Memorial University of Newfoundland
Department of Physics and Physical Oceanography
Physics 2053 Laboratory
Viscosity of Water
Introduction
When a uid ows slowly and steadily through a pipe, it may be considered to consist of
various layers which move at dierent velocities relative to each other. Fluid at the centre
of a pipe moves at maximum velocity, while the uid at the edge of the pipe is almost
stationary. This is calledLaminar Flow, illustrated in Figure (1).laminar
flow
Figure 1: Laminar ow in a pipe
Since each layer moves with a dierent velocity relative to its neighbour, a frictional force
Fwill exist between layers, which depends on the areaAof the liquid surface and on the
rate of shear strain, or
F/Arate of shear strain:
where the rate of shear strain is equal to the velocity gradient in the pipe, and hence
F=A
dv
dr
;
whereis thecoecient of viscosity, dened as
=
shear stress
rate of shear strain
:
The units ofare Nsm
2
or `dekapoise'.
Consider a cylinder of uid of radiusrcentered on the axis of the pipe of radiusa. The
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surface area of the uid cylinder (length`) is 2r`. The force exerted by the uid outside
the cylinder on the uid inside the cylinder is
F=2r`
dv
dr
(1)
Fopposes the uid motion inside the cylinder. The `' sign is necessary becausevdecreases
asrincreases. For steady ow, a driving force must be applied to counteract the resisting
viscous force. IfP1andP2are the pressures at the ends of the uid element, the net force
is given by
F= (P1P2)r
2
: (2)
Hence we have
2r`
dv
dr
= (P1P2)r
2
(3)
and thus the velocity of a cylindrical shell (radiusr) in a cylinder of radiusais given by
v(r) =
P1P2
4`
(a
2
r
2
): (4)
The total volume per unit time (i.e. the ow rate or volume ux) through the pipe is
obtained by adding up the ow due to all such shells of radiusrand thicknessdr. i.e.,
dV
dt
=
Z
a
0
2vr dr=
a
4
(P1P2)
8`
: (5)
This isPoiseuille's Law(1835), and holds as long as the ow is laminar; it does not hold
for turbulent motion.
Derivation of Poiseuille's Law by Dimensional Analysis
Poiseuille's equation can be derived on the assumption that the volume of liquid owing
through a pipe depends on,aand the pressure gradient p=`. Thus we have
volume per second =k
x
a
y
p
`
z
wherekis a constant. Use dimensional arguments to obtain values forx,yandz.
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Method
1. A sketch of the experimental setup is shown in Fig. 2. A small electric pump takes
water from the tank, and pushes it through the glass tube before being returned to the
tank. You may assume that the diameter of the glass tube is 1:20:1 mm.pump
water tank
P1 P2
Figure 2: Schematic diagram of the apparatus, including the pump and water reservoir.
2. Start with a low pump speed and record the pressuresP1andP2from the water
columns. Determine the ow rate by measuring the time required to collect 10 or 20
cm
3
of water.
3. Increase the pump speed slightly, recording the new pressures and ow rate. Repeat,
taking as many data points as you can, but do not let the voltage supplied to the
pump exceed 12 volts. Plot your data on a suitable graph and determine a value for
the viscosity of water (with its associated uncertainty) at room temperature.
4. Estimate the velocity of the water at the centre of the capillary tube when the pressure
dierence between the ends of the tube is greatest.
Turbulent Flow
Fluid ow can be described in terms of a dimensionless quantity known as theReynolds
number.
Re=
dv
(6)
wheredis the diameter of the tube andv=Q=Ais the velocity of the water in the tube.
What is the range of values ofRefor your data?
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At thecritical velocity, (denoted byvc), there is a transition in the ow from laminar to
turbulent ow. For uid ow along a cylindrical pipe,vcis given by
vc=
Rc
2a
whereis the density of the uid andRcis thecritical Reynold's number, which for most
uids is approximately 2000. Estimatevcfor your conguration.
Can you observe the onset of turbulence?
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