Measurement of the viscosity of glycerol
The viscosity of a liquid can easily be measured in the laboratory
with the apparatus shown in Figure 4. A one-litre measuring cylinder
is filled with glycerol and two rubber bands are placed around it a
known distance apart (say 20 cm). The diameter of a small steel ballbearing is measured with a micrometer and it is then released from
just above the glycerol surface and allowed to fall through the fluid,
the time for it to pass from the level of one band to that of the other
being taken. If the bands are placed sufficiently far from he surface it
can be assumed that the ball bearing has reached its terminal
velocity before passing between them. The velocity of the ballbearing between the bands can then be found, and it is assumed
that this is its terminal velocity. From Stoke's Law the viscosity of the
fluid may be found. The temperature of the glycerol should be
recorded and the experiment should be repeated with ball-bearings
of different radii.
For accurate work allowance should be made for the effects of the
walls of the container and for this reason the experiment should
always be carried out with cylinders of large radii compared with the
ball-bearings.
Figure 4
Golden syrup makes a cheaper and perfectly suitable substitution for glycerol!
Example
Calculate the terminal velocity of a raindrop of radius 0.2 cm (raindrops with radii much greater than this
will become unstable and break up). (Density of water = 1000 kg m -3 and that of air about 1 kg m -3)
Terminal velocity V =2gr2( – s) = 2 x 9.81 x (0.2 x 10-2)2 x 999 = 7.04 x 10-2
9
9 x 10-3
9 x 10-3
= 8.7ms-1
We can use Stoke's Law to calculate either the size of small spherical particles falling through a
fluid or the time it will take a particle to fall a given distance if the radius of the particle is known.
Example
Find the time taken for a particle of carbon (density 2300 kg m -3) with a radius of 0.0001 m to fall 2 m
through air (viscosity 0.001 Pa s).
Terminal velocity = 2 x 10 x 10-10 x 2300
10-2
=4.6 x 10-4ms-1
Time to fall 2 m = 2/[4.6x10-4] = 4348 s = 72 m = 1.2 hr