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I was just pondering a silly physics problem where air is made to flow though a narrow tube. For a viscous fluid and laminar flow through a pipe you have the Hagen–Poiseuille equation.

$\Delta P=\frac{8\mu L Q}{\pi R^{4}}$

I believe it should apply for air, although thinking about it I am not sure quite why. A viscous fluid like treacle or marmite would stick to the glass surface, so that the velocity at the glass surface is zero. But in the case of air, is velocity actually zero at the glass surface, or would it slip? If I am right and it is essentially zero, then why? Thinking of air as just particles of relatively long root mean square distance apart, it is hard to see how a shear sticking effect could manifest at the gas/glass interface.

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‘Viscous’ doesn’t mean ‘sticky’. It means “the fluid exerts force against adjacent slip”.

The viscosity of air or water is less than treacle, but it’s still non-zero. Air next to a non-moving surface will feel a force that’ll tend to make it stop. Smaller viscosity means less force, but that just means a less massive I.e. thinner layer is brought to a stop.

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  • $\begingroup$ Hi. I know that viscous doesn't mean sticky. That is why I ask the question. I know air has viscosity. It isn't sticky. Since it isn't stick I wonder why it would stop, as you put it. Obviously it will stop in the normal direction, but without a tendency to stick, why would it stop in the transverse direction. I suspect it is that at a microscopic level, the surface of the glass is not exactly smooth. Thus there would be a transverse element to the collisions between the air molecules and the the glass molecules. $\endgroup$ – Rory Cornish Nov 22 '19 at 23:27
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Under norm conditions there are around $2.69 \times 10^{19}$ molecules in a cubic centimeter of air. As you can imagine this number is easily sufficient such that particles close to the wall are pressed against it (don't think of it as sticking) and left unable to move in any direction (You can see this is a question of diluteness and not viscosity: As soon as there are sufficient particles this will happen!). Wherever this will be the case or not, is given by the ratio of the mean free path $\lambda$ and the characteristic length scale of a given problem $L$, the Knudsen number

$$ Kn := \frac{\lambda}{L} .$$

For an ideal gas it is proportional to $ Kn \propto \frac{Ma}{Re} $, where the Reynolds number $Re := \frac{U \, L}{\nu}$ is a measure for turbulence and the Mach number $Ma := \frac{U}{c_s}$ a measure for compressibility. As a threshold to continuum based flow generally $ Kn \lesssim 0.01$ is generally taken and thus actually almost all real-world flows can be considered continuum flows. Nonetheless for the case of very small-scale microscopic flows this assumption can be violated resulting in higher Knudsen numbers: The velocity will not fulfill the no-slip condition and there will be a slip-velocity at the solid boundary. This will though unlikely be the case for your narrow glass tube but can happen in small-scale porous media structures where the pore-scale flow is indeed dilute.

Viscosity (which is proportional to the shear rate) on the other hand is basically the ability of a continuum fluid to redistribute differences in momentum, a measure for the internal friction - it is a continuum property. The model of an inviscid flow neglects the effects of viscosity by assuming a high degree of turbulence, a large Reynolds number, and does not necessarily have to fulfill vanishing velocities at the wall. This renders this model only useful for estimating flow far from the wall by neglecting wall effects or approximating properties that might not be influenced by the precise flow near the wall. Physically this model is not correct: Even for high Reynolds number continuum flows near the wall viscosity and dissipation will prevail.

Summing up, wherever the no-slip condition will be fulfilled and the flow will be pressed against the wall or not is not a question of viscosity or material of the wall but rather of the local diluteness of the fluid.

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