Viscosity and Applications of Fluid Motion – Fluid Mechanics (Senior 5 Physics) | Ganzaa

Fluid Mechanics

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Senior 5 · Physics

Fluid Mechanics

4 lessons

Lesson 4 of 4

Viscosity and Applications of Fluid Motion

25 min

Why honey is slow

Pour water and honey from two spoons: the water runs off at once, the honey stretches into a slow, thick thread. Drop a small stone into each: it sinks quickly through water but takes its time through honey. After heavy rain, rivers run brown because fine soil particles hang in the water for hours or days, while sand settles out within seconds. The property behind all this is viscosity, the internal friction of a fluid. Bernoulli's equation in Lesson 3 ignored it; this lesson puts it back.

Viscosity

When a fluid flows past a surface, the layer touching the surface sticks to it, and each layer further out slides over the one below. Neighbouring layers drag on each other, like cards in a pack being pushed sideways. Picture a liquid of thickness $d$ between a fixed plate and a plate of area $A$ dragged along at speed $v$ .

Experiment shows that the force needed to keep the top plate moving is proportional to its area and to the velocity gradient $\frac{v}{d}$ :

F = \eta A\frac{v}{d}

The constant $\eta$ (eta) is the coefficient of viscosity of the fluid, measured in $\mathrm{Pa\,s}$ ( $= \mathrm{N\,s\,m^{-2}}$ ).

Fluid (about 20 °C)	$\eta$ / $\mathrm{Pa\,s}$
air	$1.8 \times 10^{-5}$
water	$1.0 \times 10^{-3}$
blood (at body temperature)	about $3 \times 10^{-3}$
glycerol	about 1.4
honey	several Pa s, depending on the honey

The viscosity of a liquid falls quickly as it warms (warm honey pours easily), because the molecules move apart and attract each other less. The viscosity of a gas rises slightly with temperature, because faster molecules carry more momentum between layers.

Definition

The coefficient of viscosity $\eta$ of a fluid is the tangential force per unit area needed to maintain unit velocity gradient between its layers: $\eta = \frac{F/A}{v/d}$ . Unit: $\mathrm{Pa\,s}$ . Example: oil with $\eta = 0.20\,\mathrm{Pa\,s}$ in a 0.50 mm film under a plate of area $0.040\,\mathrm{m^2}$ moving at $0.10\,\mathrm{m\,s^{-1}}$ needs $F = 0.20 \times 0.040 \times \frac{0.10}{0.50 \times 10^{-3}} = 1.6\,\mathrm{N}$ .

Stokes' law and terminal velocity

A small sphere of radius $r$ moving slowly at speed $v$ through a fluid of viscosity $\eta$ , with smooth (laminar) flow around it, feels a drag force given by Stokes' law:

F = 6\pi\eta rv

Drop a small steel ball into a tall tube of glycerol. Three forces act on it: its weight $W = \frac{4}{3}\pi r^3\rho g$ ( $\rho$ the density of the ball), the upthrust $U = \frac{4}{3}\pi r^3\sigma g$ ( $\sigma$ the density of the liquid), and the drag $F$ . At first $v = 0$ , there is no drag, and the ball accelerates. As $v$ grows, the drag grows, the resultant force shrinks, and so does the acceleration. Eventually $W = U + F$ and the ball falls at a constant terminal velocity $v_{t}$ :

\tfrac{4}{3}\pi r^3\rho g = \tfrac{4}{3}\pi r^3\sigma g + 6\pi\eta rv_t \quad\Longrightarrow\quad v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}

The terminal velocity is proportional to $r^{2}$ : halve the size of a particle and it sinks four times more slowly. Stokes' law only holds while the flow round the sphere stays smooth, which needs the Reynolds number $Re = \frac{\sigma vd}{\eta}$ (with $d = 2 r$ ) to be small, below about 1. Large, fast objects such as a skydiver or a raindrop of 2 mm leave a turbulent wake; their drag grows roughly as $v^{2}$ instead.

Worked example 1: a ball in glycerol

Find the terminal velocity of a steel ball of radius 1.0 mm ( $\rho = 7800\,\mathrm{kg\,m^{-3}}$ ) falling through glycerol ( $\sigma = 1260\,\mathrm{kg\,m^{-3}}$ , $\eta = 1.41\,\mathrm{Pa\,s}$ ), and check that Stokes' law applies.

Solution

$v_t = \frac{2 \times (1.0 \times 10^{-3})^2 \times (7800 - 1260) \times 9.8}{9 \times 1.41} = \frac{0.128}{12.7} = 1.0 \times 10^{-2}\,\mathrm{m\,s^{-1}}$ , about 1 cm per second.

$Re = \frac{1260 \times 0.010 \times 2.0 \times 10^{-3}}{1.41} = 0.018$ , well below 1, so the flow is smooth and Stokes' law holds.

Worked example 2: why muddy water stays muddy

Soil particles ( $\rho = 2650\,\mathrm{kg\,m^{-3}}$ ) settle in still water ( $\eta = 1.0 \times 10^{-3}\,\mathrm{Pa\,s}$ ). Compare the time to settle through 1.0 m for (a) a silt grain of radius 0.050 mm, (b) a clay particle of radius 2.0 $\mathrm{\mu m}$ .

Solution

(a) $v_t = \frac{2 \times (5.0 \times 10^{-5})^2 \times 1650 \times 9.8}{9 \times 1.0 \times 10^{-3}} = 9.0 \times 10^{-3}\,\mathrm{m\,s^{-1}}$ : about 110 s for 1.0 m.

(b) $v_t = \frac{2 \times (2.0 \times 10^{-6})^2 \times 1650 \times 9.8}{9 \times 1.0 \times 10^{-3}} = 1.4 \times 10^{-5}\,\mathrm{m\,s^{-1}}$ : about $7 \times 10^4$ s, roughly 19 hours, and any current stirs it up again. The radius is 25 times smaller, so the speed is $25^{2} = 625$ times smaller. Water-treatment works add a chemical coagulant so that tiny particles clump together into larger flakes, which settle much faster.

Experiment 4.4: the falling-ball viscometer

Aim: to verify that $v_t \propto r^2$ and to measure the viscosity of glycerol.

Apparatus: a tall measuring cylinder or tube of glycerol, two rubber bands, steel ball bearings of several sizes, a micrometer screw gauge, a stopwatch, a metre rule, a thermometer, a magnet.

Put band A well below the surface (at least 10 cm), so the balls reach terminal velocity before passing it, and band B a measured distance $s$ below A (here 20.0 cm).
Measure each ball's diameter with the micrometer and halve it to get $r$ .
Wet each ball with glycerol, release it at the centre of the surface, and time its fall from A to B. Repeat with the other balls. Record the temperature.
Calculate $v_t = \frac{s}{t}$ and plot $v_{t}$ against $r^{2}$ . The gradient is $\frac{2(\rho - \sigma)g}{9\eta}$ .

Illustrative results ( $s = 0.200\,\mathrm{m}$ , steel $\rho = 7800$ , glycerol $\sigma = 1260\,\mathrm{kg\,m^{-3}}$ ):

$r$ / mm	$r^{2}$ / $\mathrm{mm^2}$	$t$ / s	$v_{t}$ / $\mathrm{mm\,s^{-1}}$
0.50	0.250	80.1	2.50
0.75	0.563	34.8	5.75
1.00	1.000	20.1	9.95
1.25	1.563	12.6	15.87
1.50	2.250	8.9	22.47

The points lie on a straight line through the origin, confirming $v_t \propto r^2$ . The gradient is $1.00 \times 10^{4}\,\mathrm{m^{-1}\,s^{-1}}$ , so $\eta = \frac{2(\rho - \sigma)g}{9 \times \text{gradient}} = \frac{2 \times 6540 \times 9.8}{9 \times 1.00 \times 10^4} = 1.42\,\mathrm{Pa\,s}$ .

Warning

Safety. Wipe up spilled glycerol at once: it makes floors and benches very slippery. Recover the balls with a magnet rather than tipping out the tube, and wash glycerol off your hands. Clamp a tall tube so that it cannot be knocked over.

Applications of fluid motion

Situation	Physics
fog and clouds	droplets of radius about $10\,\mathrm{\mu m}$ have $v_t = \frac{2r^2\rho g}{9\eta_{\text{air}}} \approx 1\,\mathrm{cm\,s^{-1}}$ : gentle air currents keep them up
engine and bicycle oil	a viscous film keeps moving metal surfaces apart; oil grades are chosen so the viscosity stays suitable as the engine heats up
blood flow	for viscous flow in a tube, the flow rate depends on the fourth power of the radius, so a slightly narrowed artery greatly reduces blood flow
measuring the electron's charge	Millikan's oil-drop experiment used Stokes' law to find the radius of tiny drops from their terminal speed
aircraft, sprayers, burners, Venturi meters	streamline flow and Bernoulli's principle (Lesson 3)

Warning

Common mistakes. (1) Forgetting the upthrust: the driving force is $(\rho - \sigma)$ , not $\rho$ alone. (2) Using the diameter for $r$ in Stokes' law. (3) Timing the ball from the moment of release, before it reaches terminal velocity. (4) Applying Stokes' law to large or fast objects (a skydiver, a football) where the flow is turbulent. (5) Saying viscosity of all fluids falls with temperature: for gases it rises.

Tip

At terminal velocity the acceleration is zero, so write "forces up = forces down" and solve. The same three-force balance (weight, upthrust, drag) appears in raindrops, sediment, blood cells in a centrifuge and Millikan's experiment.

Viscosity of a liquid: A Level practical (Physics Online). Watch how the marks are placed below the surface so the ball is already at terminal velocity, and how the graph gives the viscosity.

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Practice

Define the coefficient of viscosity and give its SI unit in base units.
A plate of area $0.050\,\mathrm{m^2}$ slides at $0.20\,\mathrm{m\,s^{-1}}$ on an oil film 0.40 mm thick ( $\eta = 0.15\,\mathrm{Pa\,s}$ ). Find the force needed.
Derive $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$ .
Find the terminal velocity of a glass bead of radius 0.40 mm ( $\rho = 2500\,\mathrm{kg\,m^{-3}}$ ) in an oil of density $900\,\mathrm{kg\,m^{-3}}$ and viscosity 0.50 Pa s.
Two spheres of the same material have radii in the ratio 1 : 3. Find the ratio of their terminal velocities in the same liquid.
Sketch the velocity–time graph of a ball released from rest in glycerol, and explain its shape in terms of the forces.
A fog droplet of radius $5.0\,\mathrm{\mu m}$ falls in air ( $\eta = 1.8 \times 10^{-5}\,\mathrm{Pa\,s}$ ). Find its terminal velocity and the time it takes to fall 100 m in still air.
In Experiment 4.4, a student places band A only 2 cm below the surface. Evaluate the effect on the value of $\eta$ obtained.

Show answersHide answers

The tangential force per unit area per unit velocity gradient: $\eta = \frac{F/A}{v/d}$ . Unit Pa s $= \mathrm{kg\,m^{-1}\,s^{-1}}$ .
$F = 0.15 \times 0.050 \times \frac{0.20}{4.0 \times 10^{-4}} = 3.75\,\mathrm{N}$ .
At terminal velocity $W = U + F$ : $\frac{4}{3}\pi r^3\rho g = \frac{4}{3}\pi r^3\sigma g + 6\pi\eta rv_t$ ; rearrange for $v_{t}$ .
$v_t = \frac{2 \times (4.0 \times 10^{-4})^2 \times 1600 \times 9.8}{9 \times 0.50} = 1.1 \times 10^{-3}\,\mathrm{m\,s^{-1}}$ .
$v_t \propto r^2$ : 1 : 9.
Speed rises steeply at first (only weight minus upthrust acts), then more gently as drag grows with speed, levelling off at $v_{t}$ when drag equals weight minus upthrust.
$v_t = \frac{2 \times (5.0 \times 10^{-6})^2 \times 1000 \times 9.8}{9 \times 1.8 \times 10^{-5}} = 3.0 \times 10^{-3}\,\mathrm{m\,s^{-1}}$ ; 100 m takes about $3.3 \times 10^4$ s, roughly 9 hours.
The larger balls may still be accelerating at band A, so their average speed from A to B is less than $v_{t}$ . The measured speeds are too low, so the calculated $\eta$ is too high, and the error is largest for the biggest balls, which take longest to reach terminal velocity.

Key points

Viscosity: $F = \eta A\frac{v}{d}$ ; $\eta$ in Pa s; liquids get less viscous when warm.
Stokes' law: $F = 6\pi\eta rv$ for a small, slow sphere in laminar flow.
Terminal velocity: $v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} \propto r^2$ .
A graph of $v_{t}$ against $r^{2}$ gives $\eta$ .

1Why honey is slow

2Viscosity

3Stokes' law and terminal velocity

4Experiment 4.4: the falling-ball viscometer

5Applications of fluid motion

Why honey is slow

Viscosity

Stokes' law and terminal velocity

Experiment 4.4: the falling-ball viscometer

Applications of fluid motion