Stokes law.
Viscosity coefficient
7.1. Introduction
Real fluid has a certain amount of internal friction, which is called viscosity.
Viscosity exists in both liquids and gases, and is essentially the frictional force
between the adjacent layers of fluid as the layers move past one another. In liquids,
viscosity appears due to the cohesive forces between the molecules. In gases, it arises
from collisions between the molecules.
F
x
moving plate
fluid
v
velocity
gradient
l
static plate
Fig. 7.1 Experiment setup for obtaining of viscosity coefficient
Different fluids posses different amounts of viscosity: syrup is more viscous than
water; grease is more viscous than the engine oil; liquids in general are much more
viscous than gases. The viscosity of different fluids can be expressed quantitatively by
the coefficient of viscosity, η (the Greek lowercase letter eta), which could be defined
using the following experiment. A thin layer of fluid is placed between two flat plates.
One plate is static and the other is made to move (see Fig. 7.1). The fluid directly in
contact with each plate is held to the surface by the adhesive force between the
molecules of the liquid and those of the plate. Thus the upper surface of the fluid
moves with the same speed v as the upper plate, whereas the fluid in contact with the
stationary plate remains stationary. The stationary layer of fluid retards the flow of the
layer just above it, which in turn retards the flow of the next layer, and so on. Thus the
53
STOKES LAW. VISCOSITY COEFFICIENT
velocity varies continuously from 0 to v, as shown. The increase in velocity divided by
the distance over which the change is made - equal to v/l - is called the velocity
gradient. To move the upper plate requires a force, which you can verify by moving a
flat plate across the puddle of syrup on the table. For a given fluid, it is found that the
required force F, is proportional to the area of a fluid in contact with each plate A, and
to the speed v, but is inversely proportional to the separation l, of the plates, what
comes down to the following relation: F  vA / l . For different fluids, the more viscous
the fluid, the greater is the required force. Hence the proportionality constant for this
equation is defined as the coefficient of viscosity, η:


Av
F  
l
(7.1)
Solving for η, we find   Fl / vA . The SI unit for η is N  s / m 2  Pa  s . In the CGS
system, the unit is dyne  s / cm 2 and the unit is called a poise (P). Viscosities are often
given in centipoise* ( 1cP  10 2 P ). Viscosity is a function of the temperature – for
example a hot engine oil is less viscous than the cold one. The Table 7.1 lists the
viscosity coefficients of various fluids at the specified temperatures
Table 7.1. The viscosity coefficients of various fluids at the specified temperatures
*
Fluid
Temperature [oC]
Viscosity [Pa.s]
Water
0
1.8 .10-3
Water
20
1.0.10-3
Water
100
0.3.10-3
Ethyl alcohol
20
1.2.10-3
Engine oil
30
200.10-3
Air
20
0.018.10-3
Hydrogen
0
0.009.10-3
Water vapour
100
0.013.10-3
1[Pa*s] = 10[P] = 1000[cP]
54
STOKES LAW. VISCOSITY COEFFICIENT
In general the equation of friction force for any velocity gradient occurring during a
laminar flow is given below (Newton’s equation):


dv
F   A
dx
(7.2)
The equation is valid only for small velocities (low values of Reynolds number,
Re<1160, Re 
V
). Fluids, which obey this equation, are called Newtonian fluids.
tr
It would be rather difficult to calculate viscosity of the liquids directly from the above
equation. Especially it would be difficult to measure the velocity gradient and make
sure that the area of contact between the plates is kept constant. Instead, a Stokes
viscosimeter is used, in which small metal balls are dropped in a glass tube filled with
liquid.
liquid dragged by
the ball
metal ball
v
Fig. 7.2. Scheme of gravitationally falling ball in viscous liquid.
When an object (like a metal ball) falls gravitationally in viscous liquid it drags certain
55
STOKES LAW. VISCOSITY COEFFICIENT
amount of the liquid with itself due to the molecular interactions between surface of
the object and the molecules of the liquid. These layers situated close to the moving
object drag farther layers (shown on Fig. 7.2). Thus viscosity of the fluid slows down
the falling object and creates a velocity gradient in the fluid perpendicular to the
direction of motion of the object and the layers.
The velocity of the object v, is small enough that we can assume a laminar flow and
use the Stokes law to calculate the friction force acting on the metal ball:


F f  6rv
(7.3)
where r stands for the radius of the ball, v –velocity,  - viscosity. There are two more
forces that act on the metal ball. The first one is obviously the gravity.


4
Fg  mg  r 3  m g
3
(7.4)
where m is a density of the metal (steel). The second force is a buoyant force. It
occurs because the pressure in the fluid increases with depth. Thus the upward
pressure on the bottom surface of the submerged object is greater than the downward
pressure on its top surface (see Fig. 7.3)
h1
F1
A
h2
h=h2-h1
F2
Fig. 7.3. Determination of buoyant force
56
STOKES LAW. VISCOSITY COEFFICIENT
To see the effect of buoyancy consider a cylinder of height h whose top and bottom
have areas A and which is completely submerged in the fluid of density ρf, as shown
on Figure 7.3. The fluid exerts a pressure P1 = ρfgh1 at the top surface of the cylinder.
The force due to this pressure on top of the cylinder is F1  P1A  f gh 1A , and it is
directed downward. Similarly the fluid exerts an upward force on the bottom of the
cylinder equal to F2  P2 A  f gh 2 A . The net force due to the fluid pressure, which is a
buoyant force, FB, acts upward and has the magnitude:






FB  F2  F1   f gAh2  h1    f gAh   f gV
(7.5)
In case of a metal ball submerged in a liquid the buoyant force is equal to:


4
FB  r 3  f g
3
(7.6)
Initially, when the metal ball is dropped through the funnel to the liquid it steadily
accelerates. As the velocity increases, the opposing friction force also increases,
leading finally to the balance of forces. Hence we can assume that after some initial
time the ball moves with a constant speed and the three forces are in equilibrium:
 


F  Fg  FB  F f  0
(7.7)



Fg  FB  F f
(7.8)
Substituting all previously derived formulas (7.3., 7.4. and 7.5.) into the above
equation, we can observe that it links two quantities: velocity of the ball and viscosity
of the liquid. Thus calculating the velocity we can determine the viscosity using
equation 7.9.

2r 2 g (  m   f )


9v
(7.9)
57
STOKES LAW. VISCOSITY COEFFICIENT
The Stokes equation is accurate for infinitely large environment and does not take into
account the effect of the walls of the cylinder. A corrective term equal to
1
1  2.4 r R
is
introduced. It provides an estimate of how much the ball was additionally slowed
down due to the presence of the walls of the cylinder.
7.2. Measurements
An experiment is performed in the Stokes viscosimeter (see Fig.7.4).
funnel
falling metal ball
glass tube
glycerine
rubber cork
Fig.7.4. The Stokes viscosimeter
Follow the experimental procedure step by step:
1. Fill the cylinder with glycerine.
2. Put a funnel into the mouth of the cylinder.
3. Measure the distance between the levels, marked with blue stripes on the cylinder.
4. Drop (one by one) steel balls into the cylinder through the funnel and measure the
time of falling between the levels.
5. Repeat the previous point for every ball (about 15).
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STOKES LAW. VISCOSITY COEFFICIENT
6. Write down all results into a table in your copy-book.
7. Collect all additional data (e.g. density of the steel and glycerine).
8. Take out the balls from the cylinder (pulling carefully the rubber cork, and letting
some of the glycerine flow out). Pour the glycerine back into the cylinder.
The data should be collected in Table 7.2.
Table 7.2
No.
r [m]
l [m]
t [s]
m
f

[kg/m3]
[kg/m3]
[Pa.s]
R [m]
1.
7.3. Results, calculation and uncertainty
Calculate the viscosity of glycerine using equation 7.9. Estimate the uncertainty of the
measured value using below formula
  
   2    2   
2
2
d  
 dr  
 dl  
 dt  
 dR
 r 
 l 
 t 
 R 
2
2
2
2
(10)
The final result reads:
    d
(11)
7.4. Questions
1. What is viscosity? What kind of viscosity coefficients do you know?
2. Describe the phenomenon of raindrop falling down.
3. Derive the equation for viscosity.
4. Methods of viscosity measurement.
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STOKES LAW. VISCOSITY COEFFICIENT
5. On what depends the viscosity?
6. What is Reynolds number?
7. What kind of conditions should be fulfilled to use Stokes equation?
8. Why do we have to use corrective term?
9. Comment on Archimedes law.
10. Derive the equation for buoyant force.
7.5. References
1. Szydłowski H., Pracownia fizyczna, PWN, Warszawa, 1994
2. Bobrowski Cz., Fizyka – krótki kurs, WNT, Warszawa, 1993
3. Giancoli D.C., Physics. Principles with Applications, Prentice Hall, 2000
4. Feynman R., Feynmana wykłady z fizyki, Tom 2.2., PWN, Warszawa, 2002
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