9.3 Viscosity, Stokes' Law and Terminal Velocity

Tip a jar of honey and a glass of water at the same moment: the water is gone in a second, while the honey creeps out as a slow thick ribbon. Both are liquids of similar density, so gravity pulls them almost equally; what differs is how strongly each layer of liquid resists sliding over the next. That resistance is viscosity. It is why engines need oil of the right "grade", why blood pressure must push blood through narrow vessels, and why a cloud droplet drifts down at a centimetre a second instead of plummeting.

In Section 9.2 we ignored viscosity to get Bernoulli's principle. Here we put it back. You will define the coefficient of viscosity from the flow between two plates, meet Stokes' law for the drag on a small sphere, and use it to find the terminal velocity that a sphere reaches when the drag, the upthrust and its weight balance — the basis of the falling-ball viscometer and of every calculation about how fast dust, sediment and droplets settle.

Viscosity

Put a thin layer of liquid between a large fixed plate and a parallel plate that you drag along at constant speed $v$. The liquid touching each plate sticks to it (it has the plate's velocity), so the bottom layer is at rest and the top layer moves at $v$. In between, the liquid moves in thin layers, each sliding a little faster than the one below: this is laminar flow. To keep the top plate moving at constant speed you must keep pulling on it, because each layer drags back on the faster layer above it — a fluid version of friction.

Experiment shows that the force needed is proportional to the area $A$ of the plate and to the speed $v$, and inversely proportional to the thickness $l$ of the layer. In terms of shear (Chapter 8), the fluid is being sheared continuously, and the shear stress $\dfrac{F}{A}$ is proportional not to the shear strain but to the rate at which the strain grows, $\dfrac{v}{l}$.

The viscosity of a liquid falls steeply as it warms (warm honey pours easily, cold engine oil does not), because the molecules escape more easily from their neighbours' attraction. The viscosity of a gas, by contrast, rises with temperature: in a gas, momentum is carried between layers by molecules darting across, and hotter molecules dart faster. In a pipe, laminar flow is fastest along the axis and zero at the wall, where the fluid sticks; the velocity profile is a parabola, and a pressure difference along the pipe is needed to keep the flow going — Bernoulli's principle alone would say none is needed in a uniform horizontal pipe.

Critical velocity and Reynolds number. Laminar flow lasts only up to a certain speed, the critical velocity $v_c$ (Section 9.2); faster than that, eddies form and the flow becomes turbulent. Whether a flow is laminar is decided by a dimensionless combination, the Reynolds number

where $\rho$ is the fluid's density, $v$ its speed and $d$ the diameter of the pipe. It compares the fluid's inertia ($\rho v$, which drives eddies) with viscosity (which damps them). For flow in a pipe, roughly: $Re\lt1000$ laminar, $Re\gt2000$ turbulent, unsteady in between. Setting $Re\approx2000$ gives $v_c\approx\dfrac{2000\,\eta}{\rho d}$: the critical velocity is high for viscous fluids and narrow tubes, low for dense, runny fluids in wide pipes. For water ($\eta=1.0\times10^{-3}\ \text{Pa s}$) in a $1.0\ \text{cm}$ pipe, $v_c\approx\dfrac{2000\times1.0\times10^{-3}}{1.0\times10^3\times0.010}=0.20\ \text{m/s}$ — which is why water from a tap opened wide comes out turbulent.

Stokes' law

A sphere moving slowly through a fluid drags the fluid near it along, setting up velocity gradients all round it, so it feels a viscous force opposing its motion. What can the force depend on? The viscosity $\eta$, the radius $r$ and the speed $v$ — at low speeds the fluid's density does not enter, because the inertia of the fluid is unimportant compared with viscosity. Dimensional analysis (Section 1.3) then fixes the form. Suppose $F=k\,\eta^ar^bv^c$. With $[\eta]=[\text{M L}^{-1}\text{T}^{-1}]$:

Comparing powers: $a=1$; $-a-c=-2$ gives $c=1$; $-a+b+c=1$ gives $b=1$. So $F=k\,\eta rv$. A full calculation of the flow round the sphere, beyond this course, gives $k=6\pi$.

The drag is proportional to the speed, like the viscous force between plates. For larger bodies or higher speeds (a cricket ball in air, a parachutist, a car) the flow behind the body becomes turbulent and the drag grows roughly as $v^2$ instead; Stokes' law then underestimates it badly.

Terminal velocity

Drop a small steel ball into a tall jar of glycerine. Three forces act on it: its weight downward, the upthrust (buoyancy) of the displaced liquid upward, and the viscous drag upward, opposing its downward motion. At first the ball is slow, the drag is small and it accelerates. As it speeds up the drag grows, the net force shrinks, and so does the acceleration. Soon the drag has grown until the forces balance; from then on the ball falls at a constant speed.

For a sphere of radius $r$ and density $\rho$ falling through a fluid of density $\sigma$ and viscosity $\eta$, its weight is $\tfrac43\pi r^3\rho g$ and the upthrust is $\tfrac43\pi r^3\sigma g$. At the terminal velocity $v_t$ the Stokes drag balances their difference:

Once the terminal velocity is reached the kinetic energy no longer changes, so the gravitational potential energy lost (minus the work done by the upthrust) is all converted into heat in the fluid by the viscous force. A falling-ball viscometer turns Result 9.10 round: time a ball of known size and density falling between two marks, once it has reached its terminal velocity, and calculate $\eta=\dfrac{2r^2(\rho-\sigma)g}{9v_t}$.

Exercises

Take $g=9.8\ \text{m/s}^2$, $\rho_\text{water}=1.00\times10^3\ \text{kg/m}^3$, $\eta_\text{water}=1.0\times10^{-3}\ \text{Pa s}$ and $\eta_\text{air}=1.8\times10^{-5}\ \text{Pa s}$; assume Stokes' law holds unless stated otherwise.