CHAPTER 1

FLUID PROPERTIES

A fluid may be defined as a substance which deforms continuously when subjected to a

shear stress or in other words a fluid is a substance which cannot withstand shear stress.
Fluid Mechanics deals with the behaviour of fluids at rest and in motion. Fluids may be
of liquid or gaseous forms, Liquid is practically incompressible and occupy definite
volume and have free surfaces. Whereas gaseous fluid is compressible. Gases can expand
until it occupies all portions of any containing vessel.
Among the different properties of fluids, only the important properties are mentioned
below.
Mass Density,  :
It is the mass per unit volume. The density of water at standard pressure and temperature
is 1000 kg / m 3 .
Specific Volume, v S :
It is the volume occupied by unit mass of fluid i.e. specific volume is the reciprocal of the
1
density. Therefore, v S  
Specific Weight,  :
Specific weight may be defined as the weight of a unit volume of a substance. For water
at standard conditions specific weight is equal to 9.81 kN / m 3 . The relation between
specific weight and density can be written as,    g
Specific Gravity :
Specific gravity is a dimensionless quantity, which is the ratio of the weight of a
substance to the weight of an equal volume of water at 4 0 C . The specific gravity of
water is 1 and of mercury is 13.57.
Viscosity :
Viscosity may be defined as the property of a fluid which determines the amount of its
resistance to a shearing stress.
The viscosity of a liquid decreases with the increase of temperature due to the decrease of
molecular cohesion between molecules and this is responsible for reduction in liquid
viscosity. But in case of a gas viscosity increases with the increase of temperature due to
the greater molecular activity as the temperature increases.
The relation between shear stress and rate of deformation is given by Newton’s law of
viscosity as
du
 
dy
where   shear stress
du
 velocity gradient
dy
  absolute or dynamic viscosity
Fluids which follow this law are known as Newtonian fluids. Air, water etc. are
Newtonian fluids, whereas tar, paste, blood, glycerine etc. are Non-Newtonian fluids.
Kinematic viscosity is the ratio of absolute viscosity and mass density. Kinematic
  g
viscosity is given by,      / g  
The specific weight of gases changes with pressure (temperature constant), so the
kinematic viscosity varies inversely with the pressure. Although the viscosity of liquids is
affected by temperature but it is unaffected by pressure change.
Compressibility :
All fluids are compressible but liquids to a much smaller degree than gases. The
compressibility of a fluid is expressed by the bulk modulus of elasticity. Bulk modulus is
denoted by E , K or  T .
Let dp  increase of pressure in a unit volume of fluid
dV  decrease of volume in the volume V
  mass density
v  specific volume
1
Now specific volume, v

or,  v  1
Differentiating, we have
 dv  v d  1
d dv
or,    v
volume stress
Bulk modulus of elasticity = volume strain
dp
E
or, dV

V
dp
E
or, dv

v
dp
E
or, d

dV
Since is dimensionless, E is expressed in the units of pressure. For water at
V
ordinary temperature and pressure, E  2068 MPa .
Consider the application of 1000 kPa to 1m 3 of water.
V dp 1  1000 1
Decrease in volume,  dV    m3
E 2068  1000 2068
Therefore, the application of 1000 kPa to water under ordinary conditions causes its
volume to decrease by only 1 part in 2068.
Surface Tension :

2
The surface of contact between a liquid and a gas seems to form a film on the liquid due
to the attraction of liquid molecules below the surface. Now the force required to
maintain unit length of the film in equilibrium is called the surface tension.
The action of surface tension is to increase the pressure within a droplet of liquid. If p
denotes the pressure intensity inside the droplet of radius r and  denotes the surface
2
tension force, then, p  .
r
Capillarity :
Capillarity is the rise or fall of a liquid in a capillary tube due to surface tension and
depends on the cohesion and adhesion of the liquids. When adhesion  cohesion , liquids
rise in tubes (e.g. water). When cohesion  adhesion liquids fall in tubes (e.g. mercury).

Fig. 1.1

For a cylindrical glass tube the capillary rise or fall is given by,
2  cos 
h where   contact angle
 R
  specific weight of liquid
Problem 1.1
If a pressure of 10 MPa is applied to 2 m 3 of a liquid, find the bulk modulus of
elasticity. The decrease of volume is equal to 0.025 m 3 .
Given Data :
Initial volume of liquid, V  2 m 3
Change of volume, dV  0.025 m 3
Applied pressure  10 MPa
To Find :
- Bulk modulus of elasticity, E
Solution :
Change of pressure, dp  (10  0) MPa  10 MPa
Change of volume, dV  0.025 m 3

3
 dp
E
Now bulk modulus of elasticity, dV
V
10
E
or, 0.025

2
or, E  800 MPa Answer
Problem 1.2
A vessel having a volume of 0.01 m 3 is completely filled with a liquid having bulk
modulus of elasticity equal to 40 MPa . If the pressure in the vessel is reduced by
19.62 MPa , find the amount of liquid that will spill over.
Given Data :
Volume of liquid, V  0.01 m 3
Bulk modulus of elasticity of liquid, E  40 MPa
Reduction of pressure  19.62 MPa
To Find :
- Volume of liquid spilled over, dV
Solution :
Change of pressure, dp  p 2  p1   p1  p 2   19.62 MPa
From the definition of bulk modulus of elasticity,
dp
E
dV
V
 19.62
40  
or, dV
0.01
40
or,  dV  19.62
0.01
or, dV  4.91  10 3 m 3
The value of dV positive means an increase in volume. Initially, the vessel was
completely filled with liquid. The increased value due to pressure drop will spill over
from the vessel.
Therefore, volume of liquid spilled over, dV  4.91  10 3 m 3 Answer
Problem 1.3
Calculate the increase in pressure to produce 2% reduction in volume of water. The bulk
modulus of elasticity of water is 2.18 GPa.
Given Data :
Bulk modulus of elasticity of water, E  2.18 GPa
Reduction of volume = 2%
To Find :
- Increase in pressure, dp
Solution :
Let V  Initial volume of water

4
After 2% reduction final volume = 0.98V
Change of volume, dV  0.98V  V  0.02V
Now bulk modulus of elasticity,
dp
E
dV
V
dp
2.18  10 9  
or, 0.02V
V
or, dp  43.60 MPa Answer
Problem 1.4
Find the bulk modulus of elasticity of a liquid, if the pressure of the liquid is increased
from 72 MPa to 132 MPa. The volume of the liquid decreases by 0.20%.
Given Data :
Initial pressure, p1  72 MPa
Final pressure, p 2  132MPa
Decrease in volume = 0.20%
To Find :
- Bulk modulus of elasticity, E
Solution :
Let V = initial volume of liquid
 0.20 
After 0.20% reduction final volume  1   V  0.998V
 100 
Increase in pressure, dp  p 2  p1  (132  72) MPa  60 MPa
Change in volume, dV  0.998V  V  0.002 V
Now bulk modulus of elasticity,
dp
E
dV

V
60
E
or, 0.002 V

V
or, E  40,000 MPa Answer
Problem 1.5
The specific weight of a gas is 15 N / m 3 at 23 0 C and at an absolute pressure of
245kPa. Calculate the gas constant of the gas.
Given Data :
Specific weight of gas,   15 N / m 3
Temperature, t  230 C
Absolute pressure, p  245 kPa
To Find :
- Gas constant of the gas, R

5
Solution :
Let   density of the gas
Now specific weight,    g

or,   g
15
or,    1.53 kg / m 3
9.81
Absolute temperature, T  273  t  (273  23) K  296 K
From equation of state,
p  R T
p
or, R
T
243  1000
or, R 
1.53  296
Nm
or, R  536.57 kg K Answer
Problem 1.6
A vessel contains 0.16 m 3 hydrogen gas having mass of 1.2 kg at  30 0 C . Find the
pressure of the gas if the gas constant is 41.27NM/kgK.
Given Data :
Volume of gas, V  0.16 m 3
Mass of gas, m  1.2 kg
Gas constant, R  4127 Nm / kg K
Temperature of gas, t  30 0 C
To Find :
- Pressure of the gas, p
Solution :
Absolute temperature of gas, T  (273  30) K  243 K
From equation of state,
p   RT
m
or, p  RT
V
1.2
or, p   4127  243 N / m 2
0.16
or, p  7521458 N / m 2

kN
or, p  7521.46 Answer
m2
Problem 1.7
The velocity distribution in a pipeline of 30 mm diameter is parabolic and is given by
u  3 y  y 2 , where u denotes the velocity at a distance y from the pipe wall. Find the
wall shear stress and the total resistance for a length of 10 m. The dynamic viscosity of
fluid in the pipe is 0.035 Ns / m 2 .

6
Given Data :
Velocity distribution, u  3 y  y 2
Diameter of pipe, d  30 mm
Length of pipe considered, L  10 m
Viscosity of fluid,   0.035 N s / m 2
To Find :
- Wall shear stress,  0
- Total resistance, F
Solution :
d
Radius of pipe, r0   15 mm
2
du
Velocity gradient, dy  3  2 y
 du  1
At pipe wall,   3
 dy  y 0 s
From Newton’s law of viscosity wall shear stress,
 du 
 0    
 dy  y  0
or,  0  0.035  3 N / m 2  0.105 N / m 2 Answer
Now total resistance for 10 m length,
F   0  d L
or, F  0.105    0.03  10  98.9  10 3 N Answer
Problem 1.8
The specific weight of a liquid is 9.2 kN / m 3 . What are the values of its density, specific
volume and specific gravity?
Given Data :
Specific weight of liquid,   9.2 kN / m 3
To Find :
- Density of the liquid, 
- Specific volume of the liquid, v S
- Specific gravity of the liquid, S
Solution :

Density of the liquid, 
g
9.2  1000
or,   kg / m 3
9.81
or,   937.82 kg / m 3 Answer
1
Specific volume of the liquid, vS 

1 m3
or, v S 
937.82 kg

7
 m3
or, v S  0.0011 Answer
kg
specific weight of liquid
Specific gravity of the liquid, S 
specific weight of water
9.2
or, S 
9.81
or, S  0.94 Answer
Problem 1.9
If the specific volume of a gas is 10 m 3 / kg , what is its specific weight?
Given Data :
Specific volume of the gas, v S  10 m / kg
3

To Find :
- Specific weight of the gas, 
Solution :
The relation between density and specific volume is given by,
1

vS
1 kg
or,  
10 m 3
kg
or,   0.10 3
m
Specific weight,  g
or,   0.10  9.81 N / m 3
or,   0.981 N / m 3 Answer
Problem 1.10
A certain gas of 0.3 m 3 volume is contained in a vessel at 22 0 C and 250 kPa
absolute pressure. If the gas constant is 215 Nm / kg K , find the density and mass of the
gas.
Given Data :
Volume of gas, V  0.3 m 3
Temperature, t  22 0 C
Pressure of gas, p  250 kPa absolute
Gas constant, R  215 Nm / kg K
To Find :
- Density of gas, 
- Mass of gas, m
Solution :
Absolute temperature, T  (273  22) K  295 K
From equation of state, p   RT
p
or,   R T

8
 250  1000 kg
or,  
215  295 m 3
or,   3.94 kg / m 3 Answer
Mass of gas, m  V
or,m  3.94  0.3 kg
or,m  1.18 kg Answer
Problem 1.11
Find the viscosity of a liquid having kinematic viscosity 5.5 stokes and specific 1.75.
Given Data :
Kinematic viscosity of liquid,   5.5 stokes  5.5  10 4 m 2 / s
Specific gravity of liquid, S  1.75
To Find :
- Viscosity of liquid, 
Solution :
Let   density of liquid
density of liquid
Now specific gravity of liquid  density of water

or, S 
1000
or,   1 .75  1000 kg / m 3  1750 kg / m 3

Kinematic viscosity,  

or,    
or,   5.5  10 4  1750 Ns / m 2
Ns
or,   0.9625 2  9.625 poise Answer
m
Problem 1.12
A vessel contains 0.65 m 3 of air at 52 0 C and 295 kPa absolute pressure. If the air is
compressed to 0.35 m 3 , find the pressure and temperature considering adiabatic
process. Take k  1.4 for air.
Given Data :
Initial volume of air, V1  0.65 m 3
Final volume of air, V2  0.35 m 3
Initial temperature of air, t1  52 0 C
Initial pressure of air, p1  295 kPa ( absolute)
To Find :
- Final pressure of air, p 2
- Final temperature of air, t 2
Solution :
Absolute initial temperature, T1  ( 273  52) K  325 K
For adiabatic process, p1 V1k  p 2 V2k

9
 k
V 
or, p 2  p1  1 
 V2 
1.4
 0.65 
or, p 2  295  
 0.35 
or, p 2  701.79 kPa absolute Answer
From equation of state, pV  R T
RT
or, p 
V
Again for adiabatic process, p V k  const.
RT k
or, V  const.
V
or, T V k 1  const. (Since R  const. )
k 1
Therefore, T1 V1  T2 V2k 1
k 1
V 
or, T2  T1  1 
 V2 
1.4 1
 0.65 
or, T2  325    416.31 K or , 143.310 C Answer
 0.35 
Problem 1.13
Find the bulk modulus of elasticity of a liquid which is compressed in a cylinder from a
volume of 0.013 m 3 at 0.81 MPa pressure to a volume of 0.012 m 3 at 1.51 MPa
pressure.
Given Data :
Initial volume of liquid = 0.013 m 3
Final volume of liquid = 0.012 m 3
Initial pressure, p1  0.81 MPa
Final pressure, p 2  1.51 MPa
To Find :
- Bulk modulus of elasticity of liquid, E
Solution :
Increase of pressure, dp  (1.51  0.81) MPa  0.70 MPa
Decrease in volume, dV  (0.013  0.012) m 3  0.001 m 3
 dV 0.0001
Therefore, 
V 0.013
Bulk modulus of elasticity is given by,
dp
E
 dV
V
0.70
E  91 MPa
or, 0.0001 Answer
0.013

10
Problem 1.14
Two parallel and horizontal flat plates of square shape are placed 13 mm apart. The space
between the two plates is filled with oil and the upper plate is pulled with a force of 100
N. Length of each plate is 620 mm. The velocity of upper plate is 2.6 m/s. If the specific
gravity of the oil is 0.94, find the dynamic viscosity of oil in poise and the kinematic
viscosity of oil in stokes.
Given Data :
Distance between the plates, y  13 mm
Force acting on upper plate, F  100 N
Length of each side of plate = 620 mm
Velocity of upper plate, u  2.6 m / s
Specific gravity of oil, S  0.94
To Find :
- Dynamic viscosity of oil, 
- Kinematic viscosity of oil, 
Solution :
Thickness of oil film, dy  y  0  13 mm
Change of velocity between two plates, du  u  0  2.6 m / s
Area of plate, A  0.62  0.62 m 2
From Newton’s law of viscosity, shear stress is given by
du
 
dy
F du
or, A   dy
100 2.6
or, 
0.62  0.62 0.013
Ns
or,   1.3
m2
or,   13 poise Answer
kg
Density of oil,   0.94  1000 3
m

Kinematic viscosity of oil,   
1.3
or,  
940
m2
or,   1.38  10 3
s
or,   13.8 stokes Answer
Problem 1.15
Find the specific gravity of oil having an absolute viscosity of 0.051 poise and kinematic
viscosity 0.30 stokes.
Given Data :
Absolute viscosity of oil,   0.051 poise

11
Kinematic viscosity of oil,   0.036 stokes
To Find :
- Specific gravity of oil, S
Solution :
Let   density of oil
Ns
Absolute viscosity of oil,   0.051 poise  0.0052
m2

Kinematic viscosity of oil,   
4 0.0051
or, 0.036  10  
or,   1416.67 kg / m 3
density of oil
Specific gravity of oil, S  density of water
1416.67
or, S 
1000
or, S  1.42 Answer
Problem 1.16
The velocity profile of a fluid over a flat plate is parabolic which is given by,
u  0.25 y 2  10 y with the vertex at B and origin at O. The velocity at point B is 1.22
m/s. Find the velocity gradient and shear stress at point B which is at a vertical distance
of 0.22 m from the origin. The viscosity of oil is equal to 0.84 Ns / m 2 .

Fig. P 1.16

Given Data :
The velocity profile, u  0.25 y 2  10 y
Velocity at point B, V B  1.22 m / s
The viscosity of oil,   0.84 Ns / m 2
To Find :

12
- Velocity gradient at 0.22 m
- Shear stress at 0.22 m
Solution :
Velocity profile, u  0.25 y 2  10 y
du
Therefore,  0.5 y  10
dy
du 1
At y  0.22 m , velocity gradient, dy  0.5  0.22  10  9.88 s Answer
du
At y  0.22 m , shear stress,    dy
or,   0.84  9.88 N / m 2
or,   8.30 N / m 2 Answer
Problem 1.17
A water tank of 0.5m  0.6m  0.75m is completely filled with water at 20 0 C . If the
water is heated to 55 0 C , find the amount of water that will spill over. The specific
weight of water at 20 0 C and 55 0 C are 9.80 kN / m 3 and 9.65 kN / m 3 respectively.
Given Data :
Size of tank 0.5m  0.6m  0.75m
Specific weight of water at 20 0 C ,  1  9.80 kN / m 3
Specific weight of water at 55 0 C ,  2  9.65 kN / m 3
To Find :
- Volume of water spilled over
Solution :
Initial volume of water in the tank, V1  0.5  0.6  0.75 m 3  0.225 m 3
Weight of water at 20 0 C , W1  9.8  0.225 kN  2.205 kN
2.205 3
Volume of water at 55 0 C , V2  m  0.228 m 3
9.65
Therefore, volume of water spilled over  (0.228  0.225) m 3  0.003 m 3  3 L
Answer
Problem 1.18
For a fluid flow over a horizontal flat plate the velocity distribution is given by
u  y  2 y 2 , where u is the velocity in m/s at a distance y meter above the plate. Find
the shear stress at a distance y  0.2 m above the plate. The absolute viscosity of the
fluid is 0.9 Ns / m 2 .
Given Data :
Velocity distribution, u  y  2 y 2
Absolute viscosity of fluid,   0.9 Ns / m 2
To Find :
- Shear stress at y  0.2 m
Solution :
Velocity distribution, u  y  2 y 2 u

13
 du
Therefore, dy  1  4 y
du 1
At y  0.2 m , dy  1  40 / 2  0.2 s
From Newton’s law of viscosity shear stress,
du
 
dy
N
or,   0.9  0.2
m2
N
or,   0.18 Answer
m2
Problem 1.19
A certain oil of density 720 kg / m 3 is flowing through a pipe. At a certain point shear
stress is 0.25 N / m 2 and the velocity gradient at that point is 0.22 per second. Find the
kinematic viscosity of the oil.
Given Data :
Density of oil,   720 kg / m 3
Shear stress,   0.25 N / m 2
du
Velocity gradient, dy  0.21sec
To Find :
- Kinematic viscosity of oil, 
Solution :
Let   Absolute viscosity of oil
du
Now from Newton’s law of viscosity,  
dy
or, 0.25    0.21
Ns
or,   1.19
m2

Kinematic viscosity,  

1.19 2
or,   m /s
720
or,   1.65  10 3 m 2 / s
or,   1.65  10 3  10 4 stoes
or,   16.5 stokes Answer
Problem 1.20
A flat plate is pulled with a velocity of 0.55 m/s relative to another fixed plate located at a
distance of 0.3 mm from it. If a force of 1.95 N / m 2 is required to maintain this speed,
find the absolute viscosity of fluid in between the two plates.

14
 Fig. P 1.20

Given Data :
Velocity of upper plate, u  0.55 m / s
Distance between two plates, dy  0.3 mm  0.0003 m
Force per unit area,   1.95 N / m 2
To Find :
- Absolute viscosity of fluid, 
Solution :
Assuming linear variation of velocity across the gap,
Change of velocity, du  u  0  u  0.55 m / s
Change of distance, dy  0.3 mm
From Newton’s law of viscosity, force per unit area,
du
 
dy
0.55
or, 1.95   
0.0003
Ns
or,   1.06  10
3
Answer
m2
Problem 1.21
A vertical cylinder of 160 mm diameter rotates concentrically inside another cylinder of
161 mm diameter. The length of both the cylinders is 240 mm . The space between the
two cylinders is filled with oil. If a torque of 15 Nm is applied to rotate the inner
cylinder at 110 rpm, find the viscosity of oil.
Given Data :
Diameter of inner cylinder, d1  160 mm
Diameter of outer cylinder, d 2  161 mm
Length of both cylinders, L  240 mm
Applied torque, T  15 Nm
Speed of inner cylinder, N  110 rpm
To Find :
- Viscosity of oil, 
Solution :
Assume linear variation of velocity across the gap.

15
 d 2  d1 161  160
Thickness of oil film, dy   mm  0.0005 m
2 2
 d1 N   0.16  110
Tangential velocity of inner cylinder, u    0.92 m / s
60 60
Change of velocity, du  u  0  0.92 m / s
Shear force acting on the area, A   d1 L    0.16  0.24 m 2  0.12 m 2
Let F=shear force
From Newton’s law of viscosity, shear stress is given by
du
 
dy
F du
or, A   dy
du
or, F  A  dy
0.92
or, F  0.12  
0.005
or, F    220.8
d1
Now torque, T F
2
0.16
or, 15    220.8 
2

Ns
or,   0.85 Answer
m2
Problem 1.22
The atmospheric pressure outside a droplet of water of diameter 1.5 mm is 101.3 kPa.
Find the pressure within the droplet if surface tension is given as 0.0728 N/m of water.

Fig. P 1.22

Given Data :
Atmospheric pressure = 101.3 kPa
Let p=pressure inside the droplet above outside pressure
 d2
Now pressure force = p 
4
Surface tension force acting around the circumference =    d
These two forces are equal and opposite under equilibrium condition.

16
  d2
Therefore, p    d
4
4
or, p 
d
4  0.072
or, p  N / m2
0.0015
or, p  192 N / m 2
Now pressure inside the droplet, p1  p  atmospheric pressure
or, p1  (0.192  101.3) kPa  101.49 kPa Answer
Problem 1.23
Find the force required to lift a thin wire of 50 mm in diameter from water surface. The
surface tension of water in contact with air is 0.0728 N/m and neglect the weight of the
wire.
Given Data :
Diameter of wire ring, d  50 mm
Surface tension of water,   0.0728 N / m
To Find :
- Force required to lift the ring, F
Solution :
The surface tension will act on the inside and outside of the ring.
Therefore required force, F  2  d
50
or, F  2  0.0728    N
1000
or, F  0.0229 N Answer
Problem 1.24
The pressure difference between inside of a soap bubble and the outer atmosphere is
21 N / m 2 . The diameter of soap bubble is 60 mm. Find the surface tension in the soap
film.

Fig. P 1.24

Given Data :
Difference of pressure, p  21 N / m 2

17
Diameter of soap bubble, d  60 mm
To Find :
- Surface tension in the soap film, 
Solution :
Let p  pressure inside the soap bubble
Now pressure difference, p  p  0  p
A soap bubble has two surfaces on which surface tension  acts.
d 2
So balancing forces, p  2  (  d )
4
8
or, p 
d
8
or, 20 
0.06
or,   0.15 N / m Answer
Problem 1.25
The pressure inside a droplet of water is 196 N / m 2 greater than the outer atmospheric
pressure. The surface tension of water in contact with air is 0.0728 N/m. Find the
diameter of droplet of water.

Fig. P 1.25

Given Data :
Difference of pressure, p  196 N / m 2
Surface tension,   0.0728 N / m
To Find :
- Diameter of droplet, d
Solution :
Let p  pressure inside the droplet
Now difference of pressure, p  p  0  p
 d2
Pressure force = p 
4
Surface tension force    d
Balancing the forces, we have

18
  d2
p    d
4
4
or, p 
d
4  0.0728
or, 196 
d
or, d  1.49  10 3 m  1.49 mm Answer
Problem 1.26
The space between two horizontal flat plates is filled with a Newtonian fluid. The lower
plate is fixed and the upper plate is moving. The upper plate attains a speed of 0.5m/s
when a force of 390 N is applied. Find the speed if a force of 1960 N is applied.
Given Data :
Force, F1  390 N
Speed, u1  0.5 m / s
Force, F2  1960 N
To Find :
- Speed of the upper plate, u 2
Solution :
Let area of plate = A
Change of velocity, du  u  o  u
Clearance between the plates, dy  y  0  0
du
From Newton’s law of viscosity, shear stress,    dy
F1 u
or,  1
A y
(i)
F2 u
For force F1 we have,  2
A A
(ii)
From equations (i) and (ii),
F1 u1

F2 u 2
F2
or, u 2   u1
F1
1960
or, u 2   0.5 m / s  2.72 m / s Answer
360
Problem 1.27
A small drop of water in contact with the air has a diameter of 0.50 mm. If the pressure
within the droplet is 0.52 kPa greater than the atmospheric pressure, find the value of
surface tension.
Given Data :

19
Difference of pressure, p  0.52 kPa
Diameter of droplet, d  0.50 mm
To Find :
- Surface tension force of water, 
Solution :
Let p  pressure inside the droplet
Now difference in pressure, p  p  0  p
 d2
Pressure force = p 
4
Surface tension force     d
Balancing the forces,
 d2
p    d
4
pd
or,  
4
0.52  1000  0.5
or,  
4  1000
or,   0.065 N / m Answer
Problem 1.28
Find the capillary rise of water in between two centric glass tubes of diameters 4 mm and
8 mm respectively. The surface tension of water in contact with air is 0.0728 N/m and the
contact angle is   0 0 .

Fig. P 1.28

Given Data :
Diameter of outer tube, d o  8 mm
Diameter of inner tube, d i  4 mm
Surface tension of water,   0.028 N / m
Contact angle,   0 0

20
To Find :
- Capillary rise of water, h
Solution :
Let   specific weight of water
Force due to surface tension     d i   d o  cos 
 2
Weight of water column in the concentric tubes   d o  d i  h 
2

4
Now surface tension force = weight of water column
 2
or,    d i  d o  cos   d o  d i  h 
2

4
4  cos   d i  d o 
or, h 

 d o2  d i2 
4  cos 
or, h 
  do  di 
4  0.0728  cos 0 0
or, h 
9.81  1000  (8  4)  10 3
or, h  7.42  10 3 m
or, h  7.42 mm Answer
Problem 1.29
The capillary rise of water in a glass tube   0 0  0.5 mm. Find the diameter of tube if
the surface tension of water in contact with air is 0.0728 N/m.

Fig. P 1.29

Given Data :
Capillary rise, h  0.5 mm
Surface tension,   0.0728 N / m
Angle of contact of the water surface,   0 0
To Find :
- Diameter of glass tube, d

21
Solution :
Let d  internal diameter of glass tube
h  height of capillary rise
  specific weight of water
Now surface tension force = weight of water column in the tube
 d2
or,  d  cos    h
4
4  cos 
or, d 
h
4  0.0728  cos 0 0
or, d 
0.5  10 3  9.81  10 3
or, d  0.0593 m  59.37 mm Answer
Problem 1.30
A small circular jet of water 5 mm in diameter issues from an opening. Find the pressure
difference between the inside and outside of the jet. The surface tension of water in
contact with air is 0.0728 N/m.

Fig. P 1.30

Given Data :
Diameter of jet, d  5 mm
Surface tension of water,   0.0728 N / m
To Find :
- Pressure difference between inside and outside of jet, p
Solution :
Let L  length of jet considered
Considering the equilibrium of the half of the jet,
Force due to pressure  p  projected area of semi jet

22
  p Ld
Force due to surface tension    2 L
Equating the forces, p Ld  2L
2
or, p 
d
2  0.0728
or, p 
5  10 3
or, p  29.12 N Answer
Problem 1.31
A vessel contains 4.5 m 3 of water at a temperature of 10 0 C . If it is heated to 70 0 C
and at atmospheric pressure, find the percentage change in its volume. The specific
weight of water at 10 0 C and 70 0 C is 9.81 kN / m 3 and 9.6 kN / m 3 respectively.
Given Data :
Initial volume of water, V1  4.5 m 3
Specific weight of water at 10 0 C ,  1  9.81 kN / m 3
Specific weight of water at 70 0 C ,  2  9.60 kN / m 3
To Find :
- Percentage change in volume of water
Solution :
Let V2  volume of water at 70 0 C
Since the weight of water is constant, so we can write
 1 V1   2 V2
or, 9.81  4.5  9.6  V2
or, V2  4.60 m 3
4.6  4.5
Therefore, percentage change in volume   100  2.22% Answer
4.5
Problem 1.32
The capillary depression of mercury in the 4 mm diameter capillary glass tube is 2.9 mm.
Find the angle of contact,  . The surface tension of mercury in contact with air is
0.52 N / m .

23
 Fig. P 1.32

Given Data :
Diameter of glass tube, d  4 mm
Capillary depression, h  2.9 mm
Surface tension,   0.52 N / m
To Find :
- Angle of contact, 
Solution :
Let   specific weight of mercury
Now surface tension force = weight of mercury column in the tube
 d2
or,  d  cos 1   h
4
h d
or, cos 1 
4
0.0029  9.81  13.6  1000  0.004
or, cos 1 
4  0.52
or,  1  41.93 0
Therefore,   180 0  1  138.07 0 Answer
Problem 1.33
Calculate the distance h for mercury in the capillary glass tube of 3.5 mm diameter.
The angle of contact is equal to 130 0 . The surface tension of mercury in contact with air
is 0.51 N / m .

24
 Fig. P 1.33

Given Data :
Diameter of glass tube, d  3.5 mm
Surface tension,   0.5` N / m
Angle of contact of the mercury surface,   130 0
To Find :
- Distance, h
Solution :
Let   specific weight of mercury
Now, surface tension force = weight of mercury column in the tube
 d2
or,  d  cos    h
4
4  cos 
or, h
d
4  0.51  cos 130 0
or, h 
9.81  13.6  1000  0.0035
or, h  0.0028 m  2.81 mm Answer
The negative sign indicates the capillary depression.
Problem 1.34
A flat plate of dimension 1.1m  1.5m is pulled with a velocity of 0.5m/s relative to
another plate located at a distance of 0.2 mm from it. The space between the two plates is
filled with oil having viscosity of 1.1 poise. Find the force and power required to
maintain this speed.
Given Data :
Area of plate, A  1.1  1.5 m 2
Velocity of upper plate, u  0.5 m / s
Distance between the two plates, y  0.2 mm
Viscosity of oil,   1.1poise  0.22 Ns / m 2
To Find :
- Force required for maintain the speed, F
- Power required, P

25
Solution :
Assume linear variation of velocity across the gap.
Change of velocity, du  u  0  0.5 m / s
Change of distance, dy  y  0  0.2 mm
From Newton’s law of viscosity, shear stress is given by,
du
 
dy
0.5
or,   0.11 
0.2  10 3
or,   275 N / m 2
Now shear force, F    area
or, F  275  1.1  1.5
or, F  453.75 N Answer
Required power, P  F  u
Nm
or, P  453.75  0.5
s
or, P  226.88W Answer
Problem 1.35
A cylindrical shaft of 80 mm diameter rotates at 500 rpm inside a bearing of 80.3 mm
diameter. Both the shaft and the bearing are 0.45 m long. The annular space is filled with
oil of dynamic viscosity 0.12 N s / m 2 . Find the rate at which heat is generated at the
bearing.
Given Data :
Diameter of shaft, d 1  80 mm
Diameter of bearing, d 2  80.3 mm
Length of shaft and bearing, L  0.45 m
Speed of the shaft, N  500 rpm
Dynamic viscosity of oil,   0.12 N s / m 2
To Find :
- Rate of heat generation
Solution :
Consider linear variation of velocity across the gap.
d  d1
Radial distance between the shaft and bearing, dy  2
2
80.3  80
 mm
2
 0.15 mm
500
Peripheral velocity of shaft, u   d1 N    0.08  m / s  2.09 m / s
60
Since the outer bearing is stationary, the change of velocity, du  u  0  2.09 m / s
Let F=shear force
Area, A   d1 L    0.08  0.45 m 2  0.113 m 2

26
From Newton’s law of viscosity shear stress is given by,
du
 
dy
F du
or, A   dy
F 2.09
or,  0.12 
0.113 0.15  10 3
or, F  188.94 N
Nm
Now rate of energy loss  F u  188.94  2.09
s
Nm
 394.88
s
J
Now, rate of energy loss = rate of heat generation  394.88 Answer
s
Problem 1.36
A shaft 70 mm diameter rotates concentrically inside a fixed cylinder 72 mm diameter.
The length of the cylinder is 0.6 m . Find the rpm of the shaft if the space between the
cylinder and shaft is filled with lubricating oil of viscosity 2.5 poise and a torque of 1.2
Nm is applied. Find also the power required to rotate the shaft.

Fig. P 1.36

Given Data :
Diameter of shaft, d1  70 mm
Diameter of cylinder, d 2  72 mm
Ns
Viscosity of oil,   2.5 P  0.25
m2
Applied torque, T  1.2 Nm
Length of cylinder, L  0.6 m
To Find :
-Rpm of the shaft, N
- Power required to rotate the shaft, P
Solution :

27
Assume linear variation of velocity across the gap.
d 2  d1 72  70
Radial distance between the shaft and cylinder, h  dy   mm  1 mm
2 2
70
Shear force acting on the area, A   d1 L     0.6 m 2  0.13 m 2
1000
Let F  shear force
V  peripheral velocity of the shaft
  angular velocity
From Newton’s law of viscosity,
du
 
dy
V
or,   
h
F V
or, 
A h
V
or, F  A 
h
V  1000
or, F  0.13  0.25 
1
or, F  32.5V
(i)
d1
Now torque, T F
2
70
or, 1.2  32.5V 
2  1000
or, V  1.05 m / s
d1
Again peripheral velocity, V  
2
70 2 N
or, 1.05  
2  1000 60
or, N  286.50 rpm Answer
Required power, P T
2 N
or, P  T 
60
286.5 Nm
or, P  1.2  2  
60 s
Nm
or, P  36  36W Answer
s
Problem 1.37
A small circular jet of mercury 1 mm in diameter issues from an opening. Find the
pressure difference between the inside and outside of the jet. The surface tension of
mercury in contact with air is 0.514 N / m .

28
 Fig. P 1.37

Given Data :
Diameter of jet, d  1 mm
Surface tension of mercury,   0.514 N / m
To Find :
- Pressure difference between inside and outside of jet, p
Solution :
Let L  length of jet considered
Considering the equilibrium of the semi-jet,
force due to pressure  p  projected area of semi-jet
 p Ld
Force due to surface tension    2 L
Equating the forces, p  L  d    2L
2
or, p 
d
2  0.514
or, p 
1  10 3
or, p  1028 N / m 2 Answer
Problem 1.38
Two large horizontal and fixed parallel surfaces are 30 mm apart. A thin flat plate of
0.5 m 2 area is pulled with a velocity of 0.65 m / s parallel to the plane surfaces. The
plate is situated at the middle of the two planes surfaces. The space between the two
plane surfaces is filled with oil of viscosity 0.35 Ns / m 2 . Find the force required to
maintain this motion.

29
 Fig. P 1.38

Given Data :
Distance between two surfaces  30 mm
Surface area of this plate, A  0.5 m 2
Velocity of plate, u  0.65 m / s
Viscosity of oil,   0.35 N s / m 2
To Find :
- Required force, F
Solution :
Neglect the thickness of the plate and assume a linear variation of velocity from the fixed
surface to the plate.
Since the thin plate is in the middle of the two plane surfaces, the shear force acting on
the upper surface will be equal to the shear force on the lower surface.
Change of velocity from fixed surface to the plate, du  u  0  0.65 m / s
30
Distance between fixed surface to the moving plate, dy   0  15 mm
2
Let F1  force acting on one surface of the plate
From Newton’s law of viscosity, shear stress acting on one surface
du
 
dy
F1 du
or, 
A dy
F1 0.65
or,  0.35 
0 .5 0.015
or, F1  7.58 N
Therefore total force, F  2 F1  15.16 N Answer
Problem 1.39
Two large horizontal and fixed parallel surfaces are 30 mm apart. A thin flat plate of
0.6 m 2 is pulled with a velocity of 0.67 m / s parallel to plane surfaces. The plate is

30
situated at a distance of 20 mm from the top surface. The space between the two plane
surfaces is filled with oil of viscosity 0.6 Ns / m 2 . Find the force required to maintain
this motion.

Fig. P 1.39

Given Data :
Distance between two surfaces  30 mm
Surface area of thin plate, A  0.6 m 2
Velocity of plate, u  0.67 m / s
Viscosity of oil,   0.6 Ns / m 2
To Find :
- Required force, F
Solution :
Neglect the thickness of the plate and assume a linear variation of velocity from the fixed
surface to the plate.
Let F1  shear force acting on upper surface of plate
F2  shear force acting on lower surface of plate
From Newton’s law of viscosity shear stress acting on upper surface,
du
1  
dy
F1 du
or, 
A dy
F1 0.67
or,  0 .6 
0 .6 0.02
or, F1  12.06 N
Shear stress acting on lower surface,
du
2  
dy
F2 du
or, 
A dy

31
 F2 0.67
or,  0 .6 
0 .6 0.01
or, F2  24.12 N
Therefore, total force, F  F1  F2
or, F  12.06  24.12  N
or, F  36.18 N Answer
Problem 1.40
In a journal bearing the clearance between the shaft and bearing is 1.4 mm . The
diameter of the shaft is 110 mm and it rotates at 100 rpm . The viscosity of oil in the
bearing is 0.12 Ns / m 2 . Find the intensity of shear stress acting on the surface of shaft.
Given Data :
Thickness of oil layer, y  1.4 mm
Diameter of shaft, D  110 mm
Speed of shaft, N  110 rpm
Viscosity of oil,   0.12 N s / m 2
To Find :
- Shear stress, 
Solution :
Assume linear variation of velocity across the gap.
Distance between shaft and bearing, dy  y  0  1.4 mm
 DN
Tangential speed of shaft, u 
60
  0.11  100
or, u  m/s
60
or, u  0.58 m / s
Change of velocity, du  u  0  0.58 m / s
Now from Newton’s law of viscosity shear stress is given by
du
 
dy
0.58
or,   0.12 
1.4  10 3
or,   49.71 N / m 2 Answer
Problem 1.41
A cylindrical shaft of 95 mm rotates at 650 rpm inside a cylinder of 95.2 mm
diameter. Both the shaft and the cylinder are 0.5 m long. If the torque required to rotate
the shaft is 1 Nm , find the viscosity of the oil occupying the annular space.

32
 Fig. P 1.41

Given Data :
Diameter of shaft, d 1  95 mm
Diameter of cylinder, d 2  95.2 mm
Length of shaft and cylinder, L  0.5 m
Torque required to rotate the shaft, T  1 Nm
Speed of shaft, N  650 rpm
To Find :
- Viscosity of oil, 
Solution :
Consider linear variation of velocity across the gap.
Radial distance between the shaft and the cylinder,
d  d1 95.2  95
dy  2  mm  0.1mm
2 2
N 650
Peripheral velocity of shaft, u   d1    0.095  m / s  3.23 m / s
60 60
95
Area, A   d1 L     0.5  0.15 m 2
1000
Since the outer cylinder is stationary, the velocity gradient
du u  0 3.23 1 1
  3
 323
dy dy 0.1  10 s s
Let F  shear force
From Newton’s law of viscosity shear stress,
du
 
dy
F du
or, A   dy
du
or, F   A dy

33
 d1
Again, torque T  F 
2
du d 1
or, 1   A  
dy 2
95
or, 1    0.15  323 
2  1000
Ns
or,   0.43
m2
or,   4.3 P Answer
Problem 1.42
A rectangular flat plate weighing 300 N and having dimensions 850mm  800mm at
the bases slides down a 30 0 inclined plane. There is a film of oil having a thickness of
1.2 mm . If the velocity of the plate is 0.35 m / s , find the dynamic viscosity of oil.

Fig. P 1.42

Given Data :
Weight of plate, W  300 N
Area of plate, A  0.85  0.8 m 2  0.68 m 2
Angle of inclination,   30 0
Velocity of plate, u  0.35 m / s
Thickness of oil film, y  1.2 mm
To Find :
- Dynamic viscosity of oil, 
Solution :
Assume linear variation of velocity across the gap.
Change of velocity between the plates, du  u  o  0.35 m / s
Distance between two plates, dy  y  0  1.2 mm
Component of weight along the incline, F  W cos 60 0
or, F  100  0.5  50 N
From Newton’s law of viscosity shear stress is given by,
du
 
dy

34
 F du
or, A   dy
50 0.35
or,  
0.68 1.2  10 3
Ns
or,   0.252 2 Answer
m
Problem 1.43
A rectangular block weighing 2 kN and having dimensions 200mm  250mm at the
base slides down a 25 0 inclined plane. There is a film of oil on the inclined surface
having a thickness of 0.02 mm and viscosity 0.65 poise. Find the terminal velocity of
the block along the incline.

Fig. P 1.43

Given Data :
Weight of the block, W  2 kN
Area of base of the block, A  200mm  250mm
Ns
Viscosity of oil,   0.65 poise  0.065 2
m
Thickness of oil film, h  0.02 mm
To Find :
- Terminal velocity of block along the plane, V
Solution :
Component of weight along the incline, F  W sin 25 0
F
Let   shear stress working on the black =
A
Change of velocity, du  V  0  V
Gap between two plates, dy  h  0  0.02 mm
du V
From Newton’s law of viscosity,    dy   h
F V
or, 
A h

35
 W sin 25 0 V
or, 
A h
2  1000 sin 25 0 V
or, 6
 0.065 
200  250  10 0.02  10 3
or, V  5.20 m / s Answer
Problem 1.44
A body weighing 2.5 kN with a flat surface area of 0.25 m 2 slide down a oil
lubricated inclined plane making an angle of 30 0 with the horizontal. The speed of the
body is 1.2 m / s . If the dynamic viscosity of oil is 0.064 Ns / m 2 , find the oil film
thickness.
Given Data :
Weight of the body, W  2.5 kN
Contact surface area, A  0.25 m 2
Inclination angle,   30 0
Speed of the body, u  1.2 m / s
Viscosity of oil,   0.064 Ns / m 2
To Find :
- Thickness of oil film, y
Solution :
Assume linear variation of velocity across the gap.
Change of velocity, du  u  0  1.2 m / s
Thickness of oil film, dy  y  0  y
Component of weight along the incline, F  W sin 30 0
1
or, F  2.5  kN  1.25 kN
2
From Newton’s law of viscosity shear stress is given by,
du
 
dy
F du
or, 
A dy
F u
or,  
A y
1.25  1000 1.2
or,  0.064 
0.25 y
5
or, y  1.536  10 m
 1.536  10 2 mm Answer
Problem 1.45
Find the amount of torque required to rotate the upper part at 160 rpm . The viscosity of
oil in the gap is 7 P . The radius of circular part is 60 mm .

36
 Fig. P 1.45

Given Data :
Radius of circular part, R  60 mm  0.06 m
Speed of upper part, N  160 rpm
Thickness of oil film, h  0.02 mm
Viscosity of oil,   7 P  0.7 Ns / m 2
To Find :
- Torque (T) required to rotate the upper part
Solution :
Consider a circular strip at a radius r of thickness, dr .
2N
Angular velocity of upper part,    16.76 rad / s
60
Tangential velocity at radius r , V  r
From Newton’s law of viscosity, assuming linear variation of velocity, shear stress is
du V r
given by,    
dy h h
r 
Now elemental torque, dT   dA r    2r dr r   2r 3 dr
h h
R r 
Total torque, T  
0
dT  
0 H
2 r 3 dr
R
 r4 
  2   
h  4 0

37
  R 4
 
h 2
0.7  16.76  1000 (0.06) 4
  
0.02 2
 1.19 Nm Answer

38
 EXERCISE

1. Explain the following fluid properties.

(a) Density
(b) Specific weight
© Specific volume
(d) Compressibility
2. What is the difference between dynamic viscosity and kinematic viscosity? Give three
examples of non-Newtonian fluid.
3. Deduce and explain Newton’s law of viscosity.
4. Differentiate between real fluids and ideal fluids. Find an expression of compressibility
of fluid.
5. Explain surface tension. Show that the relationship between surface tension and
pressure inside a droplet of liquid in excess of outside pressure is given by,
4
p where the notations have their usual meanings
d
6. What is capillarity? Deduce an expression for capillarity rise or fall of liquids.
7. The pressure inside a soap bubble of 50 mm diameter is 2.45 N / m 2 above the
atmospheric pressure. //calculate the surface tension in the bubble.
(Ans. 0.0153 N/m)
8. Find the diameter of a glass tube if the capillarity rise in the tube is 2 mm. The surface
tension of water in contact with air is 0.0736 N / m .
(Ans. 15 mm)
9. The absolute viscosity of a certain oil used for lubrication between a shaft and sleeve
is 0.6 Ns / m 2 . The speed of shaft is 190 rpm and its diameter is 400 mm. Both the
shaft and sleeve is 90 mm long. If the thickness of oil film is 1.5 mm , find the
power lost in the bearing.

( Ans. 716.34 W )
10. Calculate the capillary effect in a glass tube of 4 mm diameter, when immersed in
mercury. The surface tension in contact with air is is 0.52 N / m . The angle of contact
is 130 0 .
(Ans. -2.46mm)

39