L}{\mu}.$$ This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies ...

0

The higher the fluid velocity relative to the surface, the lower the pressure. In the turbulent case the higher relative velocity leads to lower pressures near the surface such that pressure from the surrounding fluid further away from the surface (the same in each case) forces the streamlines towards the surface for a longer distance compared to the laminar ...

0

A laminar flow is certainly "nicer" than a turbulent one, but that is not always what you want when you want to reduce drag. In particular, there's no obvious reason why dragging fluid around (via a laminar wake) is going to be better than casting it off, as shed vortices go. In particular, keeping the main part of the vortex around will mean that you've ...

2

Deliberately introducing turbulence can often reduce overall drag, counterintuitive as it seems. On the wing of this aircraft, you can see vortex generators fitted along it's length. From Wikipedia Turbulent Flow and Drag In turbulent flow, unsteady vortices appear on many scales and interact with each other. Drag due to boundary layer skin friction ...

1

It's simply a definition of the properties of a fluctuating component. We require that, on large enough time scales, the fluctuating component averages out to zero: $$\frac{1}{T}\int_t^{t+T}u'(t)\,dt=0$$ so that you are left with just the bulk flow term, $U$, that contributes to the mean velocity, $\bar{u}$. I think this is more easily seen visually than ...

Top 50 recent answers are included