Chapter

12 VISCOSITY

12.1. Introduction :

Viscosity is an important property of all fluids. Liquids such as kerosene oi, alchs
water etc. which flow readily are said to be mobile. while liquids such as tar, honey, gume
which do not flow readily are said to be viscous. When such a liquid fiows. its motign
opposed by the force of friction. These forces are called víscous forces and this propery f
the liquid is called as viscosity.
12.2. Viscosity :
When there is a relative motion of one body with the other. there are frictional fores
are acting between the two surfaces of the bodies, which opposes the relative motign af
these two bodies. Such a frictional force is called as Viscous force.
Consider a flow aliquid through a capillary tube consisting of thin parallel lzyers
These liquid layers move relative
to each other with different velocities.

--__Center of tube
Flow of
liquid

F
0

Bottom of tube

Fig. 12.1
 which isclose with the bottom of 149
iayer the
velocity.V and the velocities of layer I andtube appars stationary and moving
The
from bottom to the centre of the tube layer 2is V, and V,
orceofffriction"E is acting between the two
respectively.
velocity of layer increases ie. V, V
liquid layer which
the liquid layers. This force is called as viscous force or opposes the relative
A

yeryof
liquid is called as viscosity. viscousdrag and such a
is
Thus viscosity defined as the property of fluids due to which they
between the adjacent layers.
motion
oppose relative
Gradient :
/23. Velocity
Itis defined as the ratio of change in velocity to the distance between the two layers.
Considertwo liquid layers moving with velocities V, and V,. They are at distances y, and
respectively as shown in fig. 12.2
y

---f F

Fig. 12.2
Change in velocity
Velocity Gradient = Distance between two liquid layers
dv
(12.1)
dy
where dv = change in velocity between two liquid layers
=V, - V,
dy = distance between the liquid layers = y, - y

12.4. Newtons Law of Viscosity :

the two layers is
Newtons law of viscosity states that the viscous force between
directly proportional to the
() Area of liquid layers i.e. F a A
(i) Velocity gradient i.e. Fa dv/dy
dv
Fa A.
dy
 where n isconstant of proportionality known as coefficient
the negatve sgt indicales that viscous force opposes the relative of
motion viscosty
lyers
From cquation (12.2).
Advdy (magnitude only)
btwe s t

dv
IfAIm. dy then nF
CThus. cocfficient of viscosity is defined as the viscous force per
layers per unit velocity gradienty unit a
N
SI unit of n=
mm/s/m
N.S
m

dune
CGS unit of n=
cm
It is also known as Poise.
Ifone dyne of tangential force is required to
centimeter area of liquidlayer then its viscosity is maintain
unit velocity gradient Der sn
said to be l poise.
N.S
= 10 poise
m
12.5. Flow of liquid :
Flow of liquid can be steady or non-steady. If the
liquid remains constant in time, then the flow is said to velocity V at any iven point in the
be steady otherwise it will be non.
steady.
Flow of liquid can be classified in two categories
()) Streamline flow
(ii) Turbulent flow

’ Flow of liquid

Capillary
Fig. 12.3 Streamline flow
 Streamlineflow : It is the flow of liquid in 131
flow with constant which moleculthees moves in the same
velocitv at every point within
liquid. It is a steady
streamline flow :
of
s

steady flow.
1)hisa,

Velocity of the particles at a given point

P within the liquid remains
) The
direction of the particles is the same as that of flow. constant.
The streamlines are parallel to each other.

6)In
streamline flow. molecule from one
layer do not allow to jump into
nline. another
(6) Below critical velocity of fluid the flow is
streamline.
AL flow of liquid through capillary.
Turbulent flow : It is a non-steady
flow. Velocity of the molecules of the liquid
not remain constant at a given point and molecules do not moves in the
direction of liquid
A0 Water fall, fast flow of water in streams, rivers.

’ Flow of liquid

Fig. 12.4: Turbulent flow

tiperties of turbulent flow :
() It is anon-steady flow.
constant.
14) Velocity of the particles at a given point within the liquid does not remain
D) The direction of the particles is not same as
that of flow.
layer.
(4) Molecule from the layer is allowed to jump into another
turbulent.
(5) Above of fluid, the flow is
critical velocity
S Critical Velocity
: defined as the maximum
velocity of the flow upto
Critical (V) of the flow is turbulent.
the flowvelocity
is Streamline and bevond which flow becomes
152

12.7. Reynolds number :

Applied Payi
Reynold studied the motion of fluids in detail and observe that the critical
|velocityof fhe
of fluid is related to density p of the liquid, its coefticient of viscosity n and radius V
tube r by following relation.
Rn
p.r
where "R is constant known as Reynold's number.
(124)
On the basis of no. of experiments Reynold concluded that the constant 'R
whether the flow is streamline or turbulent i.e. it determines the nature of the flowdetofermliquid
ines
through a tube.
Significance of Reynold's no. (R) :
Reynold studied the flow of liquid through a tube of radius l cm and concluded ho.
if Ris less than or equal to 2000 i.e. R< 2000, the flow is streamline and when Ris aboy.
3000, the flow is t bulent. In the transition region between 2000 to 3000 the flow is unstahl.
and may change iron one type to another.
12.8. Terminal Velocity :
It is defined as the constant velocity with which a body falls through fluid.
When a body is released. its velocity increases due to gravity. But when it moves in
a fluid, viscous force and upthrust of the fluid opposes its motion. As the velocity of the body
increases the viscous resistance also increases. A stage is reached when total upward force
(viscous force and upthrust force) becomes equal to the weight of body acting in downword
direction. The resultant force at this stage becomes zero and the body moves with aconstant
velocity called 'terminal velocity'.
12.9. Stoke's Law :
It states that, "The viscous force experienced by a spherical body, falling freely with
constant velocity in a fluid is directly proportional to
(i) the coefficient of viscosity of the fluid (n)
(ii) radius of the spherical body (r) and
(ii) the terminalvelocity attained by the spherical body.
Consider a spherical body of radius 'r falling freely through a fluid having coetticient
of viscosity 'n' with terminal velocity v'. Then according to Stokes law, the viscous fote
F° experienced by a splherical body is given by,
Fa nrV
F = constant x nrV
(12.5)
F= 6nrV
 fluid.
givenvisof
cOUs
as
The equilibrium,For Consi
fluid der
also, and
downward of Deriation fal the6 of
Upthrust viscous Total density small a of
of force force downward 'd'
spherical n' sk
ale
the (F) with Fig. Hoprt sing
fluid is acting 12.5 oonstant
is given density = force terminal body free
W=
mg on W=
F+U Free :
= m=
given 6nrVF= by, rpg W the = of
4 fal of
by, 4
4/3 r' volume mass sphere Total velocityradius proponinlty
of
of a
'r a spherical
is upward V.spherical B
given and ’Glycernine
formula)
(Stokes force Let Stokes
tube Spherical
body
by.
densi
'n' ty body body Te
is
the 'p' in see
coefficientfalling liqeid
ie
freely
(12....6) of
(12.8). . through
viscosity
(12.7)
a
 e as
found ballgauge. Keep tube 12.11.
shown 154
Find bearing
using Calculate Release Find the
method The
Lab By
the below distance
to the coefficient
using becomes.
cquation
(12.5)
formula. slope experiment
as
: r cover radius the shown
and this
of balls between
the plot the of relation
one in of
graph distancedifferent viscosity to
a fgure
graplh by Adetermine
as
r(m) one andB n 4
1 AB. small 12.3. n' for 66tnrV =
n=- betweenr V= through nrV
given
then, 12.Fig.6 Calculate atFix for U-rdg
ball least a n =r
V(m/s) ’ required Distance
Time AB the thegiven fluid 2 4/3 4 3 4
the bearings using
and 50
positions rr(p ar rpg -
the fluid fluid can 6TrV
coefficient cms.
d)g V. terminal and Stokes be
(p p 4
Slope The given -
)g 4
x candetermined - 3
measure A )g- d)g
slope >X nature and method be :
of todetermined
r'dg
velocity
viscosity you B
of the on
using
graph oftime the
of the
micrometer tube in
is required lab
given ball Applicd
Pics
a usingusing
straight
(12.-.J.0) bearing
fluid for threadStokes
each screw 29)
can line as
 Solution:
Given Ot Solution rce
heynold': s liquid viscOsity P'roblsoebedms
2. Formula: acting 1.A
:
() thee
n
The Given
cal()uTo
lated (iv)The Sample
no. is on 1.56 metal determine shape, sol efticieney
its by tle
0.8 on oppos Aylit on
and liquid :
the Ns/m plate the testing ition
x nateriViscosity
alsusared of
state 10fl3 ows plate. basis size designing
ships, standard
nih of
Ifarca of c/m laboratoryoffered
kg/m
whether through the of
and
by
and plate 0.25 nnolecular
viscosity, viscousbyreducing
hiction
Milikan's of
thc the a the to
Ns/m0.4 h r= r5 d10 cm
flow pipe m V
0,0m/5s A0.25
X0.003 nL56 Ns/m concept
a lubricatsto
m1.56
m'moved isrestsm'
0.8 10
cocfficientyis of weight aeroplanes
x
6,5 N on oil nedim
x diameter
turbulent Vale
x |0
10 x dv with a drop
10 0.25 laayer ofviscosityof
kgm' m m of10 velocity method, large ef
of 0.003 of cats avoid
viscosity cm 0,.05
streamline with oil molecules
of0.003 concept is and Weat
is a 0.05 ed oer
speed
04 m/s thick m prteins of to nd
of determine veles
Ns/m2. 0.5 Caleulate viseity e
m/s, with d
The ad
Deternine coeffkent
horizotal
(S01) (52) iscelloe inpede
density
ed
p
Solution:
Given : terminal Solution:Goven
velocity f4
(Density 4. Formula Fomls
velocity A 3.Calculate
ts
spherical
mis less
of
glycerine 0.24of throughair. the than
steel
viscous 2000
m/s
= ball
through
1.26 of force flow
Coefficient of
xdiameter is
dglycerine. V= s4
m/r2x cm
10 d4mm streamlineon
= = 10° = ==27129.6 =-27129.6 F h
6 -6x3.4x
-27129.6 -4x102cm raindrop a 50
6 kg/m', 2.71 271.3 271.3 6nnrV 2.8 cm/s400 105*x08 05
x mm 6mm
10 Find x viscosity of
and
x x x 10"
m =9.8 g 10 10° 10 x x x of
the poise diameter 04
density N N x 10" 10 10° I.8
coefficient 10 N x dyne x
m/s) 10 air
N 10 1.is
8 4
7.2 x nmm,
x N 2
of 10' x falling
x
viscosity 10² 10
kg/m poise with
400 x
offalls &
glycerine cota
with (S
 A. 6. S. 2 I. .Obtain
fficient
of
sity.velocity ANslm AmStatethroughUDain 'What
critical
StateExplainvelocity'. Define What What which On velocity' 'terminal ne Define.
State DefiPoise' ystions :
State Define (i)
force metal the applications
is
streamline
is is an stokes Newtons
of by plate a an Reynolds how turbulent
medium
expression expressionfactors Viscosity(ii)
of 9 factors
applying law
0.06 N of wil law
isrequired viscous of
m/s. area on number. flow? flow? of
of ofyou viscosity.
for viscosity.
chextent.viscosity.
a 0.2 whiinfinite fordetermine gradient
Velocity
If force
the State State force "n V02Amis
the to to m State d-126
m/
kgm9810
g
of critical
is terminal for Hence Nislm
0,485 =
its its depends? Hence
thicknessmove 10
y' its
characteristics. a 2
moved characteristics free
a N. velocity significance. in write
lab falling edefine
liquid Find over velocity the
of using coefficient
10}
a of pt formula
the over the thick liquid of a (72-
liquid veloacity a stokes spherical 024
solid layer sphere
for of 126)
tube the viscosity.
body
layersurfacegradient. depends.
of falling
liquid method. viscous (W-00,
S-02)
S-01, 10
0.0002 freely in State x98
of having liguid. force.
area its
(S-00)
m. under (W-01)(W-01) (S-00)
(1.s2/5m) N0.24 vi(1sm/00cosis) ty (S-02) (S-02) unit.
Calculate (S-01)(W-00)gravity (S-01) (S-02)
m
the with 0.5
 158

18.
Assuming Reynold's number to be 1008. calculate the critical Applied Physics
the pipe of diameter 2 cm. Density and velocity of
0.85 Ns/m respectively. viscosity of glycerine are 1.36 x 10glycerine
kg/m and
in
and
19 Find the Reynold's number for a (63 m/s)
the critical velocity 60 m/s. The liquid having coefticient of viscosity 1.4Ns/m and
density of the
of pipe is 2.5 cm.
State whether the flow is liquid is 1.2 x 10° kgm.The diameter
streamline or turbulent.
20.
Find the radius of the (643 streamline)
S. Viscosity of air raindrop falling through air with the
= 1.8 x 10
N.s/m². Density constant
of air = 1.26 x 10*velocity 1.2 cm/
21. kg/m
Calculate the air resistance (viscous
falling with a constant force) acting on a rain drop of(Ans. : 10* m)
Given : n of velocity of 7.5 m/s through air. diameter 6 mm
air = 1.8 x 10
N.s/m
(0.763 x 105 N)