A method to measure the viscosity of a gas uses two vertical coaxial cylinders, with the gas confined between them. The inner cylinder has radius $a$, whilst the outer cylindrical shell has inner radius $b$. The inner cylinder is suspended from a torsion fibre from a fixed support, whilst the outer cylinder is rotated by a motor at constant angular speed $\omega_0$.

So what happens in the steady state is, this sets up an angular velocity gradient in the gas (with the different angular velocity levels varying from zero at $r=a$, with $r$ a radial length from the centre of the inner cylinder, to $\omega_0$ at $r=b$). A torque is transmitted to the inner cylinder as momentum is transported through the gas. This translates into a tangential velocity gradient using the relation $u=\omega r$. The tangential velocity gradient is then just $$\frac{du}{dr}=\frac{d(\omega r)}{dr}=\omega+r\frac{d\omega}{dr}$$

However, at this stage, my book states that 'the $\omega$ term on the right corresponds to the velocity gradient due to rigid body rotation so doesn't add to the viscous shearing stress'. Then it omits it and so states that the viscous shearing stress (usually viscosity $\eta$ multiplied by the velocity gradient) is $\eta r\frac{d\omega}{dr}$. I don't really understand the reasoning behind omitting the first term... [The rest of this post just completes the derivation - if you can explain this bit without the need to know the rest then just skip it :)].

Then you can get the force $F$ on a cylindrical layer of the gas, because we know the shear stress, using $\frac{F}A=\eta r\frac{d\omega}{dr}$ and $A=2\pi rl$ fwhere $l$ is the length of the cylinders, so $F=2\pi r^2l\eta \frac{d\omega}{dr}$. Then the torque on this cylindrical layer is $G=rF=2\pi r^3l\eta \frac{d\omega}{dr}$. In the steady state the torque $G$ must be the same for all layers (i.e for all $r$), which I understand because the layers all move at constant speed, and the force accelerating one is the force slowing another down. So $G$ is a constant. We can then solve this differential equation giving $$\eta=\frac{G}{4\pi \omega l}(\frac{1}{a^2}-\frac{1}{b^2})$$We can measure $G$ and so we have the viscosity $\eta$

So could anybody explain that little problem I'm having in terms of the velocity gradient? Thanks...


1 Answer 1


This is fairly simple, so I hope my explanation doesn't make it overly complicated.

Imagine a rigid disk rotating with angular velocity $\omega$ in its plane about its center. Consider the tangential velocity of the points along some radius. The tangential velocity as a function of $r$ is $u(r)=r\omega$. Since the disk is rigid, $\omega$ is not a function of $r$, so the gradient of $u$ is simply $\omega$. That's why you book calls it the rigid body velocity gradient.

Now imagine two people running side-by-side around a circular track at the same angular velocity, $\omega$. One runner stays in the inside lane and the other stays in the outside lane, like the layers of the fluid. They are each holding the ends of a ribbon, which represents the fluid. As long as the runners have the same $\omega$ they can hold onto the ribbon. The ribbon does not stretch or break, which means there is no shear in the fluid when the $\omega$ is the same. In this case, the outer layer of fluid has a greater tangential velocity, but there is zero shear stress between the layers, because the velocity gradient is simply $\omega$.

When the outer runner increases his $\omega$, he pulls out ahead and that creates the shear stress. The ribbon is only stretched or broken when there is a velocity gradient due to a difference in $\omega$, just as the shear stress in the fluid only arises when there is a $d\omega /dr$.

A sketch might help. I have tried to show two velocity vectors in black. One has length $\omega r$ and the other has length $\omega (r+dr)$. If the fluid had these velocities, there would be no shear between the layers. The little red vector represents the extra speed in the outer fluid layer due to $d\omega$. The total velocity in outer layer is $(r+dr)(\omega+d\omega)$. Subtract from that the component that does not contribute to shear, which is $(r+dr)\omega$. Ignore the higher order term $(dr)(d\omega)$. The result is $du=r d\omega$ or $du/dr=r d\omega/dr$.

This sketch might help.


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