Fluid Mechanics
(Theory + Objective + Conventional)
[For GATE, UPSC-ESE, State Public Service Commission,
Recruitment tests by Public Sector Undertakings]
By
Mr. Praveen Kulkarni
KULKARNI’S ACADEMY PUBLICATION
16-11-1/1/2, 5th floor, Above Bantia Furnitures, Near Super Bazar Bus Stop,
Malakpet, Hyderabad, Telangana 500036
Phone: +91-9000770927, E-mail: kamehyderabad@gmail.com
Website: www.kulkarniacademy.com
KULKARNI’S ACADEMY PUBLICATION
16-11-1/1/2, 5th floor, Above Bantia Furnitures, Near Super Bazar Bus Stop,
Malakpet, Hyderabad, Telangana 500036
Phone: +91-9000770927, 7893190907
E-mail: kamehyderabad@gmail.com
Website: www.kulkarniacademy.com
ALL RIGHTS STRICTLY RESERVED
Copyright 2020, by KULKARNI ACADEMY OF MECHANICAL ENGINEERING. No part of this book may be reproduced, or
distributed in any form or by any means, electronic, mechanical, photocopying or otherwise or stored in a database or retrieval
system without the prior information of KAME Publication, Hyderabad. Violates are liable to be legally prosecuted.
First Edition: 2020
Price:
₹350/-
Typesetting by: Kulkarni’s Academy Publication, Hyderabad
ACKNOWLEDGEMENTS
I owe special thanks to two individuals who have influenced my thinking during the
preparation of this book- Pushpendra Jangid and Hrishikesh Kulkarni
I appreciate the help of Pushpendra Jangid, the backbone of this book who prepared
solutions in a simplified manner with neat explanations. I remain indebted to Hrishikesh
Kulkarni for editing the material.
Praveen Kulkarni Director,
Kulkarni’s Academy - Hyderabad
PREFACE TO THE FIRST EDITION
This book is designed for mechanical engineering students who are interested in
appearing for GATE, ESE and other PSU’s. The main emphasis is placed on the precise
and logical presentation of basic concepts and principles, which are essential for
better understanding of the subject. Since, this is an introductory book, care has
been taken to present questions in a gradual manner to instil confidence in the
minds of the students. Due effort has been made to keep fundamentals and
principles at a very simple level. Looking forward for constructive criticism and
suggestions, if any.
Praveen Kulkarni
ABOUT THE DIRECTOR
Praveen Kulkarni did his B.E. and subsequently M.E. (Production Engineering) from
Osmania University, Hyderabad. He has qualified GATE, IES (ESE) and other state
service examinations. He secured all India 39th and 14th rank in ESE 1999 and 2001
respectively. After joining Engineering Services in IOFS (Indian Ordnance Factory
Services) and MES (Military Engineer Services), quit due to his passion for teaching.
Mr. Praveen Kulkarni is regarded as one of the best teachers in India due to his
simplification of subjects. A great Motivator, student friendly and a humble person.
He is the recipient of National Merit scholarship (NMS) and Telugu Vignana
Paritoshakam (TVP). He handles Thermodynamics, Fluid Mechanics, RAC, Heat
Transfer, IC Engines, Power Plant, Turbo machinery, Strength of Materials and
Machine Design with equal ease.
To
My beloved students
Past and Present
Table of Contents
Sr. Chapter
Pages
1.
Fluid Properties
1 to 16
2.
Pressure Measurement (Manometry)
17 to 29
3.
Buoyancy and Flotation
30 to 41
4.
Hydrostatic Forces
42 to 52
5.
Fluid Kinematics
53 to 77
6.
Fluid Dynamics
78 to 96
7.
Laminar Flow
8.
Turbulent Flow
114 to 122
9.
Flow Through Pipes
123 to 137
10. Boundary Layer Theory
138 to 157
11. Vortex Motion
158 to 165
97 to 113
1.1 Fluid
Fluid is a substance which is capable of flowing
or deforming under the action of shear force.
[However small the shear force may be] This
definition of a fluid is also known as a classical
definition of a fluid.
As long as there is a shear force fluid flows or
deform continuously. Ex: liquids, gases, vapour
etc.
Difference between solids and fluids :
In case of solids under the action of shear force
there is a deformation and this deformation does
not change with time. Therefore deformation
(d) is important is solids when this shear force
is removed, solids will try to come back to the
original position.
In case of fluids the deformation is
continuous as long as there is a shear force and
this deformation changes with time, therefore in
fluids rate of deformation (d/dt) is important
than deformation (d). After the removal of shear
force fluid will never try come back to its original
position.
“For a static fluid, the shear force is zero.”
1.2 Fluid Properties
Any measurable characteristic is a property.
1.2.1 Density/Mass density ():
It is a defined as ratio of mass of fluid to its
volume. It actually represent the quantity of
matter present in a given volume. It’s unit is
kg/m3 and it dimensional formula is [ML3].
The density of water for all calculation purpose
is taken as 1000 kg/m3 (at 4C).
Density depends on temperature and pressure.
1.2.2 Specific weight /Weight density(w)
It is defined as the ratio of weight of the fluid to
it’s volume, its unit is N/m3 and it’s dimensional
formula [ML2T2].
Weight of the fluid
w
Vol.
mg
w
V
m
w = g
V
w = F(P, T, Location)
Note :
1. Specific weight of water
wH2o g 1000 9.81 = 9810 N/m3
2. Density is an absolute quantify where as specific
weight is not an absolute quantity because it
varies from location to location.
Fluid Mechanics
1.2.3 Specific gravity (s.g.)
It is defined as the ratio of density of fluid to the
density of standard fluid.
In case of liquid the standard fluid is water and in
case of gases the standard fluid either hydrogen
or air at a given temperature and pressure. It is
unitless and dimensionless.
[M0L0T0]
s.g of water is 1. If s.g. of liquid is less than 1 it
is lighter than water, if s.g. of liquid is greater
than 1 it is heavier than water.
2
Kulkarni’s Academy
Liquids are generally treated as incompressible
and gases are treated as compressible.
As fluid is treated as incompressible fluid if there
is no variation of density with respect to
pressure.
d
0
i.e.,
dp
(i) Isothermal compressibility of ideal gas: PV = mRT
P = RT
{ T = constant
dP
RT
d
Note:
Though terms relative density and sp. gravity are
used interchangeably, there a difference between
these two. “all specific gravities are relative
density but all relative density need not be
specific gravity”.
1.2.4 Compressibility():
It is the measure of change of volume or change
of density with respect to pressure on a given
mass of fluid. Mathematically is defined as
reciprocal of bulk modulus.
i.e.,
1
K
dP
dV
V
dV
V
dP
We know that
Mass = density volume
V = mass = constant
dV + Vd = 0
dV d
V
1 d
dP
If d = 0; = 0 (incompressible fluid)
P RT
KT = P
Isothermal bulk modulus is equal to pressure.
1
P
unit of compressibility:
m2
or pascal 1
N
(ii) Adiabatic bulk modulus of an ideal gas:
{ K = bulk modulus
K
dP
RT
d
K
PV = C1
m
P C1
P
P C
m C1 1 C
m
P = C
dP
C 1
d
K
dP
d
K C 1
K = C r
K = P
1
P
Kulkarni’s Academy
3
Note:
Properties of Fluid
dt = time
As > 1 adiabatic bulk modulus is greater than
isothermal bulk modulus.
velocity gradient =
tan d =
du
dy
dudt
dy
if d is small tan d = d
dudt
,
dθ
dy
Bulk modulus is not constant and it increases
with increase in pressure because at higher
pressure the fluid offer’s more resistance for
further compression.
k1
dp
dv
v
k2
dp
dv
v
k2 > k1
1.3 Viscosity
Need to define viscosity:
Though the densities of water and oil almost
same, their flow behavior is not same and hence
a property is required to define to flow behavior
and this property is known as viscosity.
Definition: Internal resistance offered by one
layer of fluid to the adjacent layer is known as
viscosity.
dθ du
dt dy
F
A
d
F
dt
d
dt
{A constant
d
dt
d
dt
dθ
dθ
is large
is less (small)
dt
dt
Flow is easy
Flow is not easy
is less, resistance
is more, resistance is
is less.
more.
represents the internal resistance offered by
one layer of fluid to the adjacent layer and
hence is known as coefficient of viscosity
or absolute viscosity or dynamic viscosity or
simply viscosity.
d d du
dy dt dy
du
dy
d
Here
is known as rate of angular deformation
dt
du
or rate of shear strain and
is known as
dy
velocity gradient.
Fluid Mechanics
Variation of viscosity with temperature:
In case of liquids the intermolecular distance is
small and hence cohesive forces are large with
increase in temperature cohesive forces decrease
and the resistance of the flow is also decreases,
therefore “viscosity of a liquid decreases with
increase in temperature”.
In case of gases intermolecular distance is
large and hence cohesive forces are negligible
with increase in temperature molecular
disturbance increases and hence resistance to the
flow also increases. “Therefore viscosity of gas
increase with increase in temperature”.
Unit of viscosity:
du
dy
N
m 1
. .
2
m
s m
N s
= pascal. sec. (SI unit)
m2
In MKS system:
m
kg . 2 .s
N s
kg
s
2
2
m
m
ms
Dimensional formula of = [M1L1T1]
In cgs system:
kg
gm
1
1 poise
ms
cm sec
kg
103gm
10gm
2
10poise
m-s 10 cm-sec cm.sec
N S
1 2 10poise
m
N S
1poise 0.1 2 0.1 pascal-sec
m
1
4
Kulkarni’s Academy
1.4 Classification of fluid
1.4.1 Newtonian Fluid:
Fluid which obey Newton’s law of viscosity are
known as Newtonian fluid. According to
Newtons low of viscosity shear stress is directly
proportional to rate of shear strain that is
dθ
du
du
dt
dy
dy
This equation is Valid for Newtonian fluid.
oil > water > air
We know that
dθ
dt
dθ
{ = constant
dt
y=mx
m = = slope = constant
If slope
Examples of Newtonian fluid:
Air, Water, petrol, diesel, kerosene, oil,
mercury, Gasoline etc.
Kulkarni’s Academy
Note:
For a Newtonian fluid viscosity does not change
with rate of deformation.
5
Properties of Fluid
“as the app is decreasing with rate of
deformation, these fluids is also known as shear
thinning fluid.
1.4.2 Non-Newtonian Fluids:
Fluids which do not obey Newton’ law of
viscosity are known as non – Newtonian fluid.
The general relationship between shear stress ()
du
and velocity gradient is
dy
n
du
A B
dy
Case-1: B = 0; n > 1 Di-latent fluids (non –
colloidal)
A fluid is said to be dilatant fluid for which the
apparent (similar) viscosity increases with rate of
deformation.
n
du
A 0
dy
n1
du du
A .
dy dy
app
du
app
dy
Ex: Rice starch, sugar in water.
“As the app is increasing with deformation, these
fluids is also known as shear thickening fluid.
Case2: B =0; n < 1 pseudo plastic fluids
(colloidal)
For a pseudoplastic fluid apparent viscosity
decreases with rate of deformation.
Ex: Milk, blood, colloidal solution.
Case-3: Bingham plastic fluid.
B 0; n = 1
Ex: Toothpaste
“Such fluids are comes under rheology”.
Note :
In case of Bingham plastic fluid certain min.
shear stress is required for causing the flow of
fluid below this shear stress there is no flow
therefore it acts like solid, after that it behaves
like a fluid. Such substances which behaves both
fluids and solids are known as Rheological
substances and study of these substances is
known as rheology.
1.4.3 Ideal Fluid:
A fluid which is non – viscous and
incompressible is known as an ideal fluid.
Though there is no ideal fluid it is introduced for
bringing simplicity in the analysis.
Fluid Mechanics
6
Remember:
H 2O at 20oC 1 CP Centipoise
1 102 poise
102 101
kg
103 kg/m-s .
m.s
Hg at 200C = 1.55 cp
Water is 50 – 55 times more viscous than air.
1.5 Equation for a linear velocity profile:
The velocity profile can be approximated as a
linear velocity profile if the gap between plates is
very small (narrow passages).
Kulkarni’s Academy
Unit of kinematic viscosity:
kg-m
2
sec m M o L2T 1
kg
s
ms
In CGS system:
In cgs system the unit of is
cm 2
cm 2
and
is
sec
sec
equal to stoke.
2
1cm2
4 m
1stoke
10
sec
sec
Physical significance of :
Kinematic viscosity represent the ability of fluid
to resist momentum therefore it is a measure of
momentum diffusivity.
1.7 Surface Tension ():
tan θ
Vdt
y
From triangle tan θ
dudt
dy
Vdt dudt
du V
y
dy
dy y
du
V
dy
y
F
A
AV
F
y
1.6 Kinematic viscosity ():
appears frequently
and for convenience this term is known as
kinematic viscosity.
In fluid mechanics the term
Consider the molecule a which is below the
surface of liquid this molecule is surrounded by
various corresponding molecule and hence under
the influence of various cohesive forces it will be
in equilibrium. Now consider molecule B which
is on the surface of liquid, this molecule is under
the influence of net downward force because of
this there seems to be a layer form which can
resist small tensile this phenomenon is known as
surface tension, it is a line force that is it acts
normal to the line drawn on the surface and it lies
in the plane of surface. As surface tension is
basically due to unbalanced cohesive force and
with increase in temperature cohesive force is
decrease therefore “surface tension decreases
with increases in temperature, and at critical
point surface tension is zero”.
Kulkarni’s Academy
7
Note:
This force is very small force and hence it is
neglected in further fluid mechanics analysis.
The surface tension for water air interface at 20oC
is 0.0706 N/m.
While washing cloths warm water is used
because warm water reduces surface tension and
help in cleaning.
Liquid drops assume spherical shape due to
surface tension.
Dimension Formula:
F N 1MLT 2
M 1L0T 2
L m
L
Pressure in liquid drop
atmospheric pressure:
in
excess
Properties of Fluid
Note:
1. In case of soap bubble there are two surfaces and
hence.
8
P
d
2
2. In case of liquid jet P
d
3. The pressure force tries to separate the droplet
whereas surface tension force tries the contract
the droplet. i.e., surface tension force tries to
minimise the surface area and hence droplets take
spherical shape because sphere has minimum
surface area for a given volume.
1.8 Capillarity:
of
Capillarity is the effect of surface tension and it
is not a property.
(a) Wetting liquid
Adhesion is large
(b) Non – wetting liquid
cohesion is large
Fs
L
{P
FP
; FP = PA
A
Fs L
For equilibrium
Fs = FP
L = PA
d P d 2
4
4
P
d
The rise or fall of a liquid when a small diameter
tube is introduced in it is known as capillarity.
The capillary rise is due to adhesion. Ex: Water,
and the capillary fall is due to cohesion Ex:
mercury, therefore capillarity is due to both
adhesion and cohesion.
Fluid Mechanics
Expression far capillary rise/Fall in a glass
tube:
8
Kulkarni’s Academy
Expression for capillary rise between two
parallel plates:
Where t distance between plates
[Weight of liquid = vertical component of Fs]
weight
w. d 2h F4 cos w
4
volume
Fs = 2b
Fs = L
w hbt = 2b cos
Fs = d
h
Weight = w x volume
w. d 2h cos
4
or
h
4 cos
wd
h
4 cos
gd
2 cos
wt
Note:
w g
Expression for capillary rise in the annulus of
two concentric tubes:
When a liquid surface support another liquid of
density b then rise in capillary is given as
h
4 cos
( b ) gd
Work done is stretching the surface:
Wt. of the fluid = vertical Component of surface
tension force (Fs)
w.
2
d0 d12 h d0 di cos
4
4 cos
2 cos
h
w(d 0 di ) w(r0 ri )
Work = Force distance
=Lx
Work = (increase in surface area)
Kulkarni’s Academy
Note :
The angle of contact between water and glass is
22o.
The angle of contact between pure water and
clean glass tube – 0o
The angle of contact between mercury and
glass is 130o.
If height of capillary tube is insufficient for
the possible rise the liquid will rise up to the
top and stops because for further rise as there
are no glass molecules it stops at the top.
If the top of the capillary tube is closed then
the capillary rise will decrease because the air
trapped at the top exert pressure in the
downward direction.
9
Properties of Fluid
P
1.1
1.2
Practice Questions
A fluid is one which can be defined as a
substance that
(A) Has the same shear stress at all points
(B) Can deform indefinitely under the action
of the smallest shear force
(C) Has the same shear stress in all
directions
(D) Is practically incompressible
The equation of a state for a is liquid
P = (3500 1/2 +2500) N/m 2 .
The Bulk modulus of liquid at a pressure of
100kPa is
1.9 Vapour Pressure:
1.3
(A) 3500 N/m 2
(B) 2500 N/m 2
(C) 48750 N/m 2
(D) 6250 N/m 2
A liquid compressed in a cylinder has a
volume of 0.04 m 3 at 50 kg/cm 2 and a
volume of 0.039 m3 at 150 kg/cm 2 . The bulk
modulus of liquid is
Let us consider a closed container with liquid
partially filled in it the surface molecules due to
additional energy overcome cohesive force of
fluid below the surface this process occurs until
the space above the liquid is saturated. Under
equilibrium conditions the number of molecules
leaving the surfaces is equal to number of
molecules joining the surface under these
conditions the pressure exerted by vapour on the
surface of liquid is known as vapour pressure.
Vapour pressure increases with increase in
temperature because at higher temperatures the
molecular activity is high.
Note:
Highly volatile liquid (Ex: petrol) have more
vapour pressure, mercury has least vapour
pressure and because of this it is used in
manometers.
1.4
(A) 400 kg/cm 2
(B) 4000 kg/cm 2
(C) 40 106 kg/cm 2
(D) 40 105kg/cm 2
The saturation vapour pressure of three
liquids at 200 C is as given below
Methyl Alcohol 12,500Pa
Ethyl Alcohol 5900Pa
Benzene 10,000Pa
Select the correct statement from the
following
(A) Benzene vaporizes faster than methyl
alcohol at the same temperature
(B) Methyl alcohol vaporizes faster than
ethyl alcohol at the same temperature
(C) Ethyl alcohol vaporizes faster than
benzene at the same temperature
(D) Benzene vaporizes faster than both
methyl and ethyl alcohols at the same
temperature
Fluid Mechanics
1.5
10
0
Kinematic viscosity of air at 20 C is
1.6 105 m2 /s , its kinematic viscosity at
Kulkarni’s Academy
1.10 A fluid obeying the constitutive equation
du
0 K is held between two
dy
parallel plates a distance 'd' apart. If the stress
applied to the top plate is 30 , then the
0
70 C will be approximately
(A) 2.2 105 m2 /s
1.6
(B) 1.6 105 m2 /s
(C) 1.2 105 m2 /s (D) 105 m2 /s
With increase in temperature, while keeping
the pressure constant, the dynamic viscosity
, and the kinematic viscosity , behave in
the following manner for gases.
(A) Both & increases at the same rate
velocity with which the top plate moves
relative to the bottom plate would be
(D) decreases, while increases faster
1.7
A 20cm Cubical box slides on oil (mass
density = 800 kg/m 3 ), over a large plane
surface with a steady state velocity of 0.4m/s
.The plane surface is inclined at an angle of
300 with the horizontal plane. The oil film
between the block and the plane surface is
0.4mm thick. The weight of the box is 64 N .
The kinematic viscosity of the oil is
1.8
1.9
(A) 0.8Pa.s
(B) 0.001m 2 /s
(C) 1.6Pa.s
(D) 0.002 m 2 /s
Shear stress in the Newtonian fluid is
proportional to
(A) Pressure
(B) Strain
(C) Strain Rate
(D) The inverse of the viscosity
A Bingham fluid of viscosity 10Pa s, and
yield stress 0 10 k Pa , is sheared between
flat parallel plates separated by a distance
103 m . The top plate is moving with a
velocity of 1 m/s. The shear stress on the plate
is
(A) 10kPa
(B) 20kPa
(C) 30kPa
(D) 40kPa
(B) 4 0 d
K
(C) 3 0 d
K
(D) 9 0 d
K
2
2
(B) Both & decreases at the same rate
(C) Increases, while increases faster
(A) 2 0 d
K
2
2
1.11 Consider a fluid of viscosity between two
circular parallel plates of radii R separated by
a distance h. The upper plate is rotated at an
angular velocity . Whereas the bottom plate
is held stationary. The velocity profile
between the two plates is linear. The torque
experienced by the bottom plate is
(A) R 4 / 2h
(B) R 4 / 4h
(C) 2 R 3 / 3h
(D) R 3 / h
1.12 A journal bearing has a shaft diameter of
40mm and length 40mm .The shaft is
rotating at 20rad/sec and the viscosity of
lubricant is 20mPa-S . The clearance is
0.02mm .The loss of torque due to the
viscosity of lubricant is approximately
1.13
(A) 0.04 Nm
(B) 0.252 Nm
(C) 0.4 Nm
(D) 0.652 Nm
Two infinite parallel horizontal plates are
separated by a small gap ( d 20mm ) as
shown in figure. The bottom plate is fixed
and the gap between the plates is filled with
oil having density of 890 kg/m 3 and
kinematic viscosity of 0.00033m 2 /s . A shear
flow is induced by moving the upper plate
with a velocity of 5m/s .
Kulkarni’s Academy
11
Properties of Fluid
Assume, linear velocity profile between the
plates and the oil to be a Newtonian fluid. The
shear stress ( N/m2 ) at the upper plate is____
(R) Property which explains rise of sap in a
tree
(S) Property which explains the flow of jet
of oil in a unbroken stream
List - II
1.14
1.15
Match the items between the two groups.
Choose the correct matching
Group - I
(P) Ideal fluid
(Q) Dilatant fluid
(R) Newtonian Fluid
(S) Pseudo Plastic Fluid
Group - II
1) Is the one for which shear stress is
linearly proportional to the rate of
deformation
2) Is the one for which there is no resistance
to shear
3) Is the one for which apparent viscosity
increases with increasing deformation
rate
4) Is the one for which the apparent
viscosity decreases with the increasing
deformation rate.
(A) P-2, Q-3, R-1, S-4
(B) P-2, Q-4, R-1, S-3
(C) P-3, Q-1, R-4, S-2
(D) P-4, Q-3, R-1, S-2
Match List - I (Description) with List - II
(Property of fluid) and select the correct
answer using codes given below
List - I
(P) Property which explains the spherical
shape of the liquid drop
(Q) Property which explain the phenomenon
of cavitation in a fluid flow
1)
Viscosity
2)
Surface Tension
3)
Compressibility
4)
Vapour pressure
5)
Capillarity
P
Q
R
S
(A)
1
2
4
5
(B)
2
4
5
1
(C)
4
2
5
1
(D)
1
2
3
4
1.16 Match list I with list II and select the correct
answer using codes given below the list
1.17
List - I
List - II
(A) Specific Weight 1)
L/T2
(B) Density
2)
F/L3
(C) Shear Stress
3)
F/L2
(D) Viscosity
4)
FT/L2
5)
FT2/L4
A
B
C
D
(A)
4
4
1
2
(B)
4
3
2
5
(C)
4
3
5
2
(D)
2
5
3
4
A piston of 60mm diameter moves inside
cylinder of 60.1mm diameter. The
percentage decrease in force necessary to
move the piston when the lubricant warms up
from 00 C to 1200 C . (00 C 0.0182 NS/ m 2 )
(1200 C 0.00206 NS/ m 2 )
(A) 11.32
(B) 88.68
(C) 66.67
(D) 33.33
Fluid Mechanics
1.18 A skater weighing 800 N skates at a speed of
12
Kulkarni’s Academy
15m/s on ice at 00 C . The average skating
area supporting him is 10cm 2 and the
coefficient of friction between skates and ice
is 0.02, if there is actually a thin film of water
between skates and ice, then its thickness is
( 103 N s/ m 2 )
(A) 9.375 104 m
1.19
(B) 9.375 105 m
(C) 9.375 106 m
(D) 9.375 107 m
Consider a soap film bubble of diameter D. If
the external pressure is P0 and the surface
1.23
A spherical water drop of radius 'R' splits up
in air into 'n' smaller drops of equal size the
work required in splitting up the drop
tension of the soap film is , the expression
for the pressure inside the bubble is
2
(A) P0
(B) P0
D
4
8
(C) P0
(D) P0
D
D
1.20
1.21
(A) 4r 2 n
13
(B) 4 R n 1
2
A small drop of water at 200 C in contact with
air has a diameter of 0.05mm . If the pressure
(C) 4 r R 2 n 3
within the droplet is 0.6kPa higher that of
the atmosphere, the surface tension is
2
(D) 4 R 2 n 3 1
(A) 7.5 103 N/m
(B) 7.5 102 N/m
(C) 7.5 101 N/m
(D) 7.5 10N/m
If the diameter of tube is 1mm then the
capillary rise is 3cm . What will be the
1
A
Answer Key
1.1
B
1.2
C
1.3
B
1.4
B
1.5
A
1.6
C
1.7
B
1.8
C
1.9
B
An open glass capillary tube of 2mm bore is
1.10
B
1.11
A
1.12
A
lowered into a cistern containing mercury
(density = 13600 kg/m3 ) as shown in the
1.13
73.425
1.14
A
1.15
B
1.16
D
1.17
B
1.18
D
1.19
D
1.20
A
1.21
C
1.22
5.558
1.23
B
capillary rise when diameter changes to
0.2 mm ?
(A) 3cm
(B) 0.6 cm
(C) 15cm
1.22
( Surface tension of water) is
(D) 7.5cm
figure. Given that the contact angle between
mercury and glass = 1400 , surface tension
coefficient = 0.484N/m and gravitational
2
acceleration 9.81m/s , the depression of
mercury in the capillary tube below the free
surface in the cistern, in mm, is______
Kulkarni’s Academy
E
13
Explanation
1.1
(B)
1.2
(C)
Properties of Fluid
1.4
(B)
1.5
(A)
With increase in temperature, kinematic
viscosity of air increases. So, kinematic
viscosity of air at 70C is more than
1.6 10-5 m2/s.
Given data :
1
2
P 3500
2500 N
m2
Hence, the correct option is (A).
P = 100 kPa
We know that
dP
d
Bulk modulus K
1.6
(C)
1.7
(B)
Given data :
1
2
P 3500 2500
dP
1
3500
d
2
K .
1750
1
1
2
oil = 800 kg/m3
1750
v = 0.4 m/s
1
2
= 300
1
1750 2
2
P 2500
1750.
3500
1750(100 103 2500)
3500
K = 48750 N/m2
Hence, the correct option is (C).
K
1.3
F
Given data :
V1 = 0.04 m3
P1 = 50 kg/cm2
V2 = 0.039 m3
P2 = 150 kg/cm2
We know that
dP
dV
V
150 50
100
k
0.039 0.04 0.025
0.04
2
4000 kg/cm
Hence, the correct option is (B).
Bulk modulus (K) =
(20 20 104 ) 0.4
64 sin 30
0.4 103
(B)
AV
w sin
y
N s
m2
We know that kinematic viscosity
0.8
0.8
0.001 m2 /s
800
Hence, the correct option is (B).
1.8
(C)
d
dt
Shear stress is directly proportional to rate of
deformation or stain rate.
From Newton’s law of viscosity .
Hence, the correct option is (C).
Fluid Mechanics
1.9
14
Kulkarni’s Academy
(B)
n
du
A B
dy
For Bingham plastic fluid n = 1; B 0
du
A B
dy
du V
1
10 3 10 103 pascal
10
dy y
10 103 10 103 pascal
20 103 pascal
Torque dT = dF r
= 20 kPa
Hence, the correct option is (B).
1.10
(B)
du
dy
0 k
u. 2rdr r .r
h
R
u 2rdr r r
R 4
T
h
2h
0
dF
Hence, the correct option is (A).
1.12 (A)
30 0 k
2 0
2
du 4
du
dy
du
k2
dy
02
.dy
k2
u 4 0 .d
k
2
Hence, the correct option is (B).
1.11
(A)
AV
F
h
dF
u 2rdr r
h
AV
h
V dA
dF
h
Torque = dF r
F
T
r dl r
h
20 102 20 103 20
40 40 10 20 10
6
0.02 102
T = 0.0402 N -m
Hence, the correct option is (A).
2
Kulkarni’s Academy
15
1.13 73.425
Properties of Fluid
Percentage decrease in force =
Given :
d = 20 103 m
oil = 890 kg/m2
V
(.)V
y
y
(0.00033 890) 5
73.425pascal
20 103
Hence, the correct answer is 73.425.
1.14
(A)
1.15
(B)
1.16
(D)
1.18 (D)
Given data :
W = 800 N
V = 15 m/s
A = 10 cm2
(Friction coefficient) = 0.02
(viscosity) = 103 Ns/m2
Specific weight =
density
weight F
volume L3
mass
volume
F /a
F
FT 2
3
4
L 3
L
L
.
L
2
T
Force F
Shear stress =
2
area
L
F
FT
Viscosity =
2
L
du
L
2 T
dy L L
Hence, the correct option is (D).
1.17 (B)
0o C 0.0182 Ns / m 2
120o C 0.00206 Ns / m
F
k1 k2
100
k1
0.0182 0.00206
100
0.0182
88.68%
Hence, the correct option is (B).
= 0.00033 m2/s
v = 5m/s
F1 F2
100
F1
AV
y
F
F1 = k 1; F2 = k2
N = weight of skater = 800 N
Fs = 0.02 800 N
Fs
'VA
y
0.02 800
103 15 10 103
y
y = 9.375 107 m
Hence, the correct option is (D).
1.19 (D)
2
Soap bubble
P Pi P0
8
D
8
D
Hence, the correct option is (D).
Pi P0
Fluid Mechanics
1.20
16
(A)
P
4
D
0.6 103
4
0.05 103
7.5 103 N/m
Hence, the correct option is (A).
1.21
(C)
We know that
4 cos
h
gd
1
h
d
hd = constant
h1d1 = h2d2
3cm 1 mm = h2 0.2 mm
h2 = 15 cm
Hence, the correct option is (C).
1.22
5.558
h
h
4 cos
gd
4 0.484 cos140
5.558 103 m
13600 9.81 2 10 3
= 5.558 mm
Note : ve sign shows depression of mercury in the
capillary tube.
Hence, the correct answer is – 5.558.
1.23
(B)
Work done = increase in surface area
Work = [4 r2 n 4R2]
= 4R2[n-2/3. n 1]
Work = 4R2(n1/3 1)
Hence, the correct option is (B).
Kulkarni’s Academy
NOTES
2.1 Pressure
It is defined as the external normal force exerted
per unit area. The area can be real or imaginary.
The unit of pressure is N/m2 or Pascal.
“Pressure is a representative of number of
collisions per second.”
Pressure is compressive in nature.
2.1.1 Mohr’s circle for a static fluid:
For a static fluid there is no shear stress and there
are only normal forces (pressure) therefore Mohr
circle is a point as shown in Fig.
A
1
a
A W
a F
W
1
F
W>F
As W > F that is by applying small force large
weight can be raised this does not mean the
energy conservation is violated because smaller
force moves through a larger distance whereas
larger force move through smaller distance.
Note:
2.2 Pascal’s law:
According to Pascal’s law pressure at any point
in a static fluid is equal in all directions.
Conversely if pressure is applied in static fluid it
is transmitted equally in all direction.
Applications:
Hydraulic lift, hydraulic brake etc.
Pascal’s law can be applied for flowing fluids. If
the shear force is zero. This is possible only if the
fluid is ideal.
2.3 Atmospheric Pressure (Patm.):
The pressure exerted by environmental mass is
known as atmospheric pressure.
The atmospheric pressure is around 1.01325 bar.
2.4 Gauge Pressure (Pgauge):
The pressure measured with respect to
atmospheric pressure, is known as gauge
pressure.
2.5 Absolute Pressure (Pabs.):
The pressure measured with respect to zero
pressure is known as absolute pressure.
Fluid Mechanics
18
Kulkarni’s Academy
Note:
If h is taken in upward direction as the pressure
decreases with height
Pabs. = Patm. + Pgauge
2.6 Vacuum Pressure:
The pressure less than atmospheric pressure is
known as vacuum pressure.
Patm.
𝑑𝑃
𝑑ℎ
= −𝑤
For a static fluid forces acting on static fluid are
pressure and gravity forces.
At free surface all the other forces except
atmospheric pressure force is zero.
2.7.1 Pressure at any depth h:
Assumption:
Density of fluid is constant.
Pvacuum
Pabs..
Vacuum pressure = Patm. Pabs.
Note:
There can be positive gauge or negative gauge
pressure but there cannot be negative absolute
pressure.
While calculating absolute pressure local
atmospheric pressure must be taken into account.
At free surface h = 0
P = Patm
dP
w
dh
2.7 Hydrostatic Law:
dP = wdh
P = wh + c
At
h = 0 ; P = Patm
Then Patm = c
Pabs. = wh + Patm
For gauge pressure Patm = 0
Pgauge = wh = gh
Weight = (Specific weight Volume)
PdA + gdAdh = (P +dP)dA
dP
g w hydrostatic law
dh
“hydrostatic law represents or gives variation of
pressure in the vertical direction.”
P = gh is based on the assumption that the
density is constant.
Sometime pressure is expressed in height
column because and g are almost constant
and pressure varies directly with h, therefore
it is expressed in height column.
Kulkarni’s Academy
19
Pressure Measurement
2.8 Pressure measurement device:
2.8.1 Barometer:
Barometer is used for measuring atmospheric
pressure.
s1h1 = s2h2
where, s is specific gravity.
Let us assume both are gases.
1h1 2 h2
air air
s1h1 = s2h2
2.8.3 Piezometer:
Piezometer is a device which is open at both the
end with one end connected to a point where is
the pressure is to be calculated and another end is
open to atmosphere.
0 + gh = Patm
Patm = gh
From scale we measure height h = 0.76 m
Patm = 13.6 103 9.81 0.76
= 1.01396 105 N/m2
1 bar = 105 N/m2
Patm = 1.01396 bar
Note:
Piezometer are not suitable for measuring.
If water is used instead of mercury the
corresponding height will be 10.3m of water as
this height is very large therefore “mercury is
used in barometers because of its higher density”.
2.8.2 Conversion of 1 fluid column into
another fluid column:
Very high pressures
Gas pressures
Piezometers are suitable for measuring
moderate liquid pressure.
Note:
Piezometers are
manometers.
also
known
as
simple
2.8.4 Manometers:
Manometers are used for measuring pressure and
are based on balancing of liquid column.
P1=P2
1gh1 = 2gh2
1h1 = 2h2 Generalised equation.
Let us assume both are liquids.
1h1 2 h2
H 2o H 2o
Fluid Mechanics
1. Simple U-tube manometers:
20
Kulkarni’s Academy
P
2.1
Practice Questions
In a static fluid, the pressure at a point is
(A) Equal to the weight of the fluid above
(B) Equal in all directions
(C) Equal in all directions only if its
viscosity is zero
(D) Always directed downwards
(a) Jumping of fluid technique
2.2
P gy m gx 0
P m gx gy
Where,
Three containers are filled with water upto
the same height as shown. The pressure at the
bottom of the containers are denoted asP1, P2
and P3. Which one of the following
relationships is true?
P gauge pressure
2.3
(A) P P P
3 2 1
(B) P2 P1 P3
(C) P P P
1 2 3
(D) P1 P2 P3
The pressure gauges G1 and G2 installed on
the system show pressures of PG1 = 5 and
PG2 = 1 bar.
The value of unknown pressure P is
(b) Datum line technique:
PA = PB
P + gy = PA
0 + mgx = PB
PA = PB
P + gy = mgx
P = mgxgy
2.4
(A) 1.01 bar
(B) 2.01 bar
(C) 5 bar
(D) 7.01 bar
A diver descends 200 m in a sea (where the
density of sea water is 1050kg/m3) to as
unken ship wherein a container is found with
a pressure gauge reading of 225 kPa .
Kulkarni’s Academy
21
Pressure Measurement
Taking the pressure at the surface of the sea
to be atmospheric (Patm =100kPa), the
absolute pressure in the container is (g = 10
m/s2)
2.5
(A) 225 kPa
(B) 325 kPa
(C) 2325 kPa
(D) 2425 kPa
Choose the correct combination of true
statements from the following:
For a fluid at rest in equilibrium.
P.
The pressure must be the same over any
horizontal plane
Q. The density must be the same over any
horizontal plane
R. the shear stress must have the same nonzero value over any horizontal plane
S.
2.6
2.7
dP
g
dz
(B) 4.573kPa
(C) 6.573kPa
(D) 7.573kPa
Pressures have been observed at four
different points in different units of
measurement as follows
(1) 150 kPa
(A) P, Q, R
(B) Q, R, S
(2) 1800 milli bar
(C) P, R, S
(D) P, Q, S
(3) 20 m of water
An open tank contains water to a depth of 2m
and oil over it to a depth of 1m. If the specific
gravity of oil is 0.8, then the pressure
intensity at the interface of the two fluid
layers will be
(4) 1240 mm of mercury
(A) 1,2,3, and 4
(B) 3, 2, 1 and 4
(A) 9750 N/m2
(B) 8720N/m2
(C) 3, 2, 4 and 1
(D) 2, 1, 4 and 3
(C) 9347 N/m2
(D) 7848N/m2
Then the points arranged in descending order
of pressure are
2.10
A manometer measures the pressure
differential between two locations of a pipe
carrying water. If the manometric liquid is
mercury (S = 13.6) and the manometer
showed a level difference of 20cm , then the
pressure head difference of water between
the two tappings will be
2.8
2.9
(A) 3.573kPa
The pressure at the base of the mountain is
750 mm of mercury and at the top, the
pressure is600 mm of mercury. If the density
of air is 1kg/m3 , then the height of mountain
is approximately
(B) 2 km
(C) 5 km
(D) 7 km
(A) 1.26 m
(B) 2.72 m
When can a piezometer be not used for
pressure measurement in pipes?
(C) 1.36 m
(D) 2.52 m
(A) The pressure difference is low
The tank shown in the figure is closed at top
and contains air at a pressure PA. The value
of PA for the manometer reading shown will
be
2.11
(A) 3 km
(B) The velocity is high
(C) The fluid in pipe is a gas
(D) The fluid in the pipe is highly viscous
Fluid Mechanics
2.12 Multi U- tube manometers with different
fluids are used to measure
22
2.16
(A) Low pressures
(B) Medium pressures
(C) High pressures
(D) Very low pressure
2.13
In order to increase sensitivity of U-tube
manometer, one leg is usually inclined byan
angle . What is the sensitivity of inclined
tube compared to sensitivity of Utube
(A) sin
(B)
2.17
Kulkarni’s Academy
The standard atmospheric pressure is 762
mm of mercury. At a specific location, the
barometer reads 700 mm of mercury. At this
place, what does an absolute pressure of 380
mm of mercury corresponds to
(A) 320 mm of mercury vacuum
(B) 382 mm of mercury vacuum
(C) 62 mm of mercury vacuum
(D) 700 mm of mercury vacuum
The force F needed to support the liquid of
density d and the vessel on top of figure is
1
sin
1
(D) tan
cos
Which one of the following statements is
NOT CORRECT
(A) A gauge always measures pressure
above the surrounding atmospheric
pressure
(B) At a point in a static fluid, pressure is
equal in all directions
(C) Typical actual variation of pressure with
elevation in atmosphere is more
adiabatic than isothermal
(D) Vacuum pressure at a point is always
measured above absolute zero pressure
Three immiscible liquids of densities , 2
and 3 are kept in a jar as shown in figure.
Then the ratio H/h is
(C)
2.14
2.15
(A) gd [ha ( H h) A]
(B) gdHA
(C) gdHa
(D) gd (H h) A
2.18
Refer to figure, the absolute pressure of gas
A in the bulb is
(A) 771.2mmHg
(B) 752.65mmHg
(A) 9
(B) 3.5
(C) 767.35mmHg
(C) 3
(D) 2.5
(D) 748.8mmHg
Kulkarni’s Academy
2.19
23
Pressure Measurement
Assuming 300 and the manometer fluid as
In given figure, if the pressure of gas in bulb
A
is
vacuum
and
50cmHg
oil with specific gravity of 0.86, the pressure
at A is
Patm 76cm of Hg , then height of column H
is equal to
2.20
(A) 26 cm
(B) 50 cm
(C) 76 cm
(D) 126 cm
Two pipelines, one carrying oil (mass density
900 kg/m 3 ) and the other water, are
connected to a manometer as shown in
figure. By what amount the pressure in
water pipe should be increased so that
mercury levels in both the limbs of
manometer become equal?
2.22
the
the
the
the
(mass density of mercury = 13,500 kg/m 3
difference in pressure ( kPa )between sections
P and Q is
and g 9.81m/s 2 )
2.21
(A) 43 mm water (vacuum)
(B) 43 mm water
(C) 86 mm water
(D) 100mm water
A differential U-tube manometer with
mercury as the manometric fluid is used to
measure the pressure difference between two
sections P and Q in a horizontal pipe carrying
water at steady state as shown in the figure.
If the difference in mercury levels in the two
limbs of the manometer is 0.75m , the
(A) 24.7kPa
(B) 26.5kPa
(C) 26.7kPa
(D) 28.9kPa
In the inclined manometer shown in the
figure, the reservoir is large. Its surface may
be assumed to remain at a fix edelevation. A
is connected to a gas pipe line and the
deflection noted on the inclined glass tube is
100 mm.
2.23
(A) 49.275
(B) 94.275
(C) 9.4275
(D) 492.75
A U-tube manometer, as shown in figure has
water as a manometric fluid. When an
unknown pressure 'P' acts at 5mm diameter
limb, the water rises in the limb by 100 mm
from initial level; if the other limb is open to
atmospheric (pressure Pa), the pressure
differential (P - Pa) is
Fluid Mechanics
24
2.27
Kulkarni’s Academy
Which one of the following figures correctly
represents the Mohr’s circle for a static fluid
(hydrostatic condition)
(A)
(B)
2.24
(A) 1225 N/m 2
(B) 980 N/m 2
(C) 1250 N/m 2
(D) 1000 N/m 2 s
(C)
A manometer is made of a tube of uniform
bore of 0.5cm 2 cross-sectional area, with one
limb vertical and the other limb in clinedat
300 to the horizontal. Both of its limbs are
open to atmosphere and initially, it is partly
filled with a liquid of specific gravity1.25. If
an additional volume of 7.5cm 3 of water is
added to inclined tube, calculate the rise of
the liquid in vertical tube from initial level?
2.25
(A) 4 cm
(B) 7.5 cm
(C) 12 cm
(D) 15 cm
(D)
2.28
Which property of mercury is the main
reason for its use in barometers?
gravity to be 10 m/s 2 . The gauge pressure(in
(A) High density
kN/m2 , rounded off to the first decimal
place) at a depth of 2.5m from the top of the
tank will be__
(B) Negligible capillary effect
(C) Low vapour pressure
(D) Low compressibility
2.26
Atmospheric pressure at a place is equal to
10cm of water. A liquid has a specific weight
of 12kN/m3. The absolute pressure at a point
2m below the free surface of liquid in kPa
is
(A) 2.4
(B) 12.4
(C) 24
(D) 122.1
5m deep vertical cylindrical tank, water is
filled up to a level of 3m from the bottom
and the remaining space is filled with oil of
specific gravity 0.88. Assume density of
water as 1000 kg/m3 and acceleration due to
2.29
The figure below shows the pressure
measured in a well at different depths. AB is
gas cap; B is gas oil contact and C is water oil contact. Density of gas in cap is 2kg/m 3 ,
oil density is 800 kg/m 3 and water density is
1000 kg/m3 . The difference between pressure
at point D and point B (PD-PB) is ______
105N/m2. (Take g=9.81m/s2)
Kulkarni’s Academy
25
Pressure Measurement
A
2.30
Consider the density and altitude at the base
of an isothermal layer in the standard
atmosphere to be 1 and h1 respectively. The
density variation with altitude ( versus h) in
that layer is governed by (R = gas constant; T
= temperature; g = acceleration due to
gravity)
g
(B)
( hh1 )
(C)
e g
1
( h1 h )
(D)
e g
1
RT
2.31
g
( h1 h )
e RT
1
( hh1 )
(A)
e RT
1
Answer Key
2.1
C
2.2
D
2.3
D
2.4
D
2.5
D
2.6
D
2.7
D
2.8
A
2.9
C
2.10
B
2.11
C
2.12
C
2.13
B
2.14
A
2.15
C
2.16
A
2.17
B
2.18
A
2.19
B
2.20
A
2.21
B
2.22
B
2.23
A
2.24
A
2.25
A
2.26
D
2.27
b
2.28
22.6
2.29
7.84
2.30
A
2.31
A
RT
E
Explanation
Which of the following pressure units
represent the LEAST pressure
2.1
(C)
(A) millibar
2.2
(D)
As height is same in all containers, so,
pressure is same.
(B) mm of mercury
(C) N/mm2
Hence, the correct option is (D).
(D) kg-f/cm2
2.3
(D)
P = P1 + PG1
P = S + 2.01 = 7.01 bar
Hence, the correct option is (D).
Fluid Mechanics
2.4
26
(D)
Kulkarni’s Academy
2.8
(A)
PA + oil gh + water gh m. gh = 0
PA + 750 9.81 1.5 + 1000 9.81
0.6 13600 9.81 0.1 = 0
PA = (13600 0.1 1000 0.6 750
1.5) 9.81
PA = 35800.65 Pa = 3.58 kPa.
Hence, the correct option is (A).
P = gh
1050 10 200 = 2100 kPa
Pabs = Pguage + Plocal atm
225 + (100 + 2100) 2425 kPa.
Hence, the correct option is (D).
2.5
(D)
2.6
(D)
2.9
(C)
(1) 150 kPa = 15 104 Pascal
(2) 1800 milli bar = 1800 103 bar
= 18 104 Pascal
(3) 20 m of water = 1000 10 20
= 20 104 Pascal
(4) 1240 mm of mercury = 13600 9.81
1.240 = 16.54 104 Pascal
Descending order of pressure is 3 – 2 – 4 – 1.
Hence, the correct option is (C).
2.10 (B)
Pinterface = oil g h
= 900 9.81 1
7848 N/m2
Hence, the correct option is (D).
2.7
(D)
air hair= hg hhg
150
hair 13600
m
1000
= 2040 m = 2.04 km
Hence, the correct option is (B).
2.11 (C)
Piezometers are not suitable for measuring
gas pressures.
Hence, the correct option is (C).
2.12 (C)
Multi – u tube manometer with different
fluids are used to measure high pressure.
Hence, the correct option is (C).
2.13 (B)
P1 + H + 0.2 0.2 13.6 H = P2
P1 P2 2.72 0.2
2.52m of water head
Hence, the correct option is (D).
Kulkarni’s Academy
1
l x
sin
1
is the sensitivity of inclined u – tube.
sin
Hence, the correct option is (B).
2.14
(A)
2.15
(C)
g(3h) + 2g(1.5h) 3g(H – h) = 0
g(3h) + 2g(1.5h) + 3gh 3gh = 0
gh = 3H
H
3
h
Hence, the correct option is (C).
2.16
(A)
27
Pressure Measurement
2.19 (B)
PA abs H Patm
PB abs 76 50
H Patm PA abs
= 26 cm
H = 76 – 26
H = 50 cm
Hence, the correct option is (B).
2.20 (A)
Pvacuum Plocal atm Pabs
700 – 380
320mm of Hg vacuum
Hence, the correct option is (A).
2.17
Poil + 900 9.81 3 13500 9.81 0.2
1000 9.81 1.5 = Pwater
P0 Pw = 14715 Pascal
…. (1)
(B)
Force Pressure Area
dgHA
Hence, the correct option is (B).
2.18
(A)
2
PA g (0.17) 13600 9.81
100
5
1000 9.81
100
= 101325 Pascal
PA 101325 = 13600 9.81 0.02 +
1000 9.81 0.05 1000 9.81 0.17
PA = 1598.22 + 101325 pascal
102923.22 pascal
760 102923.22
771.98 mmof Hg
101325
Hence, the correct option is (A).
P0 + 900 9.81 2.9 1000 9.81 1.6
= Pw+P
P0 + 9908.1 = Pw + P
P0 Pw + 9908.1 = P
14715 + 9908.1 = P
P = 24623.1 Pascal = 24.62 kPa
Hence, the correct option is (A).
Fluid Mechanics
2.21
(B)
28
Kulkarni’s Academy
25 100
25mm
100
125
Pa – 1000 9.81
P
1000
P – Pa = 1226 N/m2
Hence, the correct option is (A).
h
2.24 (A)
PA 50 = 0
PA = 50 mm of oil
sw hw = s0h0
hw = 0.86 50
hw = 43 mm of water (+)
Hence, the correct option is (B).
2.22
(B)
PP - H2O gH-Hg 0.75 +
H2O g (0.75 + H) = PQ
PP – PQ = 13600 9.81 0.75 – 1000
9.81 0.75
92704.5 Pa
92.7 kPa
Hence, the correct option is (B).
2.23
(A)
After adding 7.5 cm3 of water to inclined tube
7.5
y1 y2
15cm
0.5
Let, y1+y2 = h = 15 cm
…. (1)
h2 = y1
…. (2)
y1 = 2h1
…. (3)
Patm 1250 g (h1 h2 )
7.5
Patm
100
7.5
h
1250 g 2 h2 1000 g
100
2
h2 = 0.04 m
h2 = 4 cm
Hence, the correct option is (A).
1000 g
2.25 (A)
2.26 (D)
Patm = 10 m of water
= gh
= 1000 9.81 10 = 98.1 kPa
By using volume conservation
A h = a h
D2 h = d2h
Kulkarni’s Academy
Pabsolute = Patm + wh
= 98.1 + 12 2
= 122.1 kPa
Hence, the correct option is (D).
2.27
(B)
For a static fluid there is no shear stress and
there are only normal forces (pressure)
therefore Mohr circle is a point.
Hence, the correct option is (B).
2.28
22.6
29
Pressure Measurement
804616.2 N/m2
PB = air 9.81 1010
2 9.81 1010
19816.2 N/m2
PDPB = 804616.2 – 19816.2
784800 N/m2
7.84 105 N/m2
Hence, the correct answer is 7.84.
2.30 (A)
dP = gdh
P = RT (From ideal gas equation)
dP = RT. d (T = C; isothermal atmosphere)
Given :
g = 10 m/s2
h = 2.5 m (from the top)
Pgauge = ?
Pgauge oil g 2 water g 0.5
880 10 2 + 1000 10 1.5
22.6 kN/m2
Hence, the correct answer is 22.6.
2.29
7.84
Given
air = 2 kg/m3
dRT gdh
d
h dh
1
1
g
n
h h1
RT
1
h
g
hh1
e RT
1
Hence, the correct option is (A).
2.31 (A)
oil = 800 kg/m3
Millibar = 103 bar = 102pascal
water = 1000 kg/m3
1 mm of Hg = 1 torr = 138.32 pascal
PD PB = ?
N
106 pascal
2
mm
w g 40
2 9.81 1010 + 800 9.81 50 +
1000 9.81 40
kg.f
9.81104 pascal
2
cm
Hence, the correct option is (A).
PD air g 1010 oil g 50
NOTES
3.1 Archimedes Principle:
When a body is submerged either partially or
completely the net vertical upward force exerted
by the fluid on the body is known as buoyancy
force [Fb] and this buoyancy force is equal to
weight of the fluid displaced and this is known as
Archimedes principle.
Note:
[1] When a homogeneous body is completely
submerged, in a fluid then the centre of gravity
(c.g.) of the body and centre of buoyancy is
coincide.
[2] For a floating homogeneous body centre of
buoyancy is below the centre of gravity.
[3] For a non-homogeneous body (heterogeneous)
centre of buoyancy and centre of gravity may not
coincide even if it completely submerged.
3.3 Principle of Floatation:
Vfd = Volume of the body submerged
= (x2 – x1)A
Net vertical upward force exerted by fluid on the
body = f gx2A - f gx1A
FvNet = f g(x2 x1)A
FvNet = fgVfd
FvNet = weight of the fluid displaced
FBuoyancy = WtFd
For a floating body to be in equilibrium weight
of the body must be equal to the weight of the
fluid displaced and line of action of these two
forces must be same.
Note:
Buoyancy is basically due to pressure difference.
3.2 Centre of Buoyancy (B):
It is the point from which the buoyancy force is
suppose to be acting and this buoyancy force will
act at the centroid of the displaced volume.
Therefore, centre of buoyancy will lie at the
centroid of the displaced volume.
Wbody = FB
FB = Weight of displaced fluid (Wfd)
Wbody = Weight of displaced fluid
Kulkarni’s Academy
31
Buoyancy & Floatation
3.4 Type’s of Equilibrium:
[1] Stable equilibrium
[2] Unstable equilibrium
[3] Neutral equilibrium
Stability
conditions
submerged bodies:
for
completely
[3] A floating body will be in neutral equilibrium
when G & M coincide.
3.5 Metacentre (M):
It is the point of intersection of normal axis of the
body with the new axis of line of buoyancy force.
When a body is tilted. i.e. it is the point about
which the body is suppose to be oscillating.
[1] A completely submerged body will be in
stable equilibrium when centre of buoyancy
(B) is above the centre of gravity (G).
Metacentric height (GM):
[2] A completely submerged body will be in
unstable equilibrium when the centre of
buoyancy (B) is below the centre of gravity
(G).
[3] A completely submerged body will be in
neutral equilibrium when centre of gravity
(G) and centre of buoyancy (B) is coincide.
Stability conditions for partially submerged
or floating bodies:
[1] A floating body will be in stable equilibrium
when Metacentre (M) is above the centre of
gravity (G).
[2] A floating body will be in unstable
equilibrium when Metacentre (M) is below
the centre of gravity (G).
The distance between centre of gravity (G) and
Metacentre (M) measured along normal axis is
known as Metacentric height.
For stable equilibrium GM > 0 or positive
Unstable equilibrium GM < 0 or Negative
Neutral equilibrium GM = 0
Fluid Mechanics
32
3.6 Mathematical condition for stable
equilibrium:
Kulkarni’s Academy
“Oscillation about longitudinal axis are known as
rolling and oscillations about transverse axis is
known as pitching.”
[BMrolling< BMpitching]
For more stable equilibrium condition BM or
GM must be as large as possible.
BM
I
VFd
If rolling is taken care of then pitching is
automatically taken care of.
3.7 Time Period of Oscillation:
BM is known as Metacentric radius.
k g2
T 2
g (GM )
GM T i.e. more oscillation
kg – least radius of gyration I Ak g2
kg
BM L
I LL
VFd
BM t
I tt
b3
b 3
I LL ; Itt
VFd
12
12
Itt > ILL
BMt> BML
From design point of view least BM is calculated
that is BM about longitudinal axis is calculated.
As BMt> BML the body will be more stable when
it oscillates about transverse axis compare to
oscillation about longitudinal axis.
I
A
For more stable equilibrium condition
Metacentric height GM must be larger but larger
GM results in smaller time period of oscillations.
i.e., more number of oscillations in a given time
therefore passengers are not comfortable under
such conditions, therefore for passenger ships
metacentric height is not very high.
In case of war ship stability is important than
comfort therefore GM for war ships is larger than
that of passenger ship.
Note :
Metacentric height for passenger ship is 0.3 m to
1.2 m.
Metacentric height for war ship is 1.0 m to 1.5 m.
Kulkarni’s Academy
3.8 Weight loss due to buoyancy:
33
Buoyancy & Floatation
Example 1
A plastic boat with a steel ball floating in
water containing if steel ball is thrown in
water container then what will happen to the
level of water.
Sol.
Concept:
If the displaced volume of the fluid is more,
the level will rise.
If the displaced volume of the fluid is less the
level falls.
If the displaced volume of the fluid is same,
the level will remain same.
Weight loss
= T – T1
W – W + FB
= FB
Weight loss = buoyancy force
Note:
Case: 1
Case: 2
In case 1
Wbody = WFd
As the density of air is very small buoyancy
effect is negligible in air therefore the correct
weight of the body is obtained when it is
submerged in air.
Mbody g = F g VFd1
Dead body floats on water because after the death
due to biological activities gases are released
specially methane (CH4) hence the density of the
body decreases & due to buoyancy force dead
body floats.
MPb g = F g VFd, pb
(Msb + Mpb) = F VFd,1
…. (1)
In case 2
WPb = WFd
Mpb = F VFd, pb
Wsb> FB
…. (2)
Msb.g >F gVFd,sb
Msb>FVFd,sb
VFd2 = VFd,pb + VFd,sb
Add (2) + (3)
Msb + Mpb = F(VFd,sb + VFd,pb)
From equation 1:
F vFd,1>F[VFd,sb+VFd,sb]
VFd,1> VFd.2
VFd2< VFd1
Level fall.
…. (3)
Fluid Mechanics
34
Kulkarni’s Academy
Example 2
An ice block float on water, if complete ice
melt then show that level of water remains
same.
In case 2
Wwater = WFd
mwg = F g VFdice
mw = F VFd, ice
Sol.
…. (a)
Wiron > FB
m2g >Fg VFd, iron
…. (b)
m2>F VFd, iron
VFd2 = VFd, ice + VFd, iron
Wice = WFd
ice
Add equation (a) & (b)
water
(m1 + m2) >F VFd2
mice g = FgVFd m = V
Mice = FVFd
v
m
VFd1> VFd2
F
Level will go down.
vFd
m
Note:
F
Level is same because displaced volume of
fluid is same.
Example 3
If inside the ice, there is a piece of metal or
iron nail, then what will happen to the level
of water.
Sol.
Case: 1
Case: 2
In case 1
Wice block = WFd
mice block g =FgVFd
VFd1F>FVFd2
…. (1)
(m1 + m2) = vFd1.F
….(1)
A ship enters from sea water to river water, ship
will go down and level of river rises.
Kulkarni’s Academy
P
3.1
35
Buoyancy & Floatation
3.5
Practice Questions
Force of buoyancy on a floating body equals
(A) Total pressure on the vertical projection
of the body
(B) Total pressure on the
projection of the body
horizontal
(C) Weight of liquid equal to the volume of
the body
(D) Weight of the liquid equal to the
immersed volume of the body
3.2
When a ship moving on seawater enters a
river, it is expected to
(A) Rise a little
(B) Sink a little
(C) Maintain the same level of draft
(D) Rise or fall depending on whether it is
made of wood or steel
3.3
In an ice berg, 15% of the volume projects
above the sea surface. If the specific weight
of sea water is 10.5 kN/ m 3 , the specific
weight of ice berg in kN/m3 is
3.4
(A) 12.52
(B) 9.81
(C) 8.93
(D) 7.83
A metallic cube of side 10cm , density
6.8 gm/cm 3 is floating in liquid mercury
(density 13.69gm/cm 3 ) with 5 cm height of
cube exposed above the mercury level. Water
(density 1gm/cm 3 ) is filled over to
submerge cube fully. The new height of cube
exposed above mercury level is
(A) 4.6 cm
(B) 5.4cm
(C) 5.0 cm
(D) 5.8 cm
3.6
The following terms relate to floating bodies:
Centre of gravity - G; Metacentre - M;
Weight of floating body - W; Buoyant force
- FB.
Match List-I which List-II and select the
correct answer
List-I
(A) G is above M
(B) G and M coincide
(C) G is below M
(D) FB W
List-II
1) Stable equilibrium
2) Unstable equilibrium
3) Floating body
4) Neutral equilibrium
A
B
C
D
(A)
1
3
2
4
(B)
3
1
4
2
(C)
2
4
1
3
(D)
2
3
4
1
Match List I with List II and select the
correct answer
List-I
(A) Stable equilibrium
(B) Stable equilibrium
(C) Unstable equilibrium
(D) Unstable equilibrium
List-II
1) Below G of a floating body
2) M above G of a submerged body
3) B above G of a floating body
4) M below G of a submerged body
A
B
C
D
(A)
2
1
4
3
(B)
4
3
2
1
(C)
2
3
4
1
(D)
2
3
1
4
Fluid Mechanics
3.7
A body weighs 100N in air and 80 N in water.
The density of the body is
3.8
(A) 4000 kg/m 3
(B) 5000 kg/m3
(C) 8000 kg/m3
(D) 7000 kg/m3
36
Kulkarni’s Academy
A body weighs 30 N in a liquid of density
800 kg/m 3 and 15 N in a liquid of density
1200 kg/m3 . The volume of body is
3.9
3.10
(A) 3.82 103 m3
(B) 2.82 103 m3
(C) 5.76 103 m3
(D) 8.98 103 m3
The weight of a sphere is 100 N. If it floats in
water just fully submerged, the diameter of
sphere is
(A) 1962 N
(B) 981N
(C) 491N
(D) 768N
The volume of the buoy that is submerged is
(B) 213 mm
(A) 0.1m 3
(B) 0.6 m 3
(C) 269 mm
(D) 315 mm
(C) 0.8 m 3
(D) 0.2 m 3
The metacentric height for a floating
spherical ball of radius R and depth of
immersion also equal to R is
(C)
3.12
3.14
The tension in the wire is
(A) 112 mm
(A) R
3.11
3.13
6R
5
(B)
1 R2
(D) 0
3.15 A spherical balloon of diameter 15m is
supposed to lift a load of 3000 N . The lifting
of load is achieved by heating the air inside
the balloon. Assume, air to be an ideal gas
and atmospheric pressure either outside or
inside the balloon. The value of acceleration
due to gravity is 9.81m/s 2 and the values of
temperature and density of atmospheric air
are 150 C and 1.2 kg/m3 , respectively. In order
The least radius of gyration of ship is 9 m and
the meta centric height is 750 mm. The time
period oscillation of the ship is
(A) 42.41 S
(B) 75.4 S
to lift the specified load, the air in side the
balloon should be heated to a temperature (
(C) 20.85 S
(D) 85 S
0
A solid cylinder (density 600 kg/m ) of
length L and diameter D floats in water under
neutral equilibrium conditions with its axis
vertical. Then L/D is
3
Linked Answer Questions (3.13 to 3.14)
A metallic sphere of volume Vm 0.1m3 ¸
density 20000 kg/m3 and fully submerged in
water is attached by a flexible wire to a buoy
of volume VB 1m3 and density = 100 kg/m 3
3.16
C ) of___
A container of square cross section is
partially filled with a liquid of density 1 .The
cylinder is intended to float in another liquid
of density 2 as shown in the figure. The
distance between metacentre and centre of
I
buoyancy is
where I and Vsub are area
Vsub
moment of inertia of the cross-section and
submerged volume respectively. Neglect the
weight of the container.
Kulkarni’s Academy
37
Which one of the following is the correct
condition for stability?
Buoyancy & Floatation
3.18
A homogenous right circular cylinder of
length L, radius R and specific gravity SG is
floating in water with its axis vertical. If
SG 0.8 , then the minimum value of R/L
above which the body will always be stable is
(A) 0.16
(B) 0.36
(C) 0.56
(D) Cannot predict due to insufficient data
b h
(A) 2 1 1 0
61 h b 2
(B)
2 b h 1
1 0
61 h b 2
2 b h 1
(C)
1 0
61 h b 2
(D)
3.17
2 b h 1
1 0
61 h b 2
A rectangular boat 6 m wide and 15 m
long(dimension perpendicular to the plane of
the figure) has a draught of 2M. The side
view of the boat is as shown in the figure. The
centre of gravity G of the boat is at the free
surface level. The metacentric height of the
boat in m
(A) –1.0
(B) 0.5
(C) 1.5
(D) 2.0
3.19
During floods, water entered an office having
wooden tables. The position of tables, if
floating, will be
(A) Legs upwards
(B) Legs on sides
(C) Legs downwards
(D) Any position
Fluid Mechanics
A
38
3.4
Answer Key
3.1
D
3.2
B
3.3
C
3.4
B
3.5
C
3.6
C
3.7
B
3.8
A
3.9
D
3.10
D
3.11
C
3.12
B
3.13
B
3.14
D
3.15
63.5
3.16
A
3.17
B
3.18
C
3.19
A
E
(D)
3.2
(B)
As the density of sea water is more (1050
kg/m3). The buoyancy effects are more. If
ship enters from sea water to river water, as
the density of river water is less, buoyancy
forces are les therefore ship will sink a little,
as ship sinks, more volume of the fluid
displaced therefore level rises.
Hence, the correct option is (B).
(C)
Given that
Wsea water= 10.5 kN/m3
VFd = 0.85 V
(let V is the volume of the body)
By using principle of floatation
Wbody = WFd
b g Vbody = F gVFd
WbV = WF(0.85 V)
Wb = 10.5 0.85 = 8.93 kN/m3.
Hence, the correct option is (C).
(B)
Given data :
Side of metallic cube = 10 cm
Density of metallic cube = 6.8 gm/cm3
mercury = 13.69 gm/cm3
Explanation
3.1
3.3
Kulkarni’s Academy
By using principle of floatation
Weight of the body = Buoyancy force
Wbody = (FB)water + (FB)Mercury
6.8 10 10 10 = 1 10 10 x +
13.69 10 10 (10 – x)
x = 5.4 cm
Hence, the correct option is (B).
3.5
(C)
3.6
(C)
3.7
(B)
Body weight in air (W) = 100 N
W = mg = 100
100
…. (1)
m
g
Body weight in water (Ww) = 80 N
Buoyancy force (FB) = 100 – 80 = 20 N
FB = FgVFd = 20
20
VFd
1000 g
For fully submerged body volume of fluid
displaced is equal to volume of the body.
20
V
…. (2)
1000 g
From equation (1) & (2)
100
m
g
5000 kg/m 3
20
V
1000 g
Kulkarni’s Academy
39
Alternate method:
Specific gravity of body =
s.g.body
Buoyancy & Floatation
3.10 (D)
Weight in air
Weight loss
100
100
5
(100 80) 20
Then density = 5 1000 = 5000 kg/m3
Hence, the correct option is (B).
3.8
(A)
Hence, the correct option is (D).
3.11 (C)
T 2
2
K g2
g (GM )
92
20.847.sec
9.81 0.750
Hence, the correct option is (C).
3.12 (B)
1 = 800 kg/m3
2 = 1200 kg/m3
Case – 1
Case – 2
Let actual weight of body (in air) is W
then in case – 1
W – 30 = 1gVFd
…. (1)
In case – 2
W – 15 = 2gVFd
…. (2)
VFd = volume of body (V)
Equation (2) – equation (1)
15 = (2 1) g V
15
V
9.81 3.82 103 m3
400
Hence, the correct option is (A).
3.9
(D)
Weight of sphere = Buoyancy force
W = F gVFd
100 1000 9.81 d 3
6
d = 269.46 mm
Hence, the correct option is (D).
Given that
SC = 600 kg/m3
By using principle of floatation
Wcylinder = WFd
cy.Vcy = FVFd
600 D2 L 1000 D2h
4
4
3L
h
5
L h L 3L
BG
2 2 2 10
Fluid Mechanics
L
BG
5
40
Kulkarni’s Academy
3.14 (D)
T + WB = FB
4
D
I
5D 2
BM
64
VFd D 2 5L 48L
4
5
Under neutral equilibrium conditions
BM = BG
981 + BgVB = FgVFd
981 + 100 9.81 1 = 1000 9.81 VFd
VFd = Volume of buoy submerged = 0.2 m3
Hence, the correct option is (D).
3.15 (63.5)
2
5D
L
48L 5
Let after heating the density of air inside the
balloon is b2 & b1 1.2 kg/m3,
25 L2
48 D 2
Then
3
4
4 15
V r 3
3
3 2
L
25
5
D
16 3
4 3
V 1767.145 m3
Hence, the correct option is (B).
3.13
For lifting
Total downward force = Buoyancy force
Wb + 3000 = FB
(B)
Given that
Vm = 0.1 m3
V b 2·g 300 = V air·g
= 2000 kg/m3
VB= 1 m3
2
B = 100 kg/m3
3000N FB
V .g (air b2 ) 3000
(V = 1767.145 m3)
b2 1.027 kg/m 3
P = RT
For constant pressure
T + FB = Weight of metallic sphere (Wms)
1T1 = 2T2
T2 =
T + FgVFd = mgVms
T+1000gVFd = 2000 g Vms {Vms=VFd
T = (2000 g – 1000 g)Vms
T = 1000 g 0.1
T = 981 N
Hence, the correct option is (B).
1
·T1
2
1.2
288
1.027
T2 = 336.51
T2 =
T2 63.5°C
Hence, the correct answer is 63.5.
Kulkarni’s Academy
41
Buoyancy & Floatation
15 63
I
1.5 m
BM 12
V 15 6 2
Metacentric height
GM = BM – BG = 1.5 – 1.0 = 0.5m
Hence, the correct option is (B).
3.16 (A)
3.18 (C)
By using principle of floatation
Weight of container = FB
1gb2h = 2gb2x
x 1h
2
h x h h
BG 1 .
2 2 2 2 2
h 1
1
2 2
b4
2
I
12 b 2
BM
VFd b 2 . 1 h 121h
2
For stable equilibrium
GM > 0
BM > GB or BM BG > 0
b22 h 1
1 0
121h 2 2
Divided by b
2 b h 1
. 1 0
61 h b 2
Hence, the correct option is (A).
3.17
(B)
Given data : L = 15 m, Width (b) = 6m
BG = 2 – 1 = 1 m
Given data :
s.g of cylinder = 0.8
By using principal Wbody = FB
bgVb = FgvFd
0.8 R2L = R2h
h = 0.8L
L h L 0.8L
BG
2 2 2
2
BG = 0.1L
BM
I
VFd
R 2
R2
4
R 2 (0.8L) 3.2 L
Body will always be stable equilibrium is
GM 0
BM – BG 0
R2
0.1L 0
3.2 L
R2
3.2 0.1
L
R
0.5656
L
Hence, the correct option is (C).
3.19 (A)
NOTES
4.1 Hydrostatic forces on plane surfaces:
(a) Inclined surface:
Taking a small elemental area dA we can
calculate the force on this small element and total
force can be calculated by integrating therefore
total hydrostatic force.
F=wA x
w = sp. Weight of fluid
A = area of surface
x = vertical distance of c.g. from the
free surface.
Centre of pressure:
It is the point through which total hydrostatic
force is suppose to be acting.
From principle of moments the centre of
pressure can be calculated.
2
x cp x I GG sin
Ax
Case:1 Plane inclined surface
IGG is the MOI about centroidal axis which is
parallel to OO.
= angle made by the surface with respect to free
surface
2
xcp x IGG sin
Ax
Note:
The centre of pressure is below the centre of
gravity because pressure increases with depth.
Case – 2: Plane vertical surface:
Put = 90o in case (1).
Kulkarni’s Academy
43
Hydrostatic Force
F wA x
dF = PdA
xcp x
I GG 2
sin 90
Ax
xcp x
IGG
Ax
dF = gxdA
Case 3: Plane horizontal surface
Put = 00 in case (1)
dFH = dFsin
dFH = gx dA sin θ
Vertical projection area
P g x
F g xA
F wA x
The horizontal component of force on curved
surface is equal to hydrostatic force on
vertical projection area and this force acts at
the centre of pressure of corresponding area.
= 0
xcp x
S.No.
Case
Force
1
Inclined
wAx
2
Vertical
wAx
3
Horizontal
wAx
Centre of
pressure
x
IG
sin 2 θ
Ax
x
IG
Ax
Volume = xdAcos
x
dFv = dFcos
4.2 Hydrostatic force on curve surfaces:
dFv g xdA cos
vol
dFv = g x Volume.
dFv = weight of fluid
The vertical component of force on the curve
surface is equal to weight of the liquid
contained by the curved surface taken upto
free surface.
This weight will act from the centre of gravity
of the corresponding weight.
Fluid Mechanics
44
Kulkarni’s Academy
P
4.1
Practice Questions
The centre of pressure of a liquid on a plane
surface immersed vertically in a static body
of liquid, always lies below the centroid of
the surface area, because
(A) In liquids the pressure acting is same in
all directions
Special cases:
(B) There is no shear stress in liquids at rest
(1)
(C) The liquid pressure is
depth
constant over
(D) The liquid pressure increases linearly
with depth
4.2
FH FH
1
2
FH
net
F
H1 FH 2 0
A plate of rectangular shape having the
dimensions of 0.4m 0.6m is immersed in
water with its longer side vertical. The total
hydrostatic thrust on one side of the plate is
estimated as 18.3kN . All other conditions
remaining the same, the plate is turned
through 900 such that its longer side remains
vertical.
What would be the total force on one face of
the plate?
4.3
(A) 9.15kN
(B) 18.3kN
(C) 36.6kN
(D) 12.2kN
The force on the door submerged in a liquid
of density (See figure) is
(A)
(2)
(C)
2 g
g
2 2
(B)
g
2
(D)
2
g
Kulkarni’s Academy
4.4
4.5
Hydrostatic Force
A circular plate 1.5 m diameter is submerged
in water with its greatest and least depths
below the surface being 2m and 0.7m
respectively. What is the total pressure
(approximately) on one face of the plate?
(A) 12kN
(B) 16kN
(C) 24kN
(D) None of these
(R)
The point of application of a horizontal force
on curved surface submerged in liquid is
(A)
IG
h
Ah
Ah
(C)
h
IG
4.6
45
(B)
(S)
I G Ah 2
Ah
List II
I
(D) G hA
h
(Depth of centre of Pressure)
A vertical dock gate 2 metre wide remains in
position due to horizontal force of water on
one side. The gate weighs 800kg and just
starts sliding down when the depth of water
upto the bottom of the gate decreases to 4
metres. Then the coefficient of friction
between dock gate and dock wall will be
4.7
(A) 0.5
(B) 0.2
(C) 0.05
(D) 0.02
Math List I with List II and select the correct
answer
List -I
d
8
1.
5
3.
d
2
2.
3
d
4
4.
2
d
3
(A) P-1, Q-2, R-3, S-4
(B) P-4, Q-2, R-3, S-1
(C) P-4, Q-3, R-1, S-2
(D) P-1, Q-2, R-4, S-3
4.8
The figure below show a hydraulic gate PQR
whose weight is negligibly small compared
to the hydrostatic forces. The gate opens
when h exceeds
(Type of Vertical Surface)
(P)
(Q)
(A) 1.414 b
(B) 0.500 b
(C) 2.732 b
(D) 1.732 b
Fluid Mechanics
46
4.9
A vertical gate 6m x 6m holds water on one
side with the free surface at its top. The
moment about the bottom edge of the gate of
the water force will be ( w is the specific
4.13
weight of water)
(A) 18 w
(B) 36 w
(C) 72 w
(D) 216 w
4.10 A container having a square cross-section has
water filled up to a height of 0.6m . The net
force on one side and the location of the
centre of pressure from the bottom are given
respectively by
4.14
Kulkarni’s Academy
The vertical force on a submerged curved
surface is equal to the
(A) Force on the vertical projection of the
curved surface
(B) Force on the horizontal projection of the
curve surface
(C) Weight of the liquid vertically above the
curved surface
(D) Product of the pressure at the centroid
and the area of the curved surface
The horizontal and vertical hydrostatic forces
Fxand Fy on the semi-circular gate, having a
width w into the plane of figure,
Are
(A) 264.5Nand0.1m
(B) 600.5 Nand0.4m
(C) 1058.4Nand0.3m
(D) 529.2 Nand0.2m
4.11 A rectangular tank with length, width and height
in the ratio 2 :1: 2 is filled completely with
water. The ratio of hydrostatic force at the
bottom to that on any LARGER vertical
surface is
4.12
(A) 1/2
(B) 1
(C) 2
(D) 4
Choose the correct statements about
horizontal component of resultant hydrostatic
pressure on a curved submerged surface
(A) It is equal to the product of pressure at
the centroid and the curved area
(B) It is equal to the weight of the liquid
above the curved surface acting at
0.5depth of the surface
(C) It is equal to the projected area of the
surface on a vertical plane multiplied by
the pressure at the centre of gravity of
area
(D) It is equal to the weight of the liquid
above the curved surface multiplied by
the projected area on a vertical plane.
(A) Fx ghrw and Fy 0
(B) Fx 2ghrw and Fy 0
(C) Fx 2ghrw and Fy gwr 2 / 2
(D) Fx 2ghrw and Fy rgwr 2 / 2
4.15
A dam is having a curved surface as shown in
the figure. The height of the water retained by
the dam is 20m , density of water is
1000 kg/m3 .Assuming g as 9.81m/s 2 , the
horizontal force acting on the dam per unit
length is
Kulkarni’s Academy
4.16
47
(A) 1.962 102 N
(B) 2 105 N
(C) 1.962 106 N
(D) 3.924 106 N
Hydrostatic Force
4.19
surface P-Q due to the water in the tank.
Note, f z is the fore per unit width along y.
Choose the correct combination of true
statements from the following:
P.
Find the vertical hydrostatic force, f z ' on the
The surface P - Q is shaped like a quartercylinder of radius R. The atmospheric
pressure is 0
For a horizontal plane surface in a liquid
at rest, the centre of pressure is at the
centroid of the surface
Q. For an inclined plane surface submerged
in a liquid at rest, the centre of pressure
is always lower than the centroid of the
surface.
R. The horizontal component of the force
exerted on a curved surface in a liquid at
rest acts at the centroid of the curved
surface.
4.17
(A) P, Q
(B) Q, R
(C) P, R
(D) P, Q, R
(A) w g R 2 R 2
4
A circular cylinder of diameter 2m and spanwise length 3m placed in a tank of water
divides it in two parts as shown in figure. The
net vertical force on the cylinder due to the
fluid is ( g 10 m/ s 2 )
(B) P0 R w g R 2 R 2
4
(C) w g R 2
4
(D) P0 R w g R 2
4
4.20
4.18
(A) 9428N
(B) 47124 N
(C) 70686 N
(D) 23562 N
A cylindrical gate rests on the crest of a
spillway and water stands up to the top of the
gate. Diameter of the gate is 1m . The vertical
component of the pressure force per meter
length of the gate is
(A)
(C)
gkN
8
(B)
gkN
4
gkN
2
(D) gkN
In which one of the following arrangement
would the vertical force on the cylinder due
to water be the maximum
(A)
(B)
Fluid Mechanics
48
Kulkarni’s Academy
A
Answer Key
4.1
D
4.2
B
4.3
C
4.4
C
4.5
B
4.6
C
4.7
B
4.8
D
4.9
D
4.10
B
4.11
B
4.12
C
4.13
C
4.14
D
4.15
C
Assertion (A): For a vertically immersed
surface, the depth of the centre of pressure is
independent of the density of the liquid.
4.16
A
4.17
C
4.18
A
4.19
A
4.20
D
4.21
C
Reason (R): Centre of pressure lies above the
centroid of area of the immersed surface.
E
(C)
(D)
4.21
(A) Both A and R are individually true and
R is the correct explanation of A
(B) Both A and R are individually true but R
is not the correct explanation of A
Explanation
4.1
(D)
4.2
(B)
4.3
(C)
(C) A is true but R is false
(D) A is false but R is true
4.22
A semi-circular gate of radius 1 m is placed
at the bottom of a water reservoir as shown in
figure. The hydrostatic force per unit width of
the cylindrical gate in y-direction is_____
kN . The gravitational acceleration g
9.8 m/s 2 and density of water 1000 kg/m 3 .
F gAx
1
F g (1). sin 45o
2
F
g
2 2
Hence, the correct option is (C).
Kulkarni’s Academy
4.4
49
Hydrostatic Force
Weight = Frictional force
(C)
800 9.81 = F
800 9.81 = [10009.81 (42) 2)
= 0.05
Hence, the correct option is (C).
4.7
(B)
P – 4, Q – 2, R – 3, S – 1.
(P)
x 0.7
1.3
1.35m
2
F g xA
1000 10 1.35 (1.5)2 {g = 10 m/s2
4
X cp x
23.8 kN (24 kN approx.)
4.5
(B)
The point of application of horizontal force
on curve surface, at the centre of pressure of
corresponding area.
Ax
3
23856.46 N
Hence, the correct option is (C).
I cg
d
bd 12
2 bd d 2
d d 4d
2 6
6
2d
X cp
3
(Q)
Hence, the correct option is (B).
4.6
(C)
X cp x
I cg
Ax
3
2d
bd 36
3 bd 2d
2 3
F wAx
F = 1000 9.81 (4 2) 2
2d d 9d
3 12 12
3d
X cp
4
Fluid Mechanics
(R)
50
Kulkarni’s Academy
P
h
G
CP
Fpq
Q
X cp x
I cg
FPQ wAx
Ax
h w2
w h
2 2
3
bd 36
d
3 bd d
3
2
X cp
d
2
(S)
FQR wAx
FQR wAx
X cp x
I cg
=wbh
Ax
d
d 64
2 2 d
d 2
4
4
X cp
5d
8
Hence, the correct option is (B).
4.8
FPQ
(D)
h
b
FQR
3
2
wh2 h
b
wbh
2 3
2
h2
b2
3
h 3b 1.732b
Hence, the correct option is (D).
Kulkarni’s Academy
4.9
51
Hydrostatic Force
F w A x
(D)
w 2 2 = 4 w .N (bottom)
Any larger
FLVS w 2 2 1
=4w
FB 4w
1
Fvs 4w
Hence, the correct option is (B).
2h 2 6
4m
3
3
Moment = F 2
X cp
F w A x
w 6 6 3 = 108 w
Moment = 2 108 w = 216 w
Hence, the correct option is (D).
4.10
4.12 (C)
4.13 (C)
4.14 (D)
(B)
FH w (2r b) h
2gr b h 2w
w
= 2ghrw.
2 0.6
0.4m from free surface
3
Hence, the correct option is (B).
X cp
4.11
(B)
L:W:H
2:1:2
Fy = weight = gv
r 2
=
g w
2
g r 2 w
2
Hence, the correct option is (D).
=
4.15 (C)
Fluid Mechanics
FH g xA
52
Kulkarni’s Academy
4.19 (A)
1000 9.81 10 (201)
= 1.962106 N
Hence, the correct option is (C).
4.16
(A)
4.17
(C)
R 2
2
4 R g 1 P0 R 1
3
= 1000 10 (1) 2 3
4
p0 R Pw g R 2 R 2
4
= 70685.83 N
Hence, the correct option is (A).
FV = .g.V
Hence, the correct option is (C).
4.18
4.20 (D)
(A)
F Vol.
Fv = gv
Hence, the correct option is (D).
FV = .g.V
1
= g (1) 2 1
4
2
g 103 N
8
=
g kN
8
Hence, the correct option is (A).
4.21 (C)
NOTES
Introduction:
Kinematics deals with motion of fluid without
any reference to cause of the motion i.e. force.
The fluid flow is analyzed by using
(1) Lagrangian Technique
(2) Eulerian Technique
In Lagrangian technique single fluid particle is
taken and the behaviour of this particle is
analyzed at different instant of time.
In Eulerian technique certain section is taken
and the fluid flow is analyzed at that section. Due
to its simplicity Eulerian technique is mostly
used in fluid flow behaviour.
Different types of fluid flows:
(1) Steady & unsteady flow:
dV
0
d space atgiven time
Space means (x, y, z)
(3) Laminar & Turbulent flow:
In laminar flow fluid particles move in the farm
of layer with one layer sliding over other, laminar
flow generally occurs at low velocity.
Ex: Flow of blood through veins, flow through
narrow passage.
In turbulent flow fluid particles move in highly
disorganized manner leading to rapid mixing of
particle, turbulent flow generally occurs at high
velocity.
Ex. Floods, flow of water in rivers, flow of
exhaust gases from chimney.
A flow is said to be steady flow when fluid
properties do not change with respect to time at
any given section. Otherwise the flow is
unsteady.
(4) Rotational and irrotational flow:
For steady flow
Rotation is possible when there is a tangential
force these tangential forces are associated with
viscous fluids, therefore real fluid flows are
generally rotational flows and ideal fluid flows
are irrotational flows.
dv
0;
dt given sec tion
d
dt
0
given sec tion
(2) Uniform & non uniform flow:
A flow is said to be uniform when fluid
properties (specially velocity) do not change with
respect to space at any given instant of time.
Otherwise the flow is non-uniform.
For uniform flow.
A flow is said to be rotational when fluid
particles rotates about their own mass centre
otherwise the flow is irrotational.
(5) Internal and external flows:
When the fluid flows through confined passages
(Ex: fluid flow through pipe, ducts) then the flow
is known as internal flow.
When the fluid flows through unconfined
passages (Flow of fluid over aircraft wing) then
the flow is known as external flow.
Fluid Mechanics
Note:
54
Kulkarni’s Academy
Equation of stream line:
Flow can also be categorized into 1–D, 2–D, 3–D.
When fluid properties vary in 1– direction, it
is a 1–D flow.
When fluid properties vary in 2–directions, the
flow is 2–D dimensional.
When fluid properties vary in 3–directions, the
flow is 3–D dimensional flow.
Flow can never be one dimensional because of
viscosity. Flow of fluid through a pipe can be
approximated as 1–D flow. If average velocities
are taken into consideration.
V uiˆ vˆj wkˆ
V u 2 v 2 w2
In 2– D
V uiˆ vˆj
Stream line:
It is an imaginary line or curve drawn a space also
such that a tangent drawn to any point gives
velocity vector. Stream line gives the direction of
flow as there is no component of velocity in
perpendicular direction there is no flow across
the steam line, there is flow always along a
stream line.
Stream line gives instantaneous snapshot of flow
pattern, it has no time history.
No two stream lines can intersect, a single
stream line can never intersect because
velocity is unique at any given instant of time
at a particular point.
velocity
dt
distance dx
u (in x dirn);
time
dt
dx
u
In y-dirn v
dy
dt
dy
v
dx dy
(equation of stream line in 2-D)
u
v
Similarly
dx dy dz
equation of stream line in 3-D.
u
v
w
dt
Path line:
It is the locus of single fluid particle at different
instants of time it follows Lagrangian approach.
A path line can intersect itself.
Kulkarni’s Academy
Streak line:
It is the locus of different fluid particles passing
through the fixed point.
55
Fluid Kinematics
Steady flow:
Wind direction is not changing, and wind is
flowing from north to south.
Experiment:
Unsteady flow:
Let us consider a wind is flowing 11:00 am to
11:30 am from north to south. And 11:30 am to
12:00 noon from east to west.
In steady flow stream line, path line and streak
lines are identical.
Note:
Stream line can intersect at stagnation point.
Wind flow direction
11:00am–11:30 amNorth (N) to south (S)
11:30am12:00 NoonEast(E) to west (W)
Conservation of mass (continuity
equation:
Generalised continuity equation:
( u) ( v) ( w) 0
t x
y
z
This is 3–D generalized continuity equation.
Every fluid flow must satisfy mass conservation
or continuity equation. If the fluid flow does not
satisfy continuity equation then that flow is not
possible.
All three lines (Stream line, path & streak line)
are different for unsteady flow.
“This equation is applicable for any type of fluid
flow.”
Fluid Mechanics
Case-1: Steady flow
0
t
( u) ( v) ( w) 0
x
y
z
Case-2: Incompressible ( = constant)
0
t
u
v
w
0
x
y
z
u v w
0
x y z
u v w
0
x y z
This equation is applicable for any type of
incompressible flow. [Steady or unsteady
incompressible flow]
Case-3: 2-D Incompressible flow
The continuity equation for 2-D incompressible
u v
flow is
0
x y
Case-4: 1-D flow
Continuity equation for steady
1 – Dimension flow:
mass
m * volume
Volume
Mass flow rate:
56
Kulkarni’s Academy
m AV
Where V is velocity
For steady flow
m1 m2
1A1V1 = 2A2V2
Continuity equation for 1-D steady flow.
If the flow is incompressible
1 = 2
1A1V1 = 2A2V2
A1V1 = A2V2
This equation is valid for steady,1-D and
incompressible flow.
Discharge (Q)
Volume flow rate is known as discharge.
volume A L
Q
time
t
Q = AV
A1V1 = A2V2
Q1 = Q2
In a steady, 1- Dimensional, incompressible flow
discharge remains constant.
Acceleration of a fluid particle:
du
dv
ay
dt
dt
u = f(x, y, z t)
v = f(x, y, z, t)
w = f(x, y, z)
ax
m
m Volume A L
t
t
t
ax
du
dt
u = f(x, y, z, t)
az
dw
dt
Kulkarni’s Academy
57
du u x u y u dz u dt
. . . .
dt x dt y dt z dt t dt
ax u
u
u
u
v w
x
y
z
Convective acc n
u
t
local or
or temporal
acc n
ay u
v
v
v v
v w
x
y
z t
az u
w
w
w w
v
w
x
y
z t
Fluid Kinematics
A1V1 = A2V2
V1 = V2 (velocity is not changing with respect to
space) for uniform flow convective acceleration
is equal to zero.
Steady, 1-D, incompressible:
Convective acceleration:
The acceleration due to change of velocity with
space is known as convective acceleration for
uniform flow convective acceleration is zero.
Temporal or local acceleration:
The acceleration due to change of velocity with
respect to time is known as temporal acceleration
for steady flow temporal acceleration is zero.
Type of flow
Steady &
uniform
Steady &
non uniform
Unsteady &
uniform
Unsteady &
nonuniform
Convective
acceleration
0
Local
acceleration
0
Exists
0
0
Exists
Exists
Exists
Steady, 1-D, incompressible:
Stream lines are converging convective
acceleration.
A1V1 = A2V2
A1> A2
V2> V1
Stream lines are diverging deceleration.
A1V1 = A2V2
A2> A1
V2< V1
Rotational component:
Sign convention:
Counter clock wise +ve;
Clockwise ve.
Fluid Mechanics
58
v
dxdt
tan d x
dx
If d is small
tan d = d =
v
dt
x
d v
dt x
Kulkarni’s Academy
d v
dt x
d
u
dt
y
1 d d
z
2 at dy
1 dv
u
z
2 az dy
x iˆ y ˆj z kˆ
i
j
x y
u
v
k
z
w
Short trick
u
dydt
y
tan d
dy
tand = d (d is small)
1 w v
x
2 y z
d u
dt y
d
u
(Clockwise rotation)
dt
y
In fluid mechanics angular velocity is defined as
the average angular velocity of initially two
perpendicular line segments.
y
1 u w
2 z x
z
1 v u
2 x y
Condition for irrotational flow:
x iˆ y ˆj z kˆ
For irrotational flow
=0
i.e. x = 0; y = 0; + z = 0
Kulkarni’s Academy
59
Generally, in fluid mechanics we are dealing with
2 – D flow x, y, z
z = 0
1 v u
v u
0
2 x y
x y
Vorticity: Twice the rotation is known as vorticity.
i
1
2 x
u
j
y
v
k
z
w
i
j k
vorticity 2
x y z
u v w
Fluid Kinematics
Note:
For irrotational flow
circulation = 0.
z
= 0; vorticity = 0 and
Velocity potential function ()
It is a function of space and time defined in such
a manner that its negative derivative with respect
to space gives velocity in that direction.
The negative sign is taken because the flow is in
the direction of decreasing potential.
u
v
w
x
y
z
Velocity potential function can be defined for a 3
– D Flow.
u v
0
x y
Circulation ():
It is the line integral of the tangential component
of velocity taken around a closed curve.
u v
x y x x y y
2 2
u v
2 2
x y
x y
Case-1: If
2 2
0
x 2 y 2
Velocity potential function satisfies Laplace
equation.
v
u
udx v dx dy u dy dx vdy
x
y
v u
dxdy
x y
We know that
1 v u
z
2 x y
v u
Vorticity 2z
x y
Circulation () = vorticity Area
2 2
0 continuity equation is satisfied
x
y
flow is possible.
Case-2: If
2 2
0
x 2 y 2
Velocity potential function does not satisfy
Laplace equation.
2 2
0
x
y
continuity equation is not satisfied
Flow is not possible.
Fluid Mechanics
Case-3: Rotational component
60
Kulkarni’s Academy
Case-2: If not satisfies Laplace equation.
1 v u
z
2 x y
2 2
0
x 2 y 2
1
2 x y y x
1 2
2
z
2 xy yx
z = 0 irrotational flow
2 0; rotational flow.
Velocity potential function exists only for
irrotational flow whereas stream function exists
for rotational &irrotational flow. If satisfies
Laplace equation then the flow is irrotational.
Significance of stream function
Velocity potential function exist only for
irrotational flows. i.e. the existence of velocity
potential function implies the flow is irrotational.
dx dy
u
v
Note:
Sometimes irrotational flows are also known as
potential flows.
Stream function ()
It is a function of space and time define in such a
manner that it satisfies continuity equation.
u
y
v
x
vdx = udy
{u
y
vdx udy = 0
{v
x
dx
dy 0
x
y
Equation of a particular streamline.
(x, y)
Note:
…. (1)
dx
dy
x
y
….. (2)
From equation (1) and (2)
Though velocity potential can be defined for
3 – D flows, it is difficult to define stream
function in 3–D flows therefore, stream function
is generally defined for 2 – D flows.
1 v
u
z
2 x y
1
2 x x
y y
1 2 2
z 2 2
2 x
y
Case-1: If satisfies Laplace equation.
2 2
0
x 2 y 2
z 0 ; irrotational flow.
= 0
= constant
For particular stream line, stream function
remains constant.
Kulkarni’s Academy
61
Fluid Kinematics
Q=AV
= (dx 1) v
Q v.dx
dx
y
d
dx
dy
x
y
d
dx
x
----- (a)
------ (b)
From equation a and b
Q = d = difference in stream function.
“The difference in stream function gives
discharge per unit width”.
Relationship between equipotential
Equipotential and constant stream function lines
are orthogonal (perpendicular) to each other in
flow field.
Cauchy – Reimann equations:
line and constant stream function line:
= f(x, y) = Constant
d
dx dy 0
x
y
dy
x
dx
y
x y
these equations are known as
y x
Cauchy-Reimann equations.
= f(x, y) = constant
dx
dy 0
x
y
dy
x
dz
y
dy v
slope of constant stream function line
dx u
Product of slope
v
y
x
From above equations we can say that
and u
dy u
slope of equipotential line
dx
v
d
u
u v
1
v u
Fluid Mechanics
P
5.1
Practice Questions
Which of the following statements is true?
(A) Eulerian description of fluid motion
follows individual fluid particles
(B) Lagrangian description of fluid motion
is a field description
(C) Both Eulerian and Lagrangian
description follows individual fluid
particles but in different reference
frames
(D) Eulerian description is a field description
while Lagrangian description follows
individual fluid particles.
5.2
A streamline in a fluid flow is a hypothetical
line at any instant such that
(A) The fluid velocity is not varying along
it
(B) There is no flow across it
(C) Fluid can flow across it
(D) It is always perpendicular to the main
direction of the flow
5.3
The flow field represented by the velocity
vector
V axiˆ by 2 ˆj czt 2kˆ where a, b and c are
constants is
(A) Three-dimensional and steady
(B) Two-dimensional and steady
(C) Two- dimensional and unsteady
(D) Three- dimensional and unsteady
5.4
A fluid element is said to have vorticity with
respect to a reference frame if in that
reference frame
(A) it travels along a circular streamline
(B) it travels along a circular pathline
(C) it revolves about its arbitrary point in
the flow-field
(D) it rotates about its own centre of mass
as it moves
62
5.5
Kulkarni’s Academy
The shape of the streamline, passing through
the
origin,
in
a
flow
field
u cos(), v sin() for a constant is
determined as
(A)
y x3
(B)
y x cot 2 ()
(C)
y x tan()
y x sin()
5.6
Consider the following statements:
(1) Streak line indicates instantaneous
position of particles of fluid passing
through a fixed point
(2) Streamlines are paths traced by a fluid
particle with constant velocity
(3) Fluid particles cannot cross streamlines
irrespective of the type of flow
(4) Streamlines converge as the fluid is
accelerated, and diverge when retarded.
Which of these statements are correct?
(A) 1 and 4
(B) 1, 3 and 4
(C) 1, 2 and 4
(D) 2 and 3
5.7 A compressible fluid is flowing steadily through
a duct whose area reduces by 40 percent from
section (1) to section (2). It is further known
that the corresponding reduction in density of
the fluid is 15 percent. Compared to the
velocity of the fluid at section (1), the
resulting velocity at section (2) is increased
by a factor of
(A) 1.67
(B) 1.96
(C) 2.69
(D) 2.96
5.8
The velocity components in the x and y
directions
are
given
by
3
u hxy 3 x 2 y, v xy 2 y 4 . The value of
4
for a possible flow field involving an
incompressible fluid is
3
4
(A)
(B)
4
3
4
(C)
(D) 3
3
(D)
Kulkarni’s Academy
5.9
63
The velocity field for flow is given by
Fluid Kinematics
5.13
V (5 x 6 y 7 z )iˆ
(6 x 5 y 9 z ) ˆj
(3 x 2 y z )kˆ
and the density varies as 0 exp(2t ) . In
order that the mass is conserved, the value of
should be
5.10
(A) - 12
(B) - 10
(C) - 8
(D) 10
In a two-dimensional flow with velocities ‘u’
and v along the x and y directions,
respectively, the convective acceleration
along the y-direction is
v
v
v
v
(A) u v
(B) v v
x
y
x
y
u
v
u
u
(D) u v
v
x
y
x
y
For a two-dimensional incompressible
irrotational flow, the x-component of
velocity u 2x 3 y . The corresponding y(C)
5.14
u
A steady flow occurs in an open channel with
lateral inflow of qm 3 /s per unit width as
component of velocity is
(A) 2 y 3x
(B)
shown in the figure. The mass conservation
equation is
(C)
5.15
2 y 3x
(D) 2 y 3x
2 y 3x
For a given location in a flow, the rate of
change of density following a fluid particle
D
is
u v w ,
x
y
z
Dt t
2.4 kg/(m3 /s) . If the density at that point is
5.11
(A)
q
0
x
(B)
Q
0
x
(C)
Q
q 0
x
(D)
Q
q 0
x
Which of the following two-dimensional
incompressible velocity fields satisfies the
conservation of mass
(A) u x, v y
velocity field () at the point is
(A)
5.16
0.5 s1
(B)
0.5 s1
(C) 2 s1
(D) 2 s1
Water enters a pipe of cross-sectional area A1
that branches out into sections of equal areas
A2 and A3 , as shown in the figure below.
(B)
u 2x, v 2 y
At one instant, the flow velocities are
V1 2 m/s, V2 3 m/s and V3 5 m/s . At
(C)
u xy, v xy
another instant, V1 3 m/s and V2 4 m/s .
(D) u x 2 y 2 , v 0
5.12
1.2 kg/m3 , then the divergence of the
What is the value of V3 at this instant?
In a steady one dimensional flow the velocity
x
u 5 / 1
. The
3
acceleration at x 0 is given by
‘u’ is given by
(A)
14.43 m/s2
(B)
25 m/s2
(C)
+43.3 m/s2
(D)
+0.0693 m/s2
(A) 5 m/s
(C) 7 m/s
(B) 6 m/s
(D) 8 m/s
Fluid Mechanics
5.17 The velocity of an incompressible fluid flow
is
given
U ( Px Q)iˆ Ryjˆ Stkˆ m/s
64
Kulkarni’s Academy
3
where, P 3 s 1 , Q 4 m/s, R 3 s1 and
S 5 m/s2 , x and y are in m and t in s:
The local and convective acceleration
components at x 1 m, y 2 m and t 5 s ,
are respectively
(A) 5kˆ m/s2 and ( 3iˆ 18 ˆj ) m/s2
5.21
0.5 m /s is flowing in the duct and is found
to increase at a rate of 0.2 m3 /s . The local
acceleration (in m3 /s ) at x 0 will be
(A) 1.4
(B) 1.0
(C) 0.4
(D) 0.667
The relation that must hold for the flow to be
irrotational is
(A)
u v
0
y x
(B)
u v
x y
(C)
2u 2 v
0
x 2 y 2
(D)
u
v
y
x
(B) zero and ( 3iˆ 18 ˆj ) m/s2
(C)
5kˆ m/s2 and (18iˆ 3 ˆj ) m/s2
(D) 5kˆ m/s2 and (3iˆ 18 ˆj ) m/s2
5.18
5.22
In a steady flow through a nozzle, the flow
velocity on the nozzle axis is given by
v u0 (1 3 x / L)i , where x is the distance
along the axis of the nozzle from its inlet
plane and L is the length of the nozzle. The
time required for a fluid particle on the axis
to travel from the inlet to the exit plane of the
nozzle is
5.19
(A)
L
u0
(B)
L
ln 4
3u0
(C)
L
4u0
(D)
L
2.5u0
For a fluid flow through a divergent pipe of
length L having inlet and outlet radii of R1
and R2 respectively and a constant flow rate
of Q assuming the velocity to be axial and
uniform at any cross section, the acceleration
at the exit is
5.20
(A)
2Q( R1 R2 )
LR2
(B)
2Q 2 ( R1 R2 )
LR23
(C)
2Q 2 ( R1 R2 )
2 LR25
(D)
2Q 2 ( R2 R1 )
2 LR25
The area of a 2 m long tapered duct decreases
as A (0.5 02. x) where ‘x’ is the distance
in meters. At a given instant a discharge of
Choose the correct combination of true
statements from the following
P.
For a steady two-dimensional flow, a
streamline is identical to a streak line
Q.
For
a
steady
two-dimensional
irrotational flow, equipotential lines are
parallel to the streamlines.
R.
For
a
steady
two-dimensional
irrotational flow, equipotential lines are
orthogonal to the streak lines
S. For a unsteady flow, the streak lines are
identical to the streamlines at any given
instant
5.23
(A) P, R
(B) P, R, S
(C) Q, S
(D) P, Q
The differential form of the mass balance
equation V 0 is valid for
(A) Any flow
(B) Steady flows only
(C) Any incompressible flow
(D) Only incompressible flows that are
steady
5.24 For a two-dimensional flow, it is given that the
values of the steam function and potential
function are, respectively, A and A at a
point A. The corresponding values at another
point B are B and A , respectively.
Kulkarni’s Academy
65
The volume flow rate across A and B is
proportional to
5.25
5.26
(A)
A B
(B)
A B
(C)
A B
(D)
A B
5.28
(A) Laminar
(B) Incompressible
(C) Steady
(D) Irrotational
(A) 0.4 units
(B) 1.1 units
(C) 4 units
(D) 5 units
If for a flow, stream function exists and
satisfies the Lapalce equation, then which
one of the following is the correct
statements?
(A) The continuity equation is satisfied and
the flow is irrotational
(B) The continuity equation is satisfied and
the flow is rotational
(C) The flow is irrotational but does not
satisfy the continuity equation
A stream function is given by
2 x 2 y ( x 1) y 2 . The flow rate across a
(D) The flow is rotational
5.31
The stream function in xy-plane is given
below
1 2 3
x y
2
The circulation around a circle of radius 2
units for the velocity field u 2x 3 y and
The velocity vector for this stream function is
v 2 y is
(A)
3
xy 3i x 2 y 2 j (B)
2
3 2 2
x y i xy 3 j
2
(C)
3 2 2
x y i xy 3 j (D)
2
3
xy 2i x 2 y 2 j
2
(A)
6 units
(B)
12 units
(C)
18 units
(D)
24 units
5.32
Consider two-dimensional flow with stream
1
function ln x 2 y 2 . The absolute
2
Match the Group I (Condition) with Group II
(Regulating Fact) and select the correct
answer using the code given below the lists
Group I
(P)
Existence of stream function
value of circulation along a unit circle centred
at (x = 0, y = 0) is
(Q) Existence of Velocity floe potential
(A) Zero
(B) 1
(S)
(D)
Group II
(C)
5.29
5.30
A potential function can be defined for a flow
if and only if it is
line joining points A (3, 0) and B(0, 2) is
5.27
Fluid Kinematics
2
For a certain two-dimensional steady
incompressible flow, the horizontal and
vertical velocity components are given by
u 6 y, v 0 , where 'y' is the vertical
(R) Absence of temporal variations
Constant velocity
(1)
Irrotationality of flow
(2)
Continuity
(3)
Uniform flow
(4)
Steady flow
distance. The angular velocity and rate of
shear strain respectively are
(A) P – 2, Q – 1, R – 4, S – 3
(A) - 3 and 3
(B) 3 and - 3
(C) P – 1, Q – 2, R – 4, S – 3
(C) 3 and - 6
(D) - 6 and 3
(D) P – 1, Q – 2, R – 3, S – 4
(B) P – 2, Q – 1, R – 3, S – 4
Fluid Mechanics
5.33
66
Potential function is given as x y
2
2
5.37
what will be the stream function () with the
condition ( 0) at x y 0?
5.34
5.35
5.36
(A)
2xy
(B)
x2 y2
(C)
x2 y2
(D)
2x 2 y 2
Kulkarni’s Academy
Let and represent, respectively, the
velocity potential and stream function of a
flow field of an incompressible fluid. Which
of the following statements are TRUE ?
P.
exists for irrotational flows only
Q.
exists for both irrotational and
If the stream function is given by y 3xy ,
then the velocity at a point (2, 3) will be
R.
rotational flows
exists for rotational flows only
(A) 7.21 unit
(B) 10.82 unit
S.
(C) 18 unit
(D) 54 unit
In a certain 2-D potential flow the stream line
passing through a point A = (1, 1) has the
following equation,
xy 1 . In the
neighbourhood of A, the Equi-potential line
passing through A may be approximated by
(A)
x y
(B)
x 2 y 1
(C)
2x y 1
(D)
x 2 y
5.38
A pipe has a porous section of length L as
shown in the figure. Velocity at the start of
this section is V0 . If fluid leaks into the pipe
through the porous section at a volumetric
2
rate per unit area q ( x / L) , what will be the
axial velocity in the pipe at any x? Assume
incompressible one dimensional flow i.e., no
gradients in the radial direction
5.39
exists for both rotational and
irrotational flows
(A) P, R
(B) Q, S
(C) Q, R
(D) P, Q
In an incompressible irrotational fluid
motion, if the y component of velocity at any
point ( x, y) is v 6 xy x 2 y 2 , the xcomponent of velocity at that point is given
by
(A)
v 2 xy 3( x 2 y 2 )
(B)
v 3 xy 2( x 2 y 2 )
(C)
v 3 xy 2( x 2 y 2 )
(D)
v 2 xy 3( x 2 y 2 )
The stream function for a two-dimensional
incompressible
flow
is
given
by
( px 2 qy 2 ) , where p and q are non-zero
constants. A potential function for this flow
can be determined only when
q
(A) p
(B) p q
2
(C) p q
(D) p 2q
qx3
(A) Vx V0 2
LD
(B) Vx V0
qx3
3L2
(C) Vx V0
2qx3
LD
4qx3
(D) Vx V0 2
3L D
5.40
For a general 3 - dimensional
incompressible, irrotational flow, which one
of the following statements is true?
(A) Velocity potential function can be
defined but stream function cannot be
defined
(B) Velocity potential function cannot be
defined but stream function can be
defined
Kulkarni’s Academy
67
(C) Both velocity potential and stream
function can be defined
(D) Both velocity potential and stream
function cannot be defined
Fluid Kinematics
5.43
Common data For 5.41 to 5.42
The velocity field for a 2 - dimensional flow
is
u
5.41
U0 x
U 0 y
, v
L
L
The above flow can be described as
5.44
(A)
y 0, z
v
2h
(B)
y 0, z
v
h
(C)
y
y
v
, z
h
h
(D)
y
y
v
, z
h
h
(A) Rotational and compressible
The power required to keep the plate in
steady motion is
(B) Irrotational and compressible
(A) 5 104 watts
(C) Rotational and incompressible
(B) 105 watts
(D) Irrotational and incompressible
5.42
The rate of rotation of a fluid particle is given
by
(C)
If L = 0.2 m and the result of total
(D) 5 105 watts
acceleration in x - and y - directions at
( x L, y L) is 10 m/s2 ; the value of U U 0
in m/s is
(A) 1.414
(B) 2.38
(C) 1.19
(D) 11.90
Common data for 43 to 44 are given below. Solve
the problems and choose the correct answers.
The laminar flow takes place between closely spaced
parallel plates as shown in figure below. The velocity
y
profile is given by u V . The gap height, h, is 5
h
mm and the space is filled with oil (specific gravity
2.5 105 watts
Statement for Linked Answer Questions 5.45 to 5.46
The gap between a moving circular plate and a
stationary surface is being continuously reduced, as
the circular plate comes down at a uniform speed V
towards the stationary bottom surface, as shown in
the figure. In the process, the fluid contained between
the two plates flows out radially. The fluid is
assumed to be incompressible and inviscid.
= 0.86, viscosity 2 104 N-s/m2 ). The bottom
plate is
stationary and the top plate moves with a steady
velocity of V = 5 cm/s. The area of the plate is
2
0.25 m .
5.45
The radial velocity at any radius r, when the
gap width is h, is
(A)
vr
Vr
2h
(B)
vr
Vr
h
(C)
vr
2Vh
r
(D)
vr
Vh
r
Fluid Mechanics
5.46 The radial component
acceleration at r R is
of
(A)
3V 2 R
4h 2
(B)
V 2R
4h 2
(C)
V 2R
2h 2
V 2R
(D)
4h 2
5.47
5.48
68
the
fluid
During an experiment, the position of a fluid
particle is monitored by an instrument over a
time period of 10 s. The trace of the particle
given by the following figure represents
(A) Stream line
(B) Streak line
(C) Path line
(D) Timeline
Smoke is released from a tall chimney from
ABC industry. Wind blows from north to
south upto time T and there after, the
direction changes from east to west.
After time T, streak lines for smoke particles
coming out of the chimney are oriented as
(A)
(B)
(C)
(D)
5.49
Kulkarni’s Academy
In a given flow field, the velocity vector in
Cartesian coordinate system is given as:
V ( x 2 y 2 z 2 )iˆ
( xy yz y 2 ) ˆj ( xz z 2 )kˆ
What is the volume dilation rate of the fluid
at a point where x 1, y 2 and z 3 ?
5.50
5.51
(A) 6
(B) 5
(C) 10
(D) 0
A reservoir connected to a pipe line is being
filled with water, as shown in the figure. At
any time t, the free surface level in the
reservoir is h. Find the time in seconds for the
reservoir to get filled upto a height of 1 m. If
the initial level is 0.2 m _____.
Water (density = 1000 ) at 0.1 and alcohol
(specific gravity = 0.8) at 0.3 are mixed in a
T-junction as shown in
the figure. Assuming all the flows to be
steady and incompressible, average density
of the mixture of alcohol and water, in , is
(A) 340
(B) 560
(C) 680
(D) 850
Kulkarni’s Academy
5.52
69
Steady state incompressible flow through a
pipe network is shown in the figure. Inlets
marked as (1), (2), and (3) and exit marked as
(4), are shown with their respective
diameters. The exit flow rate at (4) is 0.1 . A
20% increase in flow rate through
(3) results in a 10% increase in flow rate
through (4). The original velocity through
inlet (3) is____ m/s.
5.53
A cylindrical tank of 0.8 m diameter is
completely filled with water and its top
surface is open to atmosphere as shown in the
figure. Water is being discharged to the
atmosphere from a circular hole of 15 mm
diameter located at the bottom of the tank.
The value of acceleration due to gravity is .
How much time (in seconds) would be
required for water level to drop from a height
of 1 m to 0.5 m?
(A) 188
(B) 266
(C) 376
(D) 642
Fluid Kinematics
A
Answer Key
5.1
D
5.2
B
5.3
D
5.4
D
5.5
C
5.6
B
5.7
B
5.8
D
5.9
C
5.10
C
5.11
B
5.12
A
5.13
A
5.14
C
5.15
C
5.16
D
5.17
A
5.18
B
5.19
C
5.20
C
5.21
A
5.22
A
5.23
C
5.24
A
5.25
D
5.26
C
5.27
B
5.28
D
5.29
A
5.30
A
5.31
B
5.32
A
5.33
A
5.34
B
5.35
A
5.36
D
5.37
D
5.38
A
5.39
C
5.40
A
5.41
D
5.42
C
5.43
A
5.44
C
5.45
A
5.46
C
5.47
C
5.48
B
5.49
B
5.50
20
5.51
D
5.52
17.68
5.53
C
Fluid Mechanics
E
70
5.8
Explanation
5.1
(D)
5.2
(B)
5.3
(D)
5.4
(D)
5.5
(C)
dx
dy
cos sin
sin dx cos dy
y 3 2 xy 2 xy 3 / 4 4 y 3 0
–3=0
=3
Hence, the correct option is (D).
5.9
= 0 exp (–2t)
u v w 0
t x
y
z
x. sin = y. cos
– 2 + (5) + (5) + () = 0
y = x tan
(–2+10+) = 0
Hence, the correct option is (C).
(B)
5.7
(B)
A1
(C)
u = 5x+6y + 7z
v = 6x+5y + 9z
w = 3x+2y + z
= Constant
5.6
(D)
u = xy3 – x2y
3
V xy 2 y 4
4
u v
0
x y
dx dy
u
V
Kulkarni’s Academy
= – 8
Hence, the correct option is (C).
5.10 (C)
q m3/s/m
take width = dx
q per unit width
A2
Q
dx
q dx
1A1V1 = 2A2V2
2 = 0.85 1
Q
1A1V1 = 0.851 0.6A1V2
V1 = 0.51 V2
V2 = 1.96V1
Hence, the correct option is (B).
Q qdx Q
Q
Q
dx
x
Q
d/x
x
Q
q 0
x
Hence, the correct option is (C).
Kulkarni’s Academy
5.11 (B)
(A)
u
5
x
1
3
v u
ax u
x t
5
5
1
ax
2
x
3
x
1
1
3
3
ax at x = 0
1
5x 5
14.43 m/s2
3
Hence, the correct option is (A).
5.13
(A)
5.14
(C)
Fluid Kinematics
v u
x y
f(x) = 3
f (x) = 3x
V = –2y + 3x
Hence, the correct option is (C).
2-D incompressible flow
u v
0
x y
Option (a)
x y 1 1 0
x
y
(b) 2 x 2 y 2 2 0
x
y
Hence, the correct option is (B).
5.12
71
u = 2x + 3y
2-D incompressible, irrotational flow
u v
0
x y
v
u
y
x
v
2
y
V = – 2 y
V = -2y + f(x)
v
0 f ( x)
x
As flow is irrotational. so,
z = 0
5.15 (C)
Generalised continuity equation is
u v w 0
t x
y
z
u
v
. u
t
x
x
y
w
v
w
0
y
y
z
u
v
w
2.4 (Given)
t
x
y
z
u v w
2.4 0
x y z
u v w
1.2
2.4
x y z
u v w
2s 1
x y z
Hence, the correct option is (C).
5.16 (D)
A1V1=A2V2 + A3V3
2A1 = 3A2 + 5A3
A2 = A3 (Given)
2A1 = 8A2
A1
4
A2
At another instant
V1 = 3 m/s ; V2 = 4m/s
3A1 + 4A2 + A2V3
3A1 + (4+V3) A2
A
3 1 4 V3
A2
3 4 4 V3
V3 = 12 – 4 = 8m/s
Hence, the correct option is (D).
Fluid Mechanics
5.17
72
(A)
Kulkarni’s Academy
5.19 (C)
Given : u = Px-Q, v = Ry, w = st
u
u
u u
ax u. v w
x
y
z t
Temporal Acceleration
u
v
w
0; 0;
s 5m / s 2
t
t
t
R1
R2
L
at = 0+0+5 = 5 K̂ m/s2
convective acceleration
ac x uP V (0) w(0)
Q = constant
Steady, 1-D incompressible
V uiˆ yˆj wkˆ
ac x 3u 3 px Q 3(3(1) 4) 3
ac y 4(0) VR w(0) 3V
= -3(Ry) = –3 (–32) = 18
0
ac z
3iˆ 15 ˆj m/s2
V=u
a a x iˆ
a ax u
Hence, the correct option is (A).
5.18
au
(B)
u u
x t
{
u
0 steady flow
t
u
u
v
x
x
Q = AV
L
V
Q
A
a
Q Q Q2 1
A x A
A x A
dx
x
Let us calculate a at a distance x
3x
V u0 1 iˆ
L
dis tan ce
time
velocity
dt
R
R R1 R 2
R
R1
dx
3x
u 0 1
L
T
L
dx
dt
0
0 3x
u0
L
x
L
L
1 3x L
n 1 T
u0
L 3 0
L
T
n4
3u 0
Hence, the correct option is (B).
T
R2 R1
L
Kulkarni’s Academy
73
Fluid Kinematics
A = 0.5 – 0.02x
y
x
y = x tan
= R – R1 = x tan
R = R1 + x tan
R R1
tan 2
RL
tan
x
R2 R1
L
R = R1 + Kx
A = R2
A = (R1+Kx)2
R R1
a
2
Q2
R1 kx
a
=
2
Q Au
R1 kx
2
2 R1 kx 3 k
du 1 dQ
dt A dt
du
1
0.2
dt 0.5 0.02x
at x = 0
= 20.2 = 0.4 m/s2
Hence, the correct option is (C).
5.21 (A)
1 u v
wz 0
2 y x
5
At x = L
R1 kx R1
Q
d Q
A
dt A
For irrotational flow
2Q 2 k
2 R1 kx
u
R2 R1
K
L
2
R1 kx
x
Q2
2
dQ
0.2m3 / sec
dt
1
x R1 kx 2
Q2
R1 kx
Q = 0.5 m3/sec
R2 R1
L
L
R1 kx R2
u v
y x
Hence, the correct option is (A).
2Q R2 R1 1
.
2
L R25
2
aexit
aexit
2Q 2 ( R1 R2 )
2 LR25
5.22 (A)
5.23 (C)
. V = 0
Hence, the correct option is (B).
5.20
(C)
u v w
0
x y z
valid for any incompressible flow
Hence, the correct option is (C).
5.24 (A)
2m
5.25 (D)
Fluid Mechanics
5.26
74
(C)
5.29 (A)
= 2x2y + (x+1) y2
at (3,0)
1 = 2(3)2(0) +(3+1)(0)2
1 = 0
2 at (0,2)
2 = 0+(0+1)4 = 4
2 – 1 = 4 – 0 = 4 units
Hence, the correct option is (C).
5.27
5.28
Kulkarni’s Academy
u = 6y v = 0
1 v u 1
wz 0 6 3
2 x y 2
Rate of shear strain
1 v u 1
0 6 3
2 x y 2
Hence, the correct option is (A).
5.30 (A)
(B)
U = 2x+3y
V = –2y
Radius = 2 unit
u
v u
Circulation area
x y
= (0–3)4
Hence, the correct option is (B).
v
1
2
n x 2 y 2 ; r 1
2
2
0
xy xy
Continuity equation is satisfied.
1 v u
wz 0
2 x y
U
x
u v
0
x y
(D)
y
Flow is irrotational.
rd
Hence, the correct option is (A).
5.31 (B)
Given
2
2
2
x +y = r
1
n r
2
1 1
r 2 r
2
.rd
0
r
2
1
r.d
0
2r
1 2
2
0
2
2
Hence, the correct option is (F).
1
2
x2 y3
u
v
1
3
x 2 3 y 2 x 2 y 2 ˆj
y
2
2
1
2 xy3 xy3 ˆj
x 2
3
2
x 2 y 2iˆ xy 3 ˆj
Hence, the correct option is (B).
5.32 (A)
Kulkarni’s Academy
5.33 (A)
2x
y
= 2xy + C
At x = 0; y = 0 ;
= 2xy
Hence, the correct option is (A).
(B)
Given
= 3xy
u
3x
y
at (2,3)
u = – 6 unit
V
3y
x
at (2,3)
V=9
V ulˆ vˆj
Vˆ u 2 v 2 36 81 10.82 units
Hence, the correct option is (B).
5.35
Fluid Kinematics
5.36 (D)
= x2–y2
= ? at x = y = 0
By using C-R equations
x y
5.34
75
(A)
Stream line passing through a point A (1,1)
xy = 1
xdy + ydx = 0
dy
y
1
m1
1
dx
x
1
Equation potential line m1 m2 = –1
then m2 = 1
y-y1 = m(x-x1)
y-1 = 1 (x-1)
x=y
Hence, the correct option is (A).
Q
Q0
qx 2
2
L
V0
D
x=0
dx
x=L
x
Q0 = AV0
Q0
V0
A
qx 2
dQ 2 Dxdx
L
q D x3
Q 2
C
L 3
At x=0, Q = Q0 ; C = Q0
q d x3
Q 2
Q0
L 3
Q Q0
qDx 3
Velocity =
A A 3L2 D 2
4
3
4qx
Velocity V0 2
3L D
Hence, the correct option is (D).
5.37 (D)
5.38 (A)
5.39 (C)
Given
= Px2 + qy2
Stream function is valid for both rotational
and irrotational flow, but potential function
is valid for only irrotational flow.
Then it should satisfies Laplace equation
2 2
0
x 2 y 2
2P + 2q = 0
P=–q
Hence, the correct option is (C).
5.40 (A)
Fluid Mechanics
5.41
76
(D)
5.43 (A)
2-D flow
ux
u y
u 0
v 0
L
L
u v u0 u0
0 i.e incompressible
x y L L
1 v u
wz
2 x y
1
0 0 0 i.e irrotational
2
Hence, the correct option is (D).
5.42
Kulkarni’s Academy
(C)
10 a x iˆ ayˆj
u xu
u
10 x 0 10
x
L L
u02 0.2 10
a u
y
h
Hence, the correct option is (A).
wz
5.44 (C)
F
AV
h
2 104 0.25 5 102
5 103
= 0.510–3N
Power = FV
= 0.510–3510–2 =2.510–5 Watt
Hence, the correct option is (C).
L = 0.2
ax u
1 v u
wz
2 x y
1
y
0
2
h
u0 = 1.414 m/s
u
u
u u
v w
x
y
z t
5.45 (A)
A R 2
V
a a x2 a y2
ax
u0 x u0 x u0 y
0 0
L x L L
u0 x u0 u02 x
ax
L L L2
Similarly a y
2
0
2
u x
L
2
u2 x u2 x
10 02 02
L L
2
2
u2x
2 02 10
L
2
u2x
102 2 02
L
X = L, L = 0.2
2 u04 L2
100
50 L2 u04
L4
U0 = 1.189 m/s
Hence, the correct option is (A).
A 2Rh; V Vr
h
R2V = 2Rh Vr
VR
Vr
2h
Hence, the correct option is (A).
5.46 (B)
ar u
Vr
u
r
Vr Vr V
r 2h 2h
V 2r
4h 2
Hence, the correct option is (B).
5.47 (C)
ar
5.48 (B)
Kulkarni’s Academy
77
Fluid Kinematics
0.11.1 1.2Q3 Q3 0.1
5.49 (B)
Volume dilation rate
u v w
x y z
2x+x+z+2y+x–2z
At (1,2,3)
2(1) + 1 + 3 + 2 (2) + 1 – 2 (3) = 5
Hence, the correct option is (B).
5.50
20
0.11 0.2Q3 0.1
0.01
Q3
0.05 m3 /s
0.2
By using continuity equation
A3V3 Q3
Q
0.05
17.68 m/s
V3 3
V3 (0.06) 2
4
Hence, the correct answer is 17.68.
5.53 (C)
Time
Volume to be filled (m3 )
Volume flow rate (m3 /s)
(0.5)2 0.8
4
20 sec
2
(0.1) 1
4
5.51
(D)
Given data :
water 1000 kg/m3
Qwater 0.1 m3 /s
alcohal 0.8 1000 800 kg/m3
Qalcohal 0.3 m3 /s
Average density of mixture of alcohol and
water
1000 0.1 800 0.3 (kg/s)
850 kg/m3
0.1 0.3
(m3 /s)
Hence, the correct option is (D).
5.52
17.68
Q1 Q2 Q3 Q4
Q1 Q2 1.20Q3 0.11.1
Q1 Q2 0.11.1 1.2Q3
Q va a 2 gH
dH
Velocity
dt
dH
Discharge A
= Negative sign is taken
dt
dH
because height is decreasing A
a 2 gH
dt
A dH
dt
a 2g H
t
H
A 1 2 dH
0 dt a 2 g H H
1
A 2
H 21/2 H11/2
t
a 2g
Note : From above equation, we can say that time
required to empty a tank of height H is proportional
to H 1/2 .
(0.8) 2
2
4
t
(0.5)1/2 1
(0.015) 2 2 9.81
4
t 376.1726 sec
Hence, the correct option is (C).
NOTES
Generally, forces acting on the fluid element are
pressure force (FP), gravity force (Fg) and viscous
force (Fv). In Navier–stokes equation all these
three forces are taken into consideration. In
Euler’s analysis viscous forces are neglected
only pressure and gravity forces are taken into
consideration.
6.1 Euler’ Equation:
Assumptions:
1. Flow is not viscous
2. Flow is along the stream line.
Stream wise direction
dz
ds
dz dscos
cos
From Newton second law of motion
F ma
s
PdA ( P dP )dA gdAdz
V V
dAds V
s t
V V
dP gdz ds V
s t
V V
dP gdz ds V
0
s t
In Euler equation z = vertical distance
6.2 Bernoulli’s Equation:
Bernoulli’s equation based on conservation of
energy principle.
Assumptions:
1. Flow is non – viscous
2. Flow is along a stream line
3. No energy is supplied and no energy is taken
out from the fluid during the flow.
4. steady flow
5. Incompressible flow.
Steady
flow
V V
dP gdz ds V
0
s t
a F ( s, t )
For steady flow a F ( s )
V V VdV
ds
s
V
dP gdz ds V
0
s
Kulkarni’s Academy
dP gdz VdV 0
dP
gdz VdV 0
By integrating above equation
dP
gdz VdV 0
= C incompressible
P
gz
2
V
Constant
2
This equation is known as classical Bernoulli’s
equation. In the above equation each term
represents energy of the fluid per unit mass.
6.3 Bernoulli’s theorem:
In a steady incompressible non – viscous flow
along a stream line the sum of pressure energy,
kinetic energy and potential energy is constant.
P
V2
gz const.
2
79
Fluid Dynamics
V2
2. Velocity head
:
2g
It is the height by which fluid falls in a
frictionless environment to reach a particular
velocity.
V 2 gh
h
V2
2g
3. Potential energy head (z):
It is the vertical distance with respect to same
reference line.
4. Piezometric head:
The sum of pressure head and potential
energy head is known as piezometric head.
P
z = piezometric head.
w
6.5 Bernoulli’s equation for a horizontal
stream line:
P V2
z const.
g 2g
In this equation each term represent energy per
unit weight.
6.4 Various heads in fluid mechanics:
1. Pressure head [P/w]:
The height of which the fluid rises due to
pressure when a piezometer is connected
known as pressure are head.
P1 V12
P2 V22
z
z
g 2g 1 g 2g 2
P1 V12
P V2
2 2
g 2g g 2g
Bernoulli’s equation for a real fluid problem:
o + gh = P
wh=P
P
P
h or
w g
P1 V12
P2 V22
z1
z2 hL
w 2g
w 2g
hL = head loss between 1 & 2.
Fluid Mechanics
Irrotational flow:
80
Kulkarni’s Academy
2
2
P1
P2
V2 V1
z
z
h
1
2
2g
w
w
For horizontal pipe z1= z2
P1
P1 V12
P V2
z1 2 2 z2 c12 50
w 2g
w 2g
2
3
w
V22 V12 2 gh
2
4
P3 V
P V
z3 4
z4 c34 50
w 2g
w 2g
Q = A1V1 = A2V2
V1
P V2
P V2
1 1 z1 4 4 z4 50
w 2g
w 2g
In case of irrotational flow Bernoulli’s equation
can be applied between any two points
throughout the flow field. Because stream line
constants are same for different stream lines for
irrotational flow.
Rotational flow:
2
3
2
4
P3 V
P V
z3 4
z4 hL c34 25
w 2g
w 2g
In case of rotational flow Bernoulli’s equation
must be applied only for a particular stream line
because stream line constants are different for
different stream line.
Q
A1
V2
Q
A2
Q2 Q2
h 2g
A12 A22
A2 A2
Q2 1 2 2 2 h 2 g
A1 A2
Q
P1 V12
P V2
z1 2 2 z2 c12 30
w 2g
w 2g
P2 V22 V12
h
w
2g
A1 A2 2 gh
A12 A22
As no losses is assumed while deriving this
equation this discharge is known as ideal
discharge or theoretical discharge.
Qth
A1 A2 2 gh
A12 A22
h calculation:
6.6 Application of Bernoulli’s equation:
[1] Venturimeter:
It is used for calculating discharge.
XS
P1
P
H x m H 2
w
s
w
P1 V12
P V2
z1 2 2 z 2
w 2g
w 2g
P1 P2
s
x m 1 h
w
s
Kulkarni’s Academy
81
Fluid Dynamics
6.7 Principle of Venturimeter:
By reducing area in a steady incompressible flow
velocity increasing (from continuity equation)
this results in decrease in pressure (from
Bernoulli’s equation). Due to this pressure
difference, there will be manometric fluid
deflection. When a differential manometer is
connected by measuring this deflection (x),
discharge can be calculated.
A2 A2
Q2 1 2 2 2 2 g (h hL )
A1 A2
Qact
A12 A22
…..(2)
From equation (1) and (2)
Cd A1 A2 2 gh
A12 A22
6.8 Coefficient of discharge (Cd):
Cd
It is defined as the ratio of actual discharge to the
theoretical discharge. Cd depends on type of flow
(Reynolds No) and area ratio.
A1 A2 2 g h hL
A1 A2
h hL 2 g
A12 A22
h hL
h
6.9 General proportions of a
Venturimeter:
As Venturimeter is gradually converging &
diverging device losses are less and hence Cd is
0.94 to 0.98.
Cd
Qact
Qth
1 1
d 2 to d1
3 2
Qact Cd Qth
Qact
Angle of convergence – 20 to 22o
Cd . A1 A2 2 gh
A12 A22
…. (1)
Apply Bernoulli’s equation between 1 and 2 for
real fluid flow
Angle of divergence – Less than 7o
Note:
The angle of divergence is generally kept < 7o in
order to avoid flow separation.
P1 V12 P2 V22
hL
w 2g w 2g
If d2 is very low then Pressure decreases, chances
of cavitation will be more.
P1 P2
V2 V2
hL 2 1
w w
2g 2g
6.10 Orifice meter:
h hL
V V
2g
V22 V12 2 g h hL
2
2
2
1
Q = A1V1 = A2V2
V2
Q
Q
;V1
A2
A1
Q2 Q2
2 g (h hL )
A22 A12
This device is used for finding out discharge and
it is the cheapest instrument for calculating
discharge.
It is based on the same principle as that of
Venturimeter.
Fluid Mechanics
It is a circular disc with a circular hole.
Coefficient of contraction (Cc)
Cc
a2 vena controcta area
a0
orifice area
82
Kulkarni’s Academy
6.11 Pitot tube:
It is used for finding the velocity of flow.
Case-1: Velocity in open channels.
a2 = Cc a0
apply continuity equation between 1 & 2
a1v1 = a2v2
V1
cav
a 2 v2
v1 2 0 2
a1
a1
Apply Bernoulli’s equation between 1 and 2
Stagnation point is a point at which velocity is
brought to be rest isentropically.
P1 v12 P2 v22
w 2g w 2g
P1 P2
v v
h
w
2g
2
2
2
1
v v 2 gh
2
2
2
1
C 2a 2
v22 1 c 2 0 2 gh
a1
V2
2 gh
c 2a 2
1 c 20
a1
Q
P1 V12 P2
w 2g w
h0
V12
h h0 v1 2 gh
2g
v1 2 g (dynamic head )
v1 2 g ( stagnation head static head
a2 = cca0
cc a0 2 gh
V2 0
V12
h dynomic head
2g
1 cc2 a02
a12
Discharge (Q) = a2v2
P1 V12 P2 V22
w 2g w 2g
a2
1 02
a1
1
a02
a12
cd .a1a0 2 gh
a12 a02
As the area reduction is sudden in orifice meter
losses are more and hence Cd of orifice meter is
less. (0.68 – 0.76).
Static head + Dynamic head = stagnation head
Case-2: Velocity in pipes
Kulkarni’s Academy
83
Fluid Dynamics
s
P1
P
H x m xH 2
w
s
w
P2 P1
s
x m 1
w w
s
…. (2)
From (1) and (2)
V12
s
s
x m 1 v1 2 gx m 1
2g
s
s
If the specific gravity (s.g.) of manometric fluid
is less than the s.g. of flowing fluid then the
equation for velocity is
s
v1 2 gx 1 m
s
Note:
Actual velocity = Cv theoretical velocity
Fx & Fy force exerted by the bend on the pipe in
x & y direction.
6.12 Relationship between Cc, Cv and Cd:
Apply momentum equation in x – direction
P1A + Fx P2A2 cos = Q[V2cos V1]…… (i)
Cd
Cd
Actual disch arg e
Theoretical disch arg e
Momentum equation in y – direction
Aact Vact
Ath Vth
Fy =P2 A2sin+ Q V2 sin
Cd = CcCv
6.13 Force on pipe bend:
Momentum equation
F = ma
m(v u )
t
F m(v u)
m AV Q
F = Q(v – u) Momentum equation
Fy – P2 A2sin = Q [V2 sin - 0]
……..(ii)
Fluid Mechanics
P
84
6.5
Practice Questions
Kulkarni’s Academy
Bernoulli's equation represents
(A) Momentum balance
6.1
The Euler's equation of motion is a
(B) Mechanical energy balance
(A) Statement of conservation of momentum
of a real fluid
(B) Statement of conservation of energy for
incompressible flow
(C) Statement of Newton's second law of
motion of an inviscid fluid
(D) Statement of generation of entropy
6.2
6.3
(D) Total energy balance
6.6
(A) 558Pa/m
(B) 698Pa/m
(C) 0
(D) 7960Pa/m
Match List - I (forms of Bernoulli's Equation)
with List - II (Units of these forms) and select
the correct answer using the code given
below the lists:
List I
Consider Euler's equation for onedimensional horizontal unsteady flow. In a
25cm diameter pipe, water discharge
increases from 30 to 150 liters /sec in 3.5
seconds. What is the pressure gradient that
can sustain the flow?
(P) p wz
V 2
2
(Q)
p
V2
gz
2
(R)
p
V2
z
w
2g
List II
Which of the following assumptions are
made for deriving Bernoulli's equation?
(1) Flow is steady and incompressible
(1) Total energy per unit volume
(2) Flow is unsteady and compressible
(3) Total energy per unit weight
(3) Effect of friction is neglected and flow is
along a stream line
(4) Effect of friction is taken into
consideration and flow is along a
stream line Select the correct answer
using the codes given below:
(A) 1 and 3
(C) 1 and 4
6.4
(C) Mass balance
(2) Total energy per unit mass
6.7
(A) P-1, Q-2, R-3
(B) P-1, Q-3, R-2
(C) P-2, Q-1, R-3
(D) P-2, Q-3, R-1
In the siphon shown in figure, assuming ideal
flow, pressure PB
(B) 2 and 3
(D) 2 and 4
Bernoulli's theorem
p V2
Z constant is valid
g 2 g
(A) Along different
rotational flow
(B) Along different
irrotational flow
streamlines
in
streamlines
in
(C) Only in the case of flow of gas
(D) Only in the case of flow of liquid
(A) PA
(B) PA
(C) PA
(D) PC
Kulkarni’s Academy
6.8
85
Fluid Dynamics
A Venturimeter of 20mm throat diameter is
used to measure the velocity of water in a
horizontal pipe of 40mm diameter. If the
pressure difference between the pipe and
throat sections is found to be 30kPa then,
neglecting frictional losses, the flow velocity
is
6.9
(A) 0.2m/s
(B) 1.0m/s
(C) 1.4m/s
(D) 2.0m/s
Air is inducted from atmosphere through a
bell-mouthed duct by the application of
suction at the other end. A glass tube with its
lower end immersed into a vessel containing
water is attached to the cylindrical part of the
duct (see figure). If the liquid level in the
glass tube rises by 25mm above the free
surface and the density of air is equal to
1.2 kg/m3 , the velocity of air in the
cylindrical portion is
6.10
(A) 28.6m/s
(B) 14.3m/s
(C) 40.4m/s
(D) 20.2m/s
A large tank with a nozzle attached contains
three immiscible, inviscid fluids as shown.
Assuming that the changes in are negligible,
the instantaneous discharge velocity is
(A)
h h
2 gh3 1 1 1 2 2
3 h3 3 h3
(B)
2 g (h1 h2 h3 )
(C)
h h 3h3
2g 1 1 2 2
1 2 3
h h 2 h3h1 3h1h2
2g 1 2 3
1h1 2 h2 3h3
6.11 The flow of a fluid in a pipe takes place from
(A) Higher level to lower level
(B) Small end to large end
(C) Higher pressure to lower
Pressure
(D) Higher energy to lower energy
6.12 A siphon draws water from a reservoir and
discharges it out at atmospheric pressure.
Assuming ideal fluid and the reservoir is
large, the velocity at point P in the siphon
tube is
(D)
(A)
2gh1
(B)
2gh2
(C)
2 g (h2 h1 )
(D)
2 g (h2 h1 )
Fluid Mechanics
6.13 A smooth pipe of diameter 200mm carries
86
6.15
water. The pressure in the pipe at section S1
(elevation: 10m ) is 50kPa . At section S 2
Kulkarni’s Academy
For a Venturimeter, which of the following
combination of statements will make a true
realistic description?
(P)
(elevation: 12m ) the pressure is 20kPa and
The area ratio ( AThroat / Apipe ) is very
close to unity
velocity is 2ms1 . Density of water is
(Q) The discharge coefficient is very close
to unity
1000 kgm 3 and acceleration due to gravity is
9.8m/s 2 . Which of the following is
(R) The angle of convergence is around 60
TRUE?
(S) The angle of divergence is around 60
(A) Flow is from S1 to S 2 and head loss is
0.53m
(B) Flow is from S1 to S 2 and head loss is
0.53m
(C) Flow is from S1 to S 2 and head loss is
1.06m
(D) Flow is from S 2 to S1 and head loss is
1.06m
6.14 A pipeline system carries crude oil of density
6.16
(A) P, Q
(B) Q, R
(C) Q, S
(D) R, S
Two Venturimeter of different area ratios are
connected at different locations of a pipeline
to measure discharge. Similar manometers
are used across the two Venturimeter to
register the head differences. The first
Venturimeter of area ratio 2 registers a head
difference 'h', while the second Venturimeter
registers '5h'. The area ratio for the second
Venturimeter is
800 kg/m 3 . The volumetric flow rate at point
(A) 3
(B) 4
1 is 0.28m 3 /s . The cross sectional areas of
(C) 5
(D) 6
the branches 1, 2 and 3 are 0.012, 0.008 and
0.004 m 2 respectively. All the three branches
are in a horizontal plane and the friction is
negligible. If the pressures at the points 1 and
3 are 270kPa and 240kPa respectively,
then the pressure at point 2 is
(A) 202kPa
(B) 240kPa
(C) 284kPa
(D) 355kPa
6.17 The pressure differential across a vertical
venturimeter (shown in figure) is measured
with the help of a mercury manometer to
estimate flow rate of water flowing through
it. The expression for the velocity of water at
the throat is
Kulkarni’s Academy
(A)
87
V22 V12 xSm
2g
Sw
Fluid Dynamics
6.20
(A) Ensure that the flow remains laminar
S
V22 V12
x m 1
2g
Sw
S
V2
(C) 2 H x m 1
2g
Sw
(B)
S
V22 V12
x m H
2g
Sw
Water is flowing with a volume flow rate Q
through a pipe whose diameter reduces to
half across a reducer. If the flow is
frictionless, compare the manometer reading
corresponding to the three different
(B) Avoid separation
(C) Ensure that the flow remains turbulent
(D) Avoid formation of boundary layer
6.21
(D)
6.18
The discharge coefficients of a Venturimeter
and an orifice meter, both installed on a pipe
of internal diameter 100mm , are 0.95 are
0.65, respectively. The venturi throat
diameter is the same as the orifice diameter.
If the pressure drop across the orifice meter is
measured as 300mm of water column, the
3 300 Note that only the pipe tilts, while
corresponding pressure drop for the
Venturimeter in mm of water column, is
approximately
the manometer always stays vertical.
(A) 205
(B) 80
(C) 140
(D) 66
inclinations of the pipe 1 300 , 2 00 and
6.22
(A) h1 h2 h3
In an orifice meter, if the pressure drop
across the orifice is overestimated by 5%,
then the PERCENTAGE error in the
measured flow rate is
(A) 2.47
(B) 5
(C) 2.47
(D) 5
6.23 The discharge coefficient, C d , of an orifice
(B) h1 h2 h3
meter is
(C) h1 h2 h3
(A) Greater than the C d of a Venturimeter
(D) h1 h3 and h1 h2
6.19
The diverging limb of a Venturimeter is kept
longer than the converging limb to
An orifice plate of 60 mm diameter and
discharge coefficient 0.6 is used for
measuring the flow rate of air
1.2 kg/m3 , 1.8 105 kgm 1s 1 through
a pipe of 100mm diameter. A manometer
(with water as the working liquid) connected
across the orifice plate reads 180mm . The air
(B) Smaller than the C d of a venturimeter
(C) Equal to the C d of a venturimeter
(D) Greater than one
6.24 A fluid jet is discharging from a 100mm nozzle
and then vena contracta formed has a dimeter
of 90mm . If the coefficient of velocity is
flow rate is approximately equal to
0.95, then the coefficient of discharge for the
nozzle is
(A) 0.3m3 /s
(B) 0.1m3 /s
(A) 0.855
(B) 0.81
(C) 0.01m 3 /s
(D) 0.003m3 /s
(C) 0.9025
(D) 0.7695
Fluid Mechanics
6.25
The operation of a rotameter is based on
(A) Variable flow area
(B) Rotation of a turbine
(C) Pressure drop across a nozzle
(D) Pressure at a stagnation point
6.26
Figure shows the schematic for the
measurement of velocity of air (density =
1.2 kg/m3 ) through a constant-area duct using
a pitot tube and a water-tube manometer. The
differential head of water
(density = 1000 kg/m3 ) in the two columns of
the manometer is 10mm . Take acceleration
due to gravity as 9.8 m/s 2 . The velocity of air
in m/s is
6.27
(A) 6.4
(B) 9.0
(C) 12.8
(D) 25.6
Water flow through a pipeline having four
different diameters at 4 Stations is shown in
the given figure. The correct sequence of
station numbers in the decreasing order of
pressure is
(A) 3,1,4,2
(B) 1,3,2,4
(C) 1,3,4,2
(D) 3,1,2,4
88
Kulkarni’s Academy
6.28 The following instruments are used in the
measurement of discharge through a pipe:
1. Orifice meter
2. Flow meter
3. Venturimeter
The correct sequence of the ascending order
of the head loss in these instruments is
(A) 1, 3, 2
(B) 1, 2, 3
(C) 3, 2, 1
(D) 2, 3, 1
6.29 An orifice meter being used for measuring flow
rate of a liquid in a pipe shows a pressure
differential of x meters of water column,
when the flow rate is 6. If the flow rate is
doubled the pressure differential in meters of
water column will be
(A) 2x
(B) 8x
(C) x 2
(D) 4x
6.30 Group I gives a list of devices and Group
II gives the list of uses.
Group I
(P) Pitot tube
(Q) Manometer
(R) Venturimeter
(S) Anemometer
Group II
1) Measuring
2) Measuring velocity of flow
3) Measuring air and gas velocity
4) Measuring discharge in a pipe
(A) P-1, Q-2, R-4, S-3
(B) P-2, Q-1, R-3, S-4
(C) P-2, Q-1, R-4, S-3
(D) P-4, Q-1, R-3, S-2
6.31 Water is flowing at 1m/s through a pipe (of
10cm I.D). With a right angle bend. The
force in Newton exerted on the bend by the
water is
5x
(A) 10 2x
(B)
2
5
5x
(C)
(D)
2x
2
Kulkarni’s Academy
89
Fluid Dynamics
Linked Answer Questions 32 to 33
Linked Answer questions 35 and 36
A free jet of water is emerging from a nozzle
(diameter 75mm ) attached to a pipe
Water enters a symmetric forked pipe and
discharges into atmosphere through the two
branches as shown in the figure. The cross-
(diameter 225mm ) as shown below.
sectional area of section 1 is 0.2 m 2 and the
velocity across section 1 is 3m/s . The density
of water may be taken as 1000 kg/m3 . The
viscous effects and elevation changes may be
neglected.
The velocity of water at point A is 18m/s .
Neglect friction in the pipe and nozzle. Use
g 9.81m/s 2
and
density
of
water
1000 kg/m 3 .
6.32
6.33
6.34
The velocity of water at the tip of the nozzle
(in m/s) is
(A) 13.4
(B) 18.0
(C) 23.2
(D) 27.1
The gauge pressure (in kPa ) at point B is
(A) 80.0
(B) 100.0
(C) 239.3
(D) 367.6
The figure shows a reducing area conduit
carrying water. The pressure p and velocity V
are uniform across sections 1 and 2. The
density of water is 1000 kg/m3 .
If the total loss of head due to friction is just
equal to the loss of potential head between the
inlet and the outlet, the V2 in m/s will
be___________.
6.35
6.36
The gauge pressure at section 1, in kPa , is
(A) 0.6
(B) 13.5
(C) 135
(D) 600
The magnitude of the force, in kN required
to hold the pipe in place, is
(A) 2.7
(B) 5.4
(C) 19
(D) 27
Fluid Mechanics
A
90
Kulkarni’s Academy
QV
Answer Key
6.1
C
6.2
B
6.3
A
6.4
B
6.5
B
6.6
A
6.7
B
6.8
D
6.9
D
6.10
A
6.11
D
6.12
C
6.13
C
6.14
C
6.15
C
6.16
B
6.17
B
6.18
C
6.19
B
6.20
B
6.21
C
6.22
A
6.23
B
6.24
D
6.25
B
6.26
C
6.27
D
6.28
C
6.29
D
6.30
C
6.31
D
6.32
D
6.33
D
6.34
8 m/s
6.35
B
6.36
A
dP
= pressure gradient
ds
V
dP ds
t
dP
Q
; Q AV
ds
t A
dP
Q
ds
A t
4
103
0.25
0.03428
2
698 pa/m
E
Explanation
6.1
(C)
6.2
(B)
Hence, the correct option is (B).
D = 25 cm
dQ 150 30 10
dt
3.5
3
m3/sec
6.3
(A)
6.4
(B)
6.5
(B)
Bernoulli’s equation is not the total energy
conservation equation because while
deriving Bernoulli’s equation heat transfer
and work transfer are not taken into
consideration therefore Bernoulli’s equation
is known as mechanical energy balance
equation.
120 103
0.03428 m3/sec
3.5
Hence, the correct option is (B).
Diameter is uniform so
v
0
s
6.6
(A)
0
P wz
V 2
2
N m Joule Energy
m2 m
m3
volume
A = AV
A=C
Hence, the correct option is (A).
Kulkarni’s Academy
6.7
(B)
91
Fluid Dynamics
6.8
(D)
d2 = 20 mm, d1 = 40 mm
P1 P2 = 30 kPa
A1V1 = A2V2
40
2
V1
4
4
16V1 = 4V2
V2= 4V1
V2
P1 P2 V22 V12
g
2g
VB = Vc
aAVA = acVc
aV
VA c c
aA
aA is very large = VA is negative
2 30 103
(4V1 )2 V12
1000
= 15V12 60
V12 4
Pc = PA
V1 = 2m/s
Hence, the correct option is (D).
Apply Bernoulli’s equation between B and C
2
P1 V12
P V2
2 2
g 2g g 2g
aBVB = acVc
Pc Vc2
PB VB2
zB
zc
w 2g
w 2g
20
6.9
(D)
P
PB
zB c
w
w
PB Pc
zB {Pc PA
w w
PB PA
zB
w w
PB PA
In this problem the fluid is flowing from B to
C i.e. the fluid is flowing from low pressure
to high pressure therefore pressure alone will
not decide the direction of flow it is the total
energy that decides the direction of flow.
Hence, the correct option is (B).
air = 1.2 kg/m3
V2
h
2g
1ha = whw
1000 25 103
20.83m
a 1.2
V2 = 2 9.81 20.83
V = 20.21 m/s
Hence, the correct option is (D).
ha
Fluid Mechanics
6.10
92
Kulkarni’s Academy
A1V1 = A2V2
(A)
A1 = A2
V1= V2 = 2m/s
P1 V12
P V2
z1 2 2 z2 hL
w 2g
w 2g
2 h2 3 .hour
hour
2 h2
3
2h2 3hour
hour
2 h2
3
PA = PB = Patm
VA = 0
Apply Bernoulli’s equation between A & B
P1
P
z1 2 z2
w
w
50 103
20 103
10
1000 9.81
1000 9.81
15.102 > 14.0387
As piezometric head 1 is greater than
piezometric head at 2. So, flow is from 1 to 2.
hL = 15.102 – 14.0387
hL = 1.0632m
Hence, the correct option is (C).
6.14 (C)
PA VA2
PB VB2
z
z
g 2g A w 2g B
1h1 2 h2
v2
h3 B 0
3
3
2g
h h
VB 2 g 1 1 2 2 h3
3
3
Hence, the correct option is (A).
6.11
(D)
6.12
(C)
6.13
(C)
Q1 = Q2 + Q3
= 0.28 = a2v2 + a3v3
Q = 0.28 m3/s
A1 = 0.012 m2
A2 = 0.008 m2
A3 = 0.004 m2
P1 = 270 kPa; P3 = 240 kPa; P2 = ?
For calculating v3 apply Bernoulli’s equation
between 1 and 3
Q1 = A1V1
V1
0.28
23.33 m/s
0.012
Kulkarni’s Academy
93
P V2
P1 V12
3 3
g 2g g 2g
P1 P3
Fluid Dynamics
6.17 (B)
V32 V12
2
30 2 1000
v32 5.44.44
8000
270 103
23.332 240 103 V32
1800 9.81 2 9.81 800 9.81 2 g
v3 = 24.89 m/s
0.28 = 0.008 V2 + 0.004 24.89
V2 = 22.5 m/s
By applying Bernoulli’s equation between 1
and 2, we get P2 = 284.5 kPa.
Hence, the correct option is (C).
6.15
(C)
6.16
(B)
ar
Q
a1a2 2 gh
Q
a1a2 2 gh
a1
a2
a2
a1 2 gh
a 1
2
r1
P1 P2
V 2 V12
------ (1)
H 2
w w
2g
P1
S
P
xx M H 2
w
Sw
w
S
V22 V12
x M 1
2g
Sw
Hence, the correct option is (B).
a1 2 gh
ar2 1
Q = same
P1 V12
P2 V22
z1
z2
w 2g
w 2g
S
P1 P2
x M 1 H -------- (2)
w w
Sw
From equation (1) and (2)
a12 a22
a12
1
a22
Apply Bernoulli’s equation between 1 and 2
a1 2 g ( sh)
ar22 1
1
5
ar22 15 1
2
3
ar2 1
ar 2 4
Hence, the correct option is (B).
6.18 (C)
Important points:
The discharge equation remains same
irrespective of positioning of venturimeter.
Hence, the correct option is (C).
6.19 (B)
Fluid Mechanics
Water – Manometric fluid
94
Kulkarni’s Academy
0.95 p0
2.136
0.65 pv
2
cd = 0.6
D0 = 60mm
= 1.2 kg/m3; = 1.8 105 kg/m-s.
Q cd .
a a
2
0
6.22 (A)
h x w 1 beacuse one fluid is water and
another is air.
1000
h 0.18
1 149.82m
1.2
a1
a0
4
4
(0.1) 0.00785m
2
2
(0.06)2 0.002827m2
2.20 105 0.6 0.00785 0.002827
2 9.81149.82
Q
300
140.44 mm of water column
2.136
Hence, the correct option is (C).
a1a0 2 gh
2
1
pv
0.007852 0.002827 2
Q = 0.01 m3/s
Hence, the correct option is (B).
cd a1a0 2 gh
Q
a12 a02
P
w
Q h
h
Q P
w=c
P
Q1
Q2 1.05P 1.0246 P
2.46%
Q is overestimated because P is
overestimated.
Hence, the correct option is (A).
6.23 (B)
6.20
(B)
6.24 (D)
6.21
(C)
cd
Qact
Qth
D = 100 mm
(cd)vent = 0.95 (cd)orifice = 0.65
a2 = a0
H = 300 mm of water column.
Discharge is same in both venturimeter and
orifice meter
Qv = Q0
cd a1a2 2 gh
a a
2
1
2
2
cd a1a2 2 gh
a a
2
1
cdv hv cd0 h0
0.95
pv
po
0.65
w
w
2
0
cd cc cv
Cc 4
4
90
2
100
Cd = 0.95 0.81 = 0.7695
Hence, the correct option is (D).
= 0.81
2
Kulkarni’s Academy
95
Fluid Dynamics
Given V = 1 m/s
6.25 (B)
I.D. = 10 m
Rotameter is used for calculating discharge.
Hence, the correct option is (B).
6.26
Q = AV
(C)
10 10
4
=
v 2 gx M 1
2 9.81
=
10 1000
1
100 1.2
(D)
6.28
(C)
Device
400
1
m3/s
F = ma
V u
m
m V U
t
= 12.77 m/s
Hence, the correct option is (B).
6.27
2 2
Fx AV V U 100
Shape
Losses
Cd
Cost
Venturimeter
Less
High
High
Flow nozzle or
nozzle meter
Medium
Medium
Medium
Orifice meter
High
Less
Cheap
Fy AV V U 100
400
400
0 1 =
5
2
1 0 =
5
2
F Fx2 Fy2
5
5
5 5
2
2
2
2 2
2
2
Hence, the correct option is (D).
Hence, the correct option is (C).
6.29
6.32 (D)
(D)
D = 225 mm
D0 = 75 mm
Q h
h
Q2
h
2Q
2
2
Q1
h1
Q
2
h2 = 4x
Hence, the correct option is (D).
6.30
(C)
6.31 (D)
Apply Bernoulli’s equation between A and C
P1 Vc2
P V2
zc 2 a z a
w 2g
w 2g
Vc2
182
0
21
2g
2 9.81
Vc = 27.1 m/s
Hence, the correct option is (D).
Fluid Mechanics
6.33
96
(D)
Kulkarni’s Academy
6.35 (B)
Apply equation between B and C.
ABVB = ACVC
225
VB
75
27.1
4
4
VB = 3.01 m/s
Apply Bernoulli’s equation between B and C
2
2
P V2
PB VB2
z B c c zc
w 2g
w 2g
For gauge pressure Pc = Patm = 0
PB Vc2 VB2
0.5
2
2g
Q1 = Q2 + Q3
A1V1 = A2V2 + A3V3
PB 27.12 3.012
0.5
w
2 9.81
PB = 367.576 kPa
Hence, the correct option is (D).
6.34
0.2 3
v = 6 m/s
Apply Bernoulli’s equation between 1 and 2
8 m/s
0
Given
P1 = 130 kPa
V1 = 2 m/s
P2 = 100 kPa
V2 = ?
hL = Loss of potential head between inlet and
outlet.
Apply Bernoulli’s equation between in inlet
and outlet.
2
1
v
P
2
2
1
V 2 30 4
V22 64
V2 = 8m/s
Hence, the correct option is (B).
(36 9)
2
6.36 (A)
Pressure force exerted by the fluid on the pipe
is equal to magnitude of the force required to
hold the pipe.
F=PA
P1 P2
V 2 V12
z1 z2 2
hL
g
2g
2
2
2
1000
Hence, the correct option is (B).
2
2
(130 100) 103 v22 (2) 2
1000 g
2 g
v12
P1 13.5kPa
P1 V
P V
z1 2
z2 hL
g 2g
g 2g
0.2
0.2
v
v
4
4
= 13.5 0.2= 2.7 kN
Note:
As the pipe is symmetric the net force in the
y – direction is zero, Therefore, whatever the
force required to hold the pipe it is in the x –
direction.
Hence, the correct option is (A).
NOTES
7.1 Reynolds Number:
It is the ratio of inertia force to viscous force that
VL
is, Re
Where – L is the characteristic dimension.
Significance of L:
It is such a dimension over which significant
changes in properties occur. For flow through
pipes the characteristic dimension is pipe
diameter and for a flow over a flat plate the
characteristic dimension is the distance from the
leading edge (x).
Reynolds found from his experiments for flow
through pipes:
Re < 2000 Laminar flow
2000 < Re < 4000 Transition flow
Re > 4000 Turbulent flow
The pressure decreases in the direction of flow in
order to overcome losses i.e. pressure gradient is
negative in the direction of flow.
7.2 Darcy’s Weisbach Equation:
This equation is used for calculating head loss
due to friction.
hL
FLV 2
;
2 gD
hL
4 F ' LV 2
2 gD
F = 4F
Where F = Darcy’s friction factor or Moody’s
friction factor
F = fanning friction coefficient
This equation is applicable for laminar or
turbulent flow, horizontal, inclined or vertical
pipes but the flow must be steady.
7.3 Fully developed flow:
A flow is said to be fully developed flow if the
velocity profile does not change in the
longitudinal direction and the pressure gradient
dP
remains constant.
dx
z1 = z2
a1v1 = a2v2
a1 = a2
v1 = v2
P1 V12
P V2
z1 2 2 z2 hL
w 2g
w 2g
P1 P2
hL
w
… (i)
Fluid Mechanics
98
7.4 Laminar flow through circular pipes:
Kulkarni’s Academy
Velocity distribution: -
(Hagen – Poiseuille flow)
Assumption:
(1) steady flow
(2) Fully developed flow
r+y=R
dr + dy = o
dy = dr
du
dy
du P r
.
dy
x 2
du
1 P
r.dr
2 x
Temporal (steady) = 0
F = ma
Convective (Fully developed)=0
F = ma = O
P 2
P. r 2 P
dx r .2 rdx 0
x
P
dx.r .2dx
x
P r
.
x 2
u
at pipe wall
at r = R
u=0
…. (1)
c
For fully developed flow
P
= constant
x
r
Or
1 P r 2
. c
2 x 2
As the shear stress is zero at the centre of the
pipe therefore viscous forces are zero at the
centre and hence Bernoulli’s equation can be
applied along the axis of the pipe.
In a laminar flow through pipes the shear
stress varies linearly from zero at the centre
to the maximum at pipe wall.
1 P R 2
.
2 x 2
u
1 P 2 1 P 2
r
R
4 x
4 x
u
1 P 2
2
(r R )
4 x
u
1 P 2 2
[R r ]
4 x
u
1 P 2 r 2
R 1
4 x R 2
u= umax at the centre (r = 0)
umax
1 P 2
R
4 x
r2
u umax . 1 2
R
Where u is local velocity.
…. (2)
…. (3)
Kulkarni’s Academy
99
Laminar Flow
Q
umax. R2
2
Q = AV
Q = R2.V
umax.R 2
Note:
Velocity distribution is parabolic in laminar flow
through pipes.
2
u
V max.
2
R 2 .V
…. (5)
7.5 Pressure drop in a given length ‘L’:
Discharge :
Let us calculate discharge through elemental
ring.
d = u 2rdr
umax
2
1 1 P
v R2
2 4 x
v
R
Q 2 urdr
8V
x R 2 dx
1
x2
0
r2
Q 2 umax 1 2 rdr
R
0
R
R R
Q 2 umax
4
2
2
2
R2
umax.R2
….(4)
Q 2 umax.
Q
4
2
1 P 2 2
R R
4 x
Q
2
P 4
Q
R
8 x
Average velocity (v):
p2
P
p1
8V ( x2 x1 )
{ x2 – x 1 = L
P1 P2
R2
8VL 32VL
…... (6)
P1 P2
R2
D2
Where V is average velocity.
From equation – (i)
P1 P2
hL
w
P1 – P2 = ghL
FLV 2
32VL
P1 P2 g
; P1 P2
D2
2 gD
g ( FLV 2 ) 32 VL
2 gD
D2
64
F
VD
64
…. (7)
F
Re
Fluid Mechanics
F = Darcy’s friction factor
100
7.6 Laminar flow between parallel plates:
16
Re
F = Fanning friction coefficient
F'
F = 4F
Kulkarni’s Academy
(1) Case 1: Fixed parallel plates:
Width of plate = 1 unit
In laminar flow through pipes friction factor
depends on Reynolds number only.
Shear velocity (V*):
P r
.
x 2
P R
0
x 2
P2 P1 D
0
.
x2 x1 4
F = ma 0 (fully developed & steady)
d
P
Pdy
dy dx dx P
dx dy 0
dy
x
P
Pdy dx dxdy dx Pdy
dxdy 0
dy
x
P
y x
The pressure gradient is the flow direction is
equal to shear gradient in the perpendicular
direction.
Velocity distribution:
Most of common fluid – are Newtonian fluid.
P P D
0 2 1
4L
P1 P2 whL
0
0
gFLV 2
4L
FV 2
0
F
V
8
Shear velocity =
V*
F
.V
8
P d 2u
x
dy 2
d 2u 1 P
dy 2 x
u 1 P 2
y c1 y c2
y x
…. (8)
8
0
V2
F
8
0
V *
dy
d 2u
y
dy 2
2 gD
gFLV 2 D
2 gD
du
At y = 0; u = 0; c2 = 0 (bottom plate)
y = t ; u = 0 (top plate)
o
1 dP 2
t c1t
2 dx
Kulkarni’s Academy
101
1 dP
t
2 dx
c1
2
Q umax.t
3
Average velocity:
1 dP 2 1 dP
u
y
yt
2 dx
2 dx
u
Laminar Flow
2
Q V t umax.t
3
1 dP
2
( yt y )
2 dx
For maximum velocity
2
V umax
3
u
0
y
7.7 Pressure drop in a given length ‘L’:
The velocity is maximum at centre y
t
2
2
1 P t t
umax.
t
2 x 2 2
umax.
1 P 2
t
8 x
Discharge (Q):
dQ = u.dA
t
dQ
0
2
v umax
3
1 P
2
yt y dy
2 x
2 1 P 2
v
t
3 8 x
t
1 P t. y 2 y 3
Q
2 x 2
3 0
1 P t 3 t 3
2 x 2 3
Q
Q
1 P t 3
2 x 6
Q
1 p 3
t
12 x
We know that
umax
1 P 2
t
8 x
4umax
Q
1 p 2
t
2 x
4umax.t
6
24 v(x)
P
2t 2
12V
p p t 2
1
p2
P1 P2
x2
dx
x1
12V ( x2 x1 )
t2
P1 P2 P
12VL
t2
Fluid Mechanics
P
7.1
102
Practice Questions
7.6
Flow in a pipeline of constant diameter is said
to be fully developed when
(A) The flow rate in the pipeline increase
along the length of the pipe
7.7
(B) The flow rate in the pipeline decrease
along the length of the pipe
(C) The flow rate in the pipeline does not
vary along the length of the pipe
(D) The velocity profile does not vary along
the length of the pipe
7.2
The inertia force on a fluid particle in a steady
fully developed laminar flow through a
straight pipe, at a Reynolds number of 100, is
7.8
(A) 100 times the corresponding viscous
force
(B) 0.01 times the correspond viscous force
(C) Zero
(D) Infinity
7.3
7.4
For an ideal fluid flow the Reynolds number
is
(A) 2100
(B) 100
(C) Zero
(D) Infinity
The Darcy-weisbach equation for head loss is
valid
(A) Only for laminar flow through smooth
pipes
(B) Only for laminar flow through rough
pipes
7.9
Kulkarni’s Academy
The minimum value of friction factor that can
occur in laminar flow through circular pipes
is
(A) 0.02
(B) 0.032
(C) 0.016
(D) 0.08
The mean shear stress in a fully developed
fluid flow in a pipe
(A) Is zero at the centre of the pipe and varies
linearly with distance from the centre
(B) Is constant over the cross-section
(C) Is zero at the pipe wall and increases
linearly towards the centre of the pipe
(D) Varies parabolically across the section
The velocity profile in fully developed
laminar flow in a pipe of diameter D is given
4r 2
by u0 1 2 , where 'r' is the radial
D
distance from the center. If the viscosity of
the fluid is , the pressure drop across a
length L of the pipe is
32 u0 L
4 u0 L
(A)
(B)
2
D
D2
8 u0 L
16 u0 L
(C)
(D)
2
D
D2
The velocity profile of a fully developed
laminar flow in a straight circular pipe, as
shown in the figure, is given by the
expression
u (r )
Where
(C) For laminar or turbulent flow through
smooth pipes only
R 2 dp r 2
1
4 dx R 2
dp
is a constant.
dx
(D) For laminar or turbulent flow through
smooth or rough pipes
7.5
For flow through a horizontal pipe, the
dp
pressure gradient
the flow direction is
dx
(A) ve
(B) 1
(C) Zero
(D) ve
The average velocity of fluid in the pipe is
(A)
R 2 dp
8 dx
(B)
R 2 dp
4 dx
(C)
R 2 dp
2 dx
(D)
R 2 dp
2 dx
Kulkarni’s Academy
7.10
7.11
103
Velocity for flow through a pipe, measured at
the centre is found to be 2m/s . Reynolds
number is around 800. What is the average
velocity in the pipe?
(A) 2m/s
(B) 1.7m/s
(C) 1m/s
(D) 0.5m/s
The maximum velocity of a one-dimensional
incompressible fully developed viscous flow,
between two fixed parallel plates, is 6m/s
Laminar Flow
7.16
viscosity of 0.2 Ns/m 2 of is flowing through a
pipeline of 50mm diameter at an average
7.17
The mean velocity of the flow is
7.12
(A) 2
(B) 3
(C) 4
(D) 5
The pressure drop for a relatively low
Reynolds number flow in a 600mm, 30m
7.18
long pipe line is 70kPa . What is the wall
shear stress?
7.13
(A) 0 Pa
(B) 350Pa
(C) 700Pa
(D) 1400Pa
Laminar flow developed at an average
velocity of 5m/s occurs in a pipe of 10cm
radius. The velocity at 5cm radius is
7.14
7.15
(A) 7.5m/s
(B) 10m/s
(C) 2.5m/s
(D) 5m/s
Oil having a density of 800 kg/m 3 and
7.19
velocity of 2m/s . The Darcy fraction factor
for this flow is:
(A) 3. 2
(B) 0.07
(C) 0.16
(D) 1.6
Consider a fully developed laminar flow in a
circular pipe. If the diameter of the pipe is
halved, while the flow rate and length of the
pipe are kept constant, the head loss increases
by a factor of
(A) 4
(B) 8
(C) 16
(D) 32
If laminar flow takes place in two pipes,
having relative roughnesses of 0.002 and
0.003, at a Reynolds number of 1815, then
(A) The pipe of relative roughness of 0.003
has a higher friction factor.
(B) The pipe of relative roughness of 0.003
has a lower friction factor.
(C) Both pipes have the same friction factor
(D) No comparison is possible due to
inadequate data
Flow rate of a fluid (density = 1000 kg/m3 ) in
a small diameter tube is 800 mm3 /s . The
For the laminar flow of a fluid in a circular
pipe of radius R, the Hagen- Poiseuille
equation predicts the volumetric flowrate to
be proportional to
length and the diameter of the tube are 2m
(A) R
(B) R 2
And 0.5mm , respectively. The pressure drop
in 2m length is equal to 2.0MPa . The
viscosity of the fluid is
(C) R 4
(D) R 0.5
(A) 0.025 N.s/m 2
In a laminar flow through a pipe of radius R,
the fraction of the total fluid flowing through
a circular cross-section of radius R/2 centered
at the pipe axis is
(A)
3
8
(B)
7
16
(C)
1
2
(D)
3
4
(B) 0.012 N.s/m 2
(C) 0.00192 N.s/m 2 (D) 0.00102 N.s/m 2
7.20
The pipe of 20cm diameter and 30 km
length transports oil from a tankers to the
shore with a velocity of 0.318m/s . The flow
is laminar. If 0.1Ns/m 2 the power
required for the flow would be
( 900 kg/m3 )
Fluid Mechanics
(A) 9.25kW
(C) 7.63kW
7.21
7.22
104
(B) 8.36kW
7.25
(D) 10.13kW
What is the discharge for laminar flow
through a pipe of diameter 40mm having
centre-line velocity of 1.5m/s ?
(A)
3 3
m /s
50
(B)
3
m3 /s
2,500
(C)
3
m3 /s
5,000
(D)
3
m3 /s
10,000
7.26
The resulting error in the discharge will be
(B) 18.25%
(C) 12.5%
(D) 12.5%
(A) 1.00
(B) 1.33
(C) 2.00
(D) 1.50
In a 4cm diameter pipeline carrying laminar
flow of a liquid with 1.6 centipoise, the
velocity at the axis is 2m/s . What is the shear
stress midway between the wall and the axis?
The value of friction factor is misjudged by
+25% in using Darcy-Weisbach equation.
(A) 25%
Kulkarni’s Academy
The kinetic energy correction factor for a
fully developed laminar flow through a
circular pipe is
7.27
7.23 It is desired to set up water flow with Reynolds
number of 2000 in a pipe of diameter 10mm
by controlling the inlet pressure. The pressure
difference, p (in terms mm of water
(A) 0.16 N/m 2
(B) 0.016 N/m 2
(C) 0.02 N/m 2
(D) 0.0125 N/m 2
For a steady fully developed laminar flow of
an oil of specific gravity 'S’ through two
pipes in series as shown in figure. The ratio
h1 / h2 of the manometric fluid deflections
consider only friction losses in the pipes is
column) over a 4m length of the pipe will be
(A) 10
(B) 50
(C) 25
(D) 100
7.24 Choose the correct combination of true
statements from the following
P.
The energy correction factor for
turbulent flow is less than that for
laminar flow but is greater than unity.
Q. The energy correction factor for
turbulent flow is greater than that for
laminar flow and is also greater than
unity
R. The momentum correction factor for a
given flow is less than the corresponding
energy correction factor
S.
Both the momentum correction factor
and energy correction factor are always
greater than unity
(A) P, R
(B) Q, S
(C) P, R, S
(D) Q, R, S
4
4
D L
(A) 1 2
D2 L1
2
D L
(C) 2 2
D1 L1
7.28
D L
(B) 2 1
D1 L2
2
2
D L
(D 1 2
D2 L1
2
Water flows downwards through a vertical
straight circular pipe of 1m diameter. Assume
that the flow is laminar and fully developed
and that there is no pressure gradient. The
frictional force acting on the pipe wall, over
a length of 1m , is nearly ( g 10 m/s 2 )
(A) 3925 N
(B) 5890 N
(C) 7850 N
(D) 15,700 N
Kulkarni’s Academy
7.29
105
Laminar Flow
Assertion(A): For a fully developed viscous
flow through a pipe the velocity distribution
across any section is parabolic in shape.
Reason (R): The shear stress distribution
from the centre line of pipe upto the pipe
surface increases linearly.
Common Data Questions 7.32 to 7.33
Consider fully-developed, laminar flow in a
circular pipe of radius R. The centre-line
velocity of the flow is U.
7.32
(A) Both A and R are individually true and
R is the correct explanation of A
r
(A) 1 U
R
(B) Both A and R are true but R is not the
correct explanation of A
r 2
(C) 1 U
R
(C) A is true but R is false
(D) A is false but R is true
The velocity u at a radial distance r from the
centre-line of the pipe is given by
7.33
Common data Questions 7.30 to 7.31
A syringe with a frictionless plunger contains
water and has at its end a 100mm long needle
of 1mm diameter. The internal diameter of
the syringe is 10mm . Water density is
r
(B) 1 U
R
(D
r 2
1 U
R
The value of the radial distance from the
centre-line of the pipe at which the velocity
be equal to the average velocity of the flow is
(A) 0.666 R
(B) 0.696 R
(C) 0.707 R
(D) 0.727 R
Common data Questions 7.34 to 7.35
3
7.30
7.31
1000 mg/m . The plunger is pushed in at
An upward flow of oil (mass density
10mm/s and the water comes out as a jet.
800 kg/m 3 ,dynamic viscosity
0.8kg/m-s )
takes places under laminar conditions in an
inclined pipe of 0.1m diameter as shown in
the figure. The pressures at sections 1 and 2
are measured as p1 435kN/m 2
and
Assuming ideal flow, the force F in newtons
required on the plunger to push out the water
is
p2 200kN/m2
(A) 0
(B) 0.04
(C) 0.13
(D) 1.15
Neglect losses in the cylinder and assume
fully developed laminar viscous flow
throughout the needle; the Darcy friction
factor is 64/Re. Where Re is the Reynolds
number. Given that the viscosity of water is 1
centipoise the force F in newtons required on
the plunger is
7.34
The discharge in the pipe is equal to
(A) 0.13
(B) 0.16
(A) 0.100 m 3 /s
(B) 0.127 m 3 /s
(C) 0.3
(D) 4.4
(C) 0.144 m 3 /s
(D) 0.161m3 /s
Fluid Mechanics
106
7.35
If the flow is reversed, keeping the same
discharge, and the pressure at section 1 is
maintained as 435kN/m 2 , the pressure at
section 2 is equal to
Kulkarni’s Academy
6
Q 1.1310 m /s
3
(A) 1.86 104 Pa.s
(B) 3.38 104 Pa.s
(A) 488kN/m 2
(B) 549 kN/m 2
(C) 6.75 104 Pa.s
(C) 586 kN/m 2
(D) 614 kN/m 2
(D) 7.43104 Pa.s
7.36
A viscous, incompressible and Newtonian
fluid flowing through the main branch of a
circular pipe bifurcates into two daughter
branches whose radii are 4cm and 2cm .
respectively. The flow in both the daughter
branches are laminar and fully developed. If
the pressure gradients in both the daughter
branches are same, then fraction of total
volumetric flow rate (rounded off to the
second decimal place) coming out from the
branch with 4cm diameter is__________.
7.37
A liquid of specific weight 9 kN/m3 flows by
gravity through a 0.3m tank and 0.3m
capillary tube at a rate of 1.13106 m3 /s as
shown in the figure. Top of the tank and
outlet of the capillary are open to the
atmosphere. If the flow is laminar, fully
developed and incompressible, then the
viscosity of the liquid, neglecting entrance
effect, is numerically closest to
Kulkarni’s Academy
A
107
Laminar Flow
7.6
Answer Key
(B)
64
64
0.032
Re 2000
Hence, the correct option is (B).
F
7.1
D
7.2
A
7.3
D
7.4
D
7.5
D
7.6
B
7.7
A
7.8
D
7.9
A
7.7
(A)
7.10
C
7.11
C
7.12
B
7.8
(D)
7.13
A
7.14
C
7.15
B
7.16
C
7.17
C
7.18
C
7.19
C
7.20
C
7.21
D
7.22
C
7.23
C
7.24
C
7.25
C
7.26
A
7.27
B
umax = u0
7.28
C
7.29
A
7.30
B
V
7.31
C
7.32
C
7.33
C
7.34
B
7.35
D
7.36
C
7.37
D
E
7.1
7.2
Explanation
(D)
(A)
Re=
Inertia Force
Viscous Force
Inertial Force = 100 Viscous Force
Hence, the correct option is (A).
7.3
(D)
Re=
Inertia Force
Viscous Force
P1 P2
32VL
D2
4r 2
u umax. 1 2
D
umax u0
2
2
P1 P2
32u0 L
2D2
P1 P2
16u0 L
2D2
Hence, the correct option is (A).
7.9
(A)
U (r )
R 2 dP r 2
1
4 dx R 2
1 P 2
R
U max.
4 x
V
2
2
V
R 2 P
8 x
Hence, the correct option is (A).
7.10 (C)
For ideal fluid flow viscous force = 0
Umax = 2 m/s
Re = Infinity
Re = 800 (laminar)
Hence, the correct option is (D).
V=?
7.4
(D)
7.5
(D)
V
U max 2
1 m/s
2
2
Hence, the correct option is (C).
Fluid Mechanics
7.11
(C)
108
Kulkarni’s Academy
7.15 (B)
For parallel plates
Q
R 2U max
2
2
V U max
3
2
6 4 m/s
3
R /2
Q1
Hence, the correct option is (C).
2 rdr.u
0
7.12
(B)
R /2
Q
1
P r
.
x 2
w
0
R /2
r2
r4
Q 2 U max 2
2 4R 0
1
P R
.
x 2
8R 2 R 2
Q1 2U max
64
70 103 0.6
30
4
7 R2
Q 2U max
64
1
= 350 pascal
Hence, the correct option is (B).
7.13
(A)
r2
U U max 1 2
R
V
U max
U max 10
2
25
3
U 10 1
10 7.5 m/s
100
4
Hence, the correct option is (A).
7.14
(C)
Q
P
4
R
8 x
Q R4
Hence, the correct option is (C).
r2
2U max 1 2 rdr
R
Q1
7
R 2U max
32
Q1
7 R 2U max
16
2
7
Q
16
Hence, the correct option is (B).
Q1
7.16 (C)
Given :
oil = 800 kg/m3
N S
m2
D = 2 m/s
0.2
Re
VD 800 2 0.05
400
0.2
64 64
0.16
Re 400
Hence, the correct option is (C).
F
Kulkarni’s Academy
109
7.17 (C)
Laminar Flow
7.20 (C)
hL
P1 P2
w
Power required to overcome the losses in the
pipe
P wQhL
2
Q 4 D V
V 4Q
D2
hL
32VL
gD 2
hL
32 L 4Q
D2 g D2
hL
1
D4
P = gQhL
FLV 2
hL
2 gD
F
So, if diameter is half, hL is increases by 16
times.
(C)
P = gQhL
In laminar flow, friction factor depends only
on Reynolds number.
= 900 9.81 (0.2) 2 0.318 86.4
4
Hence, the correct option is (C).
7.19 (C)
= 7.63 kW.
Hence, the correct option is (C).
Given data :
= 1000 kg/m
7.21 (D)
3
V
Q = 800 mm3/s
L = 2m
Q
D = 0.5 mm
P = 2 MPa
P P1 P2
32 4.074 2
0.5 10
3 2
U max 1.5
0.75
2
2
4
0.04
2
0.75
0.0012
4
12
3
m3/s
40000
10000
Hence, the correct option is (D).
=?
2 106
64 0.1
900 0.318 0.2
F = 0.111
0.111 30000 (0.318)2
hL
2 9.81 0.2
hL = 86.4 m
Hence, the correct option is (C).
7.18
64 64
Re VD
32VL
D2
Q AV
2
800 (0.5) V
4
V 4.074m / s
= 0.001917 N-S/m2
Hence, the correct option is (C).
7.22 (C)
F2 F1
dF
100
100 25
F1
F
dF
0.25
F
FLV 2 FL 16Q 2
hL
2 gD
2 gD D 4
Fluid Mechanics
1
F 2
Q
110
7.26 (A)
Given data :
FQ2 = C
D = 4 cm
ln F + ln Q = const
= 1.6 103 N – S/m2
lnF + 2 lnQ = C
Umax = 2 m/s
2
dF
dQ
2
0
F
Q
dQ
1 dF
Q
2 F
1
(0.25) 0.125 12.5%
2
Hence, the correct option is (C).
7.23
Kulkarni’s Academy
(C)
1
0
2
0 = 2
Given :
Re = 2000
D = 10 mm = 0.01 m
Re
V
VD
Re. 2000 103
D 1000 0.01
0
P
32VL
D2
32 103 0.2 4
(0.01) 2
= 26.02 mm of water.
Hence, the correct option is (C).
7.24
(C)
7.25
(C)
1 P 2
R
4 x
P U max 4
x
R2
2 4 1.6 103
(0.02) 2
P
32
x
0 32
P = 256 N/m2
10.3
256 0.02602 m of water
101325
column
P R
.
x 2
U max
V = 0.2 m/s
P
P r
.
x 2
0
2
0.02
0.32
2
0.32
0.16 N/m2
2
Hence, the correct option is (A).
7.27 (B)
h1
32V1 L1
D12
h2
32V2 L1
D22
Kulkarni’s Academy
111
(1) Q = AV
Q
4
(2) Q = A2V2
D12 V1
V2
V
4Q
D12
h1
32 4Q.L1
D14
h2
32 4Q.L2
D24
Laminar Flow
7.30 (B)
4Q
D22
L1
2
h1 D14 D2 L1
.
h2 L2 D1 L2
D24
4
0.01
For P1
Apply Bernoulli’s equation
P1 V12
P V2
2 2 { P2 = 0
g 2g g 2g
Hence, the correct option is (B).
7.28
F1 P1
F = P1A1
A1V1 = A2V2
(C)
D12V1 D22V2
V1 = 10 mm/s
= 10 103 m/s
V2 = 1 m/s
P1 V22 V12
g
2g
P1
P1
(V22 V12 )
2
1000 1 (10 103 ) 2
2
P1 = 495 N/m
P1 – P2 = 0
FF = w = gV
1000 10 (1)2 1
4
= 7850 N
Hence, the correct option is (C).
7.29
2
(A)
The parabolic distribution of velocity is
obtained from linear shear stress distribution.
Hence, the correct option is (A).
F1 495 (0.01)2
4
= 0.038 N = 0.04 N
Hence, the correct option is (B).
7.31 (C)
FLV 2
64 V2 D2
hL
F
2 gD
Re
2
Fluid Mechanics
112
Head loss due to friction in needles only b/c
in problem neglect losses in the cylinder
0.064 0.11
hL
2 9.81103
hL = 0.326 m
Apply Bernoulli’s equation between 1 and 2
Kulkarni’s Academy
7.34 (B)
P1 V12
P V2
2 2 hL
g 2g g 2g
P1 V22 V12
hL
g
2g
Apply Bernoulli’s equation between 1 and 2
V V
P1 g
hL
2g
2
2
2
1
12 (10 103 )2
P1 1000 9.81
0.326
2 9.81
2
P1 = 3698 N/m
P1 V12
P V2
z1 2 2 z2 hL
w 2g
w 2g
hL = 26.4 m
F P1 A1 3698 (10 103 )2
4
= 0.3 N
Hence, the correct option is (C),
435 103
200 103
5sin 450 hL
800 9.81 800 9.81
FLV 2 64 LV 2
hL
.
2 gD
VD 2 gD
7.32 (C)
26.4
32 0.8 V 5
800 9.81 0.12
V = 16.18 m/s
Q
4
D2V =
4
(0.10) 2 16.18
Q = 0.127 m3/s
r2
U U max 1 2
R
Hence, the correct option is (C).
7.33
(C)
Hence, the correct option is (B).
7.35 (D)
Flow is reversed;
r2
U 2V 1 2
R
1
r2
r2 1
1 2 2
2
R
R
2
R2 = 2r2
R 2r
R
r
0.707 R
2
Hence, the correct option is (C).
P2
P
z2 1 z1 hL
w
w
P2
435 103
0
+26.4
5sin 45
500 9.81
800 9.81
P2 = 614 kN/m2
Hence, the correct option is (D).
Kulkarni’s Academy
113
Laminar Flow
V12
V2
0.6 2 0 hL
2g
2g
7.36 (C)
Q = Q1 + Q2
r1 = 4 cm; r2 = 2 cm
d1 = 8 cm; d2 = 4cm
hL 0.6
hL
32VL
gD2
0.549
32 0.3
917 9.81 (1.2 103 ) 2
μ
Q2
?
Q Q2
Q
Q
U max R 2
2
R 2 1 p 2 1 p 4
. R
R
2 4 x
8 x
Q R4
R24
24
0.058
R14 R24 44 24
Hence, the correct option is (C).
7.37
(D)
Q = A2V2
1.13 106
1.2 10 V
4
3 2
2
V2 = 1 m/s
V1 0 (Large reservoir)
P1 V12
P V2
z1 2 2 z2 hL
w 2g
w 2g
V22
1
0.6
0.549 m
2g
2 9.81
0.00711
7.43 104 Pa.s
9.6
Hence, the correct option is (D).
NOTES
In turbulent flow as there is continuous mixing of
fluid
particle,
the
velocity fluctuates
continuously and hence no turbulent can be
purely steady flow.
In turbulent flow the shear stress is due to
fluctuation of velocity in the flow direction as
well as in transverse direction.
The head loss in turbulent flow is proportional
y1.75 to y2.
u = actual velocity
u = average velocity
u = fluctuation velocity component
u u u
u u u'
du
l
dy
1
u ' dt
T 0
Boussinesq developed for turbulent shear stress
as
du
dy
Mixing length is that length in the transverse
direction where in the particle after colliding
loose excess momentum and reach momentum of
new region. It is similar to mean free path in
gases. According to Prandtl’s mixing length
l = 0.4 y
0.4 is Carmany constant
‘y’ is distance from the pipe wall
At the pipe wall (y = 0) Prandtl’ mixing length is
zero.
du
Prandtl’s found u ' v ' l
dy
2
2
T
8.1 Prandtl’s mixing length theory:
= uv
du du
l .l
dy dy
u u u'
u'
As it is difficult to find this equation is not used
in practice. Reynolds developed the equation for
turbulent shear stress as
= uv
where u and v are fluctuating component in x
and y direction respectively
du
dy
= fluid characteristic
= eddy viscosity, flow characteristic
= turbulent shear stress
Velocity distribution is turbulent flow
du
l
dy
2
du
l2
dy
2
2
Kulkarni’s Academy
du
l
dy
115
Turbulent Flow
vk
l = 0.4 y
du
v 0.4 y
dy
v dy
du
0.4 y
2.5 v lny = u + c
…. (i)
u = actual velocity at distance y from wall
Note: The velocity distribution in turbulent flow
is logarithmic in nature.
8.2 Hydrodynamically smooth &
= Reynold’s roughness number
vk
< 4 Smooth boundary
vk
4<
> 100 Rough boundary
vk
< 100 Transition boundary
8.3 Nikuradse’s Experiment:
It is found from experiments laminar sublayer
thickness
hydrodynamically Rough boundary :
'
11.6
v
= kinematic viscosity
v = shear velocity
from equation (i)
2.5 v lny = u + c
At the center y = R, u = umax
2.5 v ln R = umax + c
umax u = 2.5 v lnR – 2.5 v lny
R
umax u
2.5ln
v
y
At y = y, u is taken as zero
Were y is very small distance from pipe wall
= laminar sublayer thickness (height)
k = average height of thickness roughness
Conditions for hydro-dynamically smooth and
hydro-dynamically rough boundaries.
From Nikuradse’s experiment.
k
0.25 smooth boundary
'
k
6 Rough boundary
'
k
0.25 6 Transition
'
Roughness criterion according to Reynold
2.5 v lny = u + c
At
y = y , u = 0
2.5 v lny = 0 + c
C = 2.5 vlny
2.5 vlny = u + 2.5vlny
u = 2.5 v[lny - lny]
u
y
2.5ln
v
y'
y
u
5.75log10 valid for rough and smooth
v
y'
pipe
Fluid Mechanics
From experiment it is found that
y1
1
sooth pipe
107
k
Rough pipe
y1
30
Velocity for smooth pipes
y
u
5.75log10
v
y'
y1
1
107
y1
1 11.6
107 v
u
y 107v
5.75log10
v
11.6
u
v y 107
5.75log10
v
11.6
u
v y
107
5.75log10
5.75log10
v
11.6
u
v y
5.75log10
5.5
v
This equation is valid for smooth pipe.
Velocity for rough pipe
y
u
5.75log10
v
y'
k
30
k = average height of roughest
y1
116
Kulkarni’s Academy
8.4 Average velocity:
Following the same procedure as we have done
in laminar flow we have
v
vR
5.75log10
1.75
v
This equation is valid for smooth pipe.
Friction factor in turbulent flow
For smooth pipe
f
0.3164
upto Re 105
1/4
Re
f 0.0052
0.221
(Re)0.232
Re = 105 to 4 107
For rough pipe
1
R
2log10 1.74
f
k
In laminar flow friction factor depends on
Reynold’s number were as in turbulent flow
friction factor depends not only on Reynold’s
number but also on average height of roughness.
If the pipe is smooth friction factor depends upon
Reynold’s number were as for rough pipe friction
factor depends on average height of roughness
‘k’.
8.5 Moody’s diagram:
u
30 y
5.75log10
v
k
u
y
5.75log10 30
v
k
u
y
5.75log10 30 5.75log10
v
k
u
y
8.5 5.75log10
v
k
u
y
5.75log10 8.5
v
k
This equation is valid for rough pipe.
The equation 0
turbulent flow also.
fv 2
8
is applicable for
Kulkarni’s Academy
117
Turbulent Flow
umax v
1.33 f
v
umax
1 1.33 f
v
Example 1
Find the distance from the pipe wall at which
local velocity is equal to average velocity in
turbulent flow.
Sol.
u v
y
5.75log10 3.75
v
R
y = distance from pipe wall
local velocity u = average velocity v
u=v
y
0 5.75log10 3.75
r
y
5.75log10 3.75
R
y 3.75
log10
R 5.75
3.75
y
10 5.75
R
y
0.223
R
y = 0.223R
Example 2
Show that for a turbulent flow in pipe the
ratio of maximum velocity to the average
velocity is given by
umax
1 1.33 f
v
Sol.
u v
y
5.75log10 3.75
v
R
at center u = umax ; y = R
umax v
R
5.75log10 3.75
v
R
We know that
f 2
0
v
8
0 fv 2
8
v
f
.v
8
umax v 3.75
f
v
8
Example 3
A rough pipe carrying water has an average
height of roughness of 0.48 mm. The
diameter of pipe is 0.68 m and length is 4.5m.
the discharge of water is 0.6 m3/sec. Find the
power required to maintain this flow.
(Take viscosity of water 1 centipoises, treat
the pipe as rough)
Sol.
Given :
K = 0.48 mm = 0.48 103 m
D = 0.68 m, Q = 0.6m3/sec
L = 4.5 m
N s
103 2
m
Power =?
Q AV
V
4
Re
4
0.68
0.6
(0.68)
2
V
1.652 m/sec
2
VD 1000 1.652 0.68
183
Re = 1.123 106
Flow is turbulent for rough pipe
1
R
2log10 1.74
f
k
1
0.34
2log10
1.74
3
f
0.48 10
f = 0.01806
hL
fLV 2 0.01806 4.5 1.6522
2 gd
2 0.68 9.81
hL =0.01662 m
Power = wQhL
= 9810 0.6 0.01662 = 97.86 kJ
Fluid Mechanics
118
Example 4
Rough pipe of 0.1 m diameter carries water
at the rate of 50 liter/sec. The average height
of roughness is 0.15 mm. Find
(i) friction factor
(ii) shear stress at the pipe surface
(iii) shear velocity
(iv) maximum velocity
Take density of water as 1000 kg/m3 and
v = 106 m2/s.
Sol.
Kulkarni’s Academy
Example 4
Using Reynold’s roughness criterion
established the type of boundary for the
following data = k = 0.01 mm, shear stress
(0) = 4.9 N/m2, w = 0.001 Ns/m2, = 1000
kg/m3.
Sol.
v
Q = AV
V
0.07 0.01103
0.7
1106
vk
VD VD 6.36 0.1
636618.28
106
Hence flow is turbulent. Rough pipe
R
1
2 log10
1.74
f
k
1
0.05
2log10
1.74
3
f
0.15 10
0
49
0.07
1000
Reynold’s criteria
Q 50 103
6.36m / s
A (0.1)2
4
Re
1106 m2 / s
vk
4 0.7 Hence, smooth boundary.
v
Example 5
A rough pipe has a diameter of 0.08m the
velocity at point 3 cm from the pipe wall is
25% more than velocity at 1cm from pipe
wall find the average height of roughness.
Sol.
f = 0.02171
0
fV 2
8
1000 0.02171 6.362
8
0 = 109.806 N/m2
v
u2 = 1.2541
0
0.3313m / s
u
y
5.75log10 8.5
v
k
u1
y
5.75log10 1 8.5 ----(1)
v
k
Maximum velocity
umax
1 1.33 f
v
umax 1 1.33 0.02171 6.36
umax 7.6063 m/s
u2
y
5.75log10 2 8.5 ----(2)
v
k
Dividing equation 2 by equation 1
k = 0.00371 m
Kulkarni’s Academy
P
8.1
119
Turbulent Flow
8.6
Practice Questions
Turbulent flow generally occurs
(A) At very low velocities
(B) In flows of highly viscous fluids
(C) In flows through very narrow passages
(D) In flows at high velocities through large
passages
8.2
8.3
8.7
Flow in a pipe can be expected to be turbulent
when the Reynolds number based on mean
velocity and pipe diameter is
(A) = 0
(B) < 2000
(C) > 4000
(D) > 100
of kg/s. Take viscosity 0.001 Ns/m 2
and 1000 kg/m3 . The Darcy friction
factor is given as f D 64 / Re d for fully
developed
laminar
for
f D 0.316Re0.25
d
Shear stress in a turbulent flow is due to
(A) The viscous property of the fluid
(B) The fluid
(C) Fluctuation of velocity in the direction of
flow
(D) Fluctuation of the velocity in the
direction of flow as well as transverse to
it
8.4
Using the Prandtl's mixing length concept,
how is the turbulent shear stress expressed?
(A) l
du
dy
du
(C) l
dy
8.5
(B) l 2
2
du
dy
du
(D) l
dy
Consider a steady, fully developed turbulent
flow in a pipe of circular cross-section at high
Reynolds number.
If the pipe diameter is doubled at a constant
flow rate, by what factor does the pressure
drop decrease?
(A) 2
(B) 16
(C) 8
(D) 32
Water flows steadily through a smooth
circular tube of 5 cm diameter at a flow rate
2
2
Friction factor in laminar and turbulent flow
in a circular pipe varies as Re1 & Re0.25
respectively. If V is the average velocity, the
pressure drop for laminar and turbulent flow
respectively will be proportional to
(A) V and V 0.8
(B) V 0.5 and V 2
(C) V 0.5 and V 1.75
(D) V and V 1.75
8.8
flow
and
fully developed
turbulent flow. The approximate pressure
drop per unit length in the fully developed
region of the tube is
(A) 20 Pa/m
(B) 120 Pa/m
(C) 480 Pa/m
(D) 960 Pa/m
Match the flow conditions in a circular pipe
of diameter D and surface roughness k to the
corresponding functional relationships of
friction factor. f. Choose the correct matching
Flow in a circular pipe
(P) Laminar flow in smooth pipe
(Q) Turbulent flow in smooth pipe
(R) Turbulent flow in rough pipe (at high Re)
(S) Turbulent flow is rough pipe (at low Re)
Friction factor f
(1) f f (Re, k / D)
(2)
f f (k / D)
(3)
f f (Re)
(4)
f 64 / Re
Here, Re is the Reynolds number
(A) P – 4, Q – 1, R – 2, S – 3
(B) P – 2, Q – 3, R – 1, S – 4
(C) P – 3, Q – 4, R – 2, S – 1
(D) P – 4, Q – 3, R – 2, S – 1
Fluid Mechanics
8.9
The flow
120
of
1000 kg/m )
3
water
kinematic
density
viscosity
(B) P – 3, Q – 2, R – 4, S – 1
10 m /s ) in commercial pipe, having
equivalent roughness k s a 0.12 mm, yields an
(C) P – 2, Q – 3, R – 1, S – 4
6
and
(mass
2
average shear stress at the pipe boundary
600 N/m 2 . The value of ks / ' ( ' being
the thickness of laminar sub-layer) for this
pipe is
8.10
8.11
8.12
Kulkarni’s Academy
(A) P – 2, Q – 3, R – 4, S – 1
(A) 0.25
(B) 0.50
(C) 6.0
(D) 8.0
A steady flow of water takes place through a
pipe of 100 mm internal diameter and 10 m
length. The average velocity of the flow is 5
m/s and the wall shear tress is 250 N/m2 . The
pressure drop for the given pipe length is
(A) 2.5 105 N/m2
(B) 2.0 105 N/m2
(C) 5.0 104 N/m2
(D) 105 N/m 2
Velocity measurement of flow through a
rough circular pipe indicate that the average
velocity is 2.6 m/s and the centre line velocity
is 3.17 m/s. What is the friction factor for the
pipeline?
(A) 0.027
(B) 0.020
(C) 0.015
(D) 0.010
Match the following flow patterns with their
characteristics
(P) Turbulent flow
(Q) Boundary layer separation
(R) Laminar flow
(S) Steady flow
(1) No change in flow properties at any
point in the flow field
(2) Highly irregular and rapid fluctuations
of flow velocities
(3) Wake formation
(4) Smooth flow without mixing of layers
(D) P – 3, Q – 2, R – 1, S – 4
8.13
The velocity profile in turblent flow through
a pipe is approximated as
u
umax
1/7
y
,
R
where umax is the maximum velocity, R is the
radius and y is the distance measured normal
to the pipe wall towards the centerline. If u av
denotes the average velocity, the ratio
is
(A)
2
15
(B)
1
5
(C)
1
3
(D)
49
60
uav
umax
Kulkarni’s Academy
A
121
8.7
Answer Key
8.1
D
8.2
C
Turbulent Flow
8.3
(C)
D = 5 102 m
D
m kg/s
8.4
D
8.5
D
8.6
B
8.7
C
8.8
D
8.9
D
= 1000 kg/m3
8.10
D
8.11
A
8.12
A
FD
8.13
D
= 0.001 Ns/m2;
64
(For fully developed)
Re
F = 0.316 Re0.25
E
m A V
Explanation
1000
0.05
8.1
(D)
8.2
(C)
8.3
(D)
P1 P2
hL
w
8.4
(D)
P1 P2 whL
8.5
(D)
8.6
(B)
4
2
V
V = 1.6 m/s
P V 2
Q
V
4
d 2 V
4Q
D2
P
16Q2
2d 4
If d = 2d, Q = constant
P decreases by 16 times.
Hence, the correct option is (B).
wFLV 2
2 gD
P1 P2 wFV 2 g F V 2 FV 2
L
2 gD
2 gD
2D
Re
VD
1000 1.6 0.05
80000
0.001
(Re > 4000 so flow is turbulent)
F = 0.316 (80,000)0.25
= 0.0187
P1 P2 1000 0.0187 1.62
L
2 0.05
P1 P2
481 Pa/m
L
Hence, the correct option is (C).
Fluid Mechanics
8.8
(D)
8.9
(D)
122
Kulkarni’s Academy
8.11 (A)
V = 2.6 m/s
Umax = 3.17 m/s
Given that
U max
1 1.33 F
V
= 1000 kg/m3
= 106 m2/s
KS = 0.12 mm
3.16
1 1.33 F
2.6
0 = 600 N/m2
F = 0.026
600
Vx 0
0.7745
1000
'
11.6 11.6 106
1.497 105 m
V*
0.7745
= 1.497
102
Hence, the correct option is (A).
8.12 (A)
8.13 (D)
u
mm
umax
K
0.12
8.01
' 1.497 102
1/ 7
y
R
u = V at y = 0.223R
1/ 7
Hence, the correct option is (D).
V
0.223R
umax R
V
49
U max 60
8.10 (D)
Di = 100 mm
L = 10 m
Vavg = 5 m/s
0 = 250 N/m2
P
FLV 2
hL
w
2 gD
P
FLV 2
FV 2
. g {w g 0
2 g .D
8
FV2 = 80
P
FLV 2
8 L 8 250 10
0
2D
2D 2 100 103
105 N/m2
Hence, the correct option is (D).
0.8016
Hence, the correct option is (D).
NOTES
When fluid flows though pipes it encounters
various losses. These losses are classified into
major loss and minor loss.
9.1 Major loss:
m 4
D2
D
D
m
4
The head loss due to friction is known as major
loss. This major loss is given by
Darcy-weisbach equation:
hL
FLV 2
2 gD
Q AV
V
4
D 2V
4Q
D2
2
hL
i = hydraulic slope
h
tan L i
L
FL 4Q
16 FLQ
2
2 gD D
2 g 2 D5
2
FLQ 2
hL
2 g 2 5
D
16
FLQ2
{Flow should be steady
hL
12D5
Major losses are also calculated Chezy’s
formula.
According to Chezy’s formula
V c mi
m = hydraulic mean depth or diameter
A
area of flow
m
P wetted perimeter
V c mi
V c
D hL
.
4 L
4 LV 2
hL 2
cD
FLV 2
hL
2 gD
4 LV 2 FLV 2
c2D
2 gD
8g
c2
F
8g
c
F
Unit of Chezy’s constant:
c
m
sec.
Fluid Mechanics
124
9.2 Minor losses:
Losses due to
Kulkarni’s Academy
From continuity equation
Sudden expansion
Sudden contraction
Bend loss
Entrance loss
Exit loss are known as minor losses.
P1 P2 V2 V2 V1
g
g
V2 (V2 V1 ) (V12 V22 )
hL e p
g
2g
hL e p
hL e p
2
2
2V22 2VV
1 2 V1 V2
2g
(V1 V2 ) 2
2g
In deriving this equation Bernoulli’s equation,
momentum equation and continuity equations are
used.
(1) From Bernoulli’s equation:
P1 V12 P2 V22
hL exp
w 2g w 2g
P1 P2 V12 V22
w
2g
…. (2)
From equation 1 and 2
1. Minor losses due to sudden expansion:
hL exp
Q
V2
V2
Q = A2V2 ;
hL e p
…. (1)
Assumption:
The pressure in the eddy region is assumed to be
equal to upstream pressure.
V2 V
1 1 2
2 g V1
2
AV
1 1 A2V2
V2 A1
V1 A2
hL e p
V2 A
1 1 1
2 g A2
2
(2) Exit loss:
It is similar to sudden expansion with A2 is
infinite.
F m(v u)
F Q(v u) [momentum equation]
P1A1 + P1(A2 A1) – P2A2 = Q[V2 – V1]
(P1 P2)A2 = Q[V2 – V1]
P1 P2
Q
V2 V1
A2
hL e p
V2 A
1 1 1
2g
hL e p
V12 V 2
2g 2g
2
Kulkarni’s Academy
125
(3) Sudden contraction loss:
Flow Through Pipes
Entrance loss
hentrance
0.5V 2
2g
Where V is velocity in pipe.
(5) Bend loss:
Bend losses are given by hbend
hcont
V V
c 2
cc
2
Where V is velocity in pipe and K depends on
angle of bend and radius of curvature on bend.
2g
V22 Vc
1
2 g V2
KV 2
2g
2
Example 1
Water flows from a reservoir through a series
of pipe joined as shown in Fig.,
ac
a2
From continuity equation
acvc = a2v2
Vc a2 1
V2 ac cc
V2 1
hcontr . 2 1
2 g cc
2
Note : If the coefficient of contraction cc is not given
then sudden contraction losses are taken as
hcont
Find the percentage error in discharge when
minor losses are neglected. Assume K = 1 for
bends and friction factor F = 0.02 for all pipe
the available head of 20 m is used in
overcoming losses.
0.5V22
2g
Where V2 = velocity in smaller diameter pipe.
(4) Entrance loss:
It is similar to sudden contraction.
Sol.
Discharge when all losses are taken into
account (actual discharge).
a1v1 = a2v2 = a3v3 (discharge is same)
4
d12v1
4
d 22v2
4
d32v3
0.12 v1 = 0.22v2 = 0.12v3
v1 = 4v2 = v3
0.5V 2
2g
0.5v12 kv12 FL1v12 kv12
20
2g
2 g 2 gd1 2 g
Fluid Mechanics
126
2
2 2
2
3
2
3 3
2
3
energy line
2
3v1
2
2
2
2
0.5v1 v1 0.02 100v1 v1 4
2g
2g
2 g (0.1)
2g
2g
v12
2
2
16 0.5v1 0.02 100 v1
2 g 0.2
2g
2 g 0.1
0.02 200.
v12
20
2g
0.02548v12 0.05096v12 2.0387v12
0.02866v12 0.06371v12 0.02548v12
1.019v12 0.5096v12 20
4
0.12 2.926
Qa = 0.0229 m3/s
FL1v12 FL2v22 FL3v32
20
2 gD1 2 gD2 2 gD3
When minor losses are neglected
0.02 100 v12 0.02 200 v12
20
2 g 0.1
16 2 g 0.2
0.02 100 v12
2 g 0.1
V = 3.084 m/s
Qb AV
1 1
4
d12 V1
= 0.02422 m3/s %error
%error
Qb Qa
100
Qb
0.02422 0.0229
100
0.02422
= 5.456%
P
The line joining piezometric heads z at
w
various points in a flow is known as hydraulic
gradient line.
If the pipe is horizontal and of uniform diameter
hydraulic gradient line represent pressure
variation.
P
V2
TEL: The line joining total head z
2g
w
at various in a flow is known as total energy line
Distance between TEL and HGL gives velocity
head.
v1 =2.9242 m/s
HGL:
Note:
3.7022v12 20
Qa AV
1 1
Kulkarni’s Academy
9.3 Hydraulic gradient line and total
(v1 v2 ) FL v 0.5v FL v
v
2g
2 gd 2
2g
2 gd3 2 g
2
Graphical representation of Bernoulli’s
equation:
Kulkarni’s Academy
127
Flow Through Pipes
Example 3
A horizontal pipe of given diameter d
suddenly enlarges to D. Find the ratio D/d
such that the rise in pressure for a given
discharge post the enlargement shall be
maximum.
Note:
Sol.
In a flow HGL can rise or fall but the total energy
line will never rise as long as there is external
energy input. i.e. Total energy line will rise in the
case of pumps and compressors.
Example 2
At a sudden expansion of a water pipe line
from a diameter of 0.24 m to 0.48 m the HGL
rises by 10 mm, find the discharge through
pipe.
Apply Bernoulli’s equation between 1 & 2
P1 V12 P2 V22
hL exp
w 2g w 2g
Sol.
P1 V12
P V2
z1 2 2 z2 hL exp.
w 2g
w 2g
V12 V22
P
P
hL exp. 2 z2 1 z1
2g
w
w
V12 V22 V1 V2
3
10 10 m
2
g
2
g
2
3
VV
1 2 V2 10 10 g
Q = A1V1 = A2V2
0.242V1
0.482V2
4
4
V1 = 4V2
4V22 V22 0.01 g
v = 0.18 m/s
Q
d22 V2
4
4
3
Q = 0.3272 m /s.
0.482 0.18
P2 P1 V12 V22 (V1 V2 ) 2
w
2g
2g
P2 P1 P V1V2 V22
g
g
2g
2
P = [v1v2 v2 ]
P
0 for max pressure rise
V2
V1 2V2 = 0
V1 = 2V2
D2V1
D2V2
4
4
2
V
D
1 2
2
d
V2
D
2
d
Pipes in series:
Assumption:
(1) Minor losses are neglected.
(2) Friction factor is same.
Fluid Mechanics
Q1= Q2 = Q3 = Q4 = Q
Neglect minor losses b/c pipe length is more
hL hL1 hL2 hL3 ......
FL Q 2 FL Q 2 FL Q 2
hL 1 51 2 52 3 53
12d1
12d 2
12d3
Q1 = Q2 = Q3 = Q
Equivalent pipe:
A pipe of uniform diameter is said to be
equivalent to a compound pipe when the
discharge and head losses are same in both pipes.
hLe
FLeQ 2
12de5
128
Kulkarni’s Academy
2
a
2
b
Pa V
P V
za b
za hL1
w 2g
w 2g
…. (1)
Pa Va2
Pb Vb2
za
zb hL2
w 2g
w 2g
…. (2)
hL1 hL2
Note:
In case of parallel connection as the ends of pipes
are connected between same points therefore the
energy loss is same for all parallel pipes.
Equivalent pipe (Parallel):
Assumption:
1) All parallel pipes are assumed to be similar.
i.e. the length and diameters of each pipes is
same.
2) Friction factor is assumed to be same in
all pipes.
FLeQ 2 FL1Q 2 FL2Q 2 FL3Q3
...
12de5
12d15
12d 25
12d35
Le L1 L2 L3
.......
d e5 d15 d 25 d35
In Dupuit’s equation minor losses are neglected.
Pipes in parallel:
Parallel connection is used for increasing
discharge.
Q = Q1 + Q2
‘n’ number of similar pipes are connected in
parallel.
hLe
FLeQ 2
12de5
2
1
Q
hL FL .
5
n 12d
FLQ 2
hL
12n2d 5
FLQ 2 FLeQ 2
12n2d 5 12de5
L
L
e5
if Le = L
2 5
nd
de
de5 n2d 5
de n2/5d
Kulkarni’s Academy
129
Flow Through Pipes
Power transmission through pipes:
P
Practice Questions
P ( gh) A V
9.1
Pth = WQH
Pact. = WQ(H – hL)
Pact
Pth
9.2
H hL
H
The Reynolds number for flow of a certain
fluid in a circular tube is specified as 2500.
What will be the Reynolds number when the
tube diameter is increased by 20% and the
fluid velocity is decreased by 40% keeping
fluid the same?
(A) 1200
(B) 1800
(C) 3600
(D) 200
A pipeline is said to be equivalent to another,
if in both
(A) Length and discharge are the same
(B) Velocity and discharge are the same
Condition for max. power transmission
FLQ 2
Pact WQ H
12d 5
(C) Discharge and frictional head loss are
the same
FLQ3
Pact W QH
12d 5
(D) Length and diameter are the same
9.3
For max. efficiency
dPact .
3FLQ 2
H
0
dQ
12d 5
(A) Head loss due to friction is equal to the
head loss in eddying motion
H = 3hL
This is condition for max. power transmission.
Max. Efficiency =
max
2
3
While deriving an expression for loss of head
due to a sudden expansion in a pipe, in
addition to the continuity and impulse momentum equations, one of the following
assumptions is made
H hL 3hL hL
H
3hL
(B) The mean pressure in eddying fluid is
equal to the downstream pressure
(C) The mean pressure in eddying fluids is
equal to the upstream pressure
66.67%
(D) Head lost in eddies is neglected
9.4
Water steadily flowing from a 100 mm
diameter pipe abruptly enters a 200 mm
diameter pipe. If the velocity in the 100 mm
diameter pipe is 5 m/s, the head loss due to
abrupt expansion in terms of height of water
is
(A) 1.276 m
(B) 0.717 m
(C) 0.562 m
(D) 1.5 m
Fluid Mechanics
130
9.5
The hydraulic diameter of an annulus of inner
9.10
and outer radii Ri and RO respectively is
9.6
(A)
4( R0 R1 )
(B)
R0 R1
(C)
2( R0 Ri )
(D)
R0 R1
Two reservoirs that differ by a surface
elevation of 40 m, are connected by a
commercial steel pipe of diameter 8 cm. If the
desired flow rate is 200 N/s of water at 20 0C
, determine the length of the pipe. Assume
fluid properties of water at 20 0C as
1000 kg/m3 and
9.7
9.8
0.001 kg/m-s . The value of friction
factor (f) = 0.0185 may be chosen if
(A) 20.5 m
(B) 205 m
(C) 2050 m
(D) 20500 m
A farmer uses a long horizontal pipeline to
transfer water with a 1 hp pump and the
discharge is Q litres per min. If he uses 5 hp
pump in the same pipe line and assuming the
friction factor is unchanged the discharge
will be approximately
(A) 5 Q
(B) 51/2 Q
(C) (5Q)1/2
(D) 51/3 Q
The head loss due to a sudden contraction in
a pipeline is given by
1
V 2
(A) 2 1
CC 2 g
(B)
1 CC2
9.11
9.12
V2
2g
2
2
1
V2
2V
1
(C)
(D) CC 1
2g
CC
2g
Here CC is the contraction coefficient and V
9.9
Kulkarni’s Academy
A fire protection system is supplied from a
water tower with a bent pipe as shown in the
figure. The pipe friction 'f' is 0.03. Ignoring
all minor losses, the maximum discharge, Q,
in the pipe is
is the average velocity of flow in the
contracted section of the pipeline.
An elbow in a pipeline of cross-sectional area
0.01 m2 , has a loss coefficient of 2.0. If the
flow rate of water, through the pipeline is
360 m3 /hr , the head loss due to the elbow in
metres of water column is:
(A) 5
(B) 2
(C) 10
(D) 1
(A) 31.7 lit/sec
(B) 24.0 lit/sec
(C) 15.9 lit/sec
(D) 12.0 lit/sec
A 12 cm diameter straight pipe is laid at a
uniform downgrade and flow rate is
maintained such that velocity head in the pipe
is 0.5 m. If the pressure in the pipe is
observed to be uniform along the length when
the down slope of the pipe is 1 in 10, what is
the friction factor for the pipe?
(A) 0.012
(B) 0.024
(C) 0.042
(D) 0.050
A liquid is pumped at the flow rate Q through
a pipe of length L. The pressure drop of the
fluid across the pipe is P Now a leak
develops at the mid-point of the length of the
pipe and the fluid leaks at the rate of Q/2.
Assuming that the friction factor in the pipe
remains unchanged, the new pressure drop
across the pipe for the same inlet flow rate
(Q) will be
1
(A) P
2
5
P
8
3
(D) P
P
4
A single pipe of length 1500 m and diameter
60 cm connects two reservoirs having a
difference of 20 m in their water levels. The
pipe is to be replaced by two pipes of the
same length and equal diameter d to convey
25% more discharge under the same head
loss.
(C)
9.13
(B)
Kulkarni’s Academy
9.14
131
If the friction factor is assumed to be the same
for all the pipes, the value of d is
approximately equal to which of the
following options?
(A) 37.5 cm
(B) 40.0 cm
(C) 45.0 cm
(D) 50.0 cm
Two water carrying circular pipes are
connected in parallel. The length L1 ,
Flow Through Pipes
9.18
diameter d1 and friction factor f1 for the first
pipe are 200m, 0.5 m and 0.025 m
respectively, while L2 100 m, d 2 1.0 and
1.
f 2 0.02 . The velocity ratio V2 / V1 is
(A) 4.0
9.15
9.16
HW
9.19
HW
2( H W )
HW
2 HW
(C)
(D)
2
4( H W )
(H W )
The energy grade line (EGL) for steady flow
in a uniform diameter pipe is shown in figure.
Which of the following items is contained in
the box?
(A)
9.17
(B) 2.0
5
(C) 5.0
(D)
In a pipe flow, the head lost due to friction is
6m. If the power transmitted through the pipe
has to be the maximum, then the total head at
the inlet of the pipe will have to be
maintained at
(A) 36 m
(B) 30 m
(C) 24 m
(D) 18 m
The hydraulic diameter for flow in a
rectangular
duct
of
cross-sectional
dimensions H, W is
Match the items between the following two
groups concerning flow in a pipeline. Choose
the most suitable matching
List I
(P) Head loss due to friction
(Q) Head loss at entrance from a reservoir
to a pipeline
(R) Head loss due to sudden expansion
(S) Head loss due to a pipe bend
List II
V2
KL
2g
2.
2
L V
f
D 2g
V2
(V1 V2 ) 2
3.
4.
0.5
2g
2g
(A) P – 3, Q – 4, R – 1, S – 2
(B) P – 2, Q – 4, R – 1, S – 3
(C) P – 2, Q – 1, R – 3, S – 4
(D) P – 2, Q – 3, R – 4, S – 1
Three reservoirs A, B and C
interconnected by pipes as shown in
figure. Water surface elevations in
reservoirs and the piezometric head at
junction J are indicated in the figure.
are
the
the
the
(B)
Discharge Q1 , Q2 and Q3 are related as
(A) Q1 Q2 Q3
(B)
Q1 Q2 Q3
(C)
Q2 Q1 Q3
(D) Q1 Q2 Q3 0
(A)
(B)
(C)
(D)
A pump
A turbine
A partially closed valve
An abrupt expansion
9.20
The phenomenon of water hammer in pipe
flow originates from
(A) The microscopic form of all matter
(B) The non-Newtonian behaviour of water
Fluid Mechanics
132
(C) The critical point singularity of the
phase diagram
(D) The compressibility of water when
subjected to suddenly applied high
pressure
9.21 'n' identical pipes of length L, diameter d and
friction factor f are connected in parallel
between two reservoirs. What is the size of a
pipe of length L and of the same friction
factor f equivalent to the above pipe?
(A) n1/2d
(B) n1/5d
9.22
(C) n2/5d
(D) n1/3d
A centrifugal pump is used to pump water
through a horizontal distance of 150 m and
then raised to an overhead tank 10 m above.
The pipe is smooth with an I.D. of 50 mm.
What head (m of water) must the pump
generate at its exit (E) to deliver water at a
flow rate of 0.001 m3 /s ?
The Fanning friction factor, f is 0.0062.
Kulkarni’s Academy
Passage 24 - 25
A pipeline (diameter 0.3 m, length 3 km)
carries water from point P to point R (see
figure). The piezometric heads at P and R are
to be maintained at 100 m and 80 m,
respectively. To increase the discharge, a
second pipe is added in parallel to the existing
pipe from Q to R. The length of the additional
pipe is also 2 km Assume the friction factor,
f = 0.04 for all pipes and ignore minor losses.
9.24
9.25
9.23
(A) 10 m
(B) 11 m
(C) 12 m
(D) 20 m
A pipe carrying a discharge of 500 litres per
minute branches into two parallel pipes, x
and y, as shown in the figure. The length and
diameter of pipes x and y are shown in figure.
The friction factor f, for all pipes is 0.03. The
ratio of flow in pipes x and y is
(A) 0.36
(B) 0.44
(C) 0.67
(D) 1.00
What is the increase in discharge if the
additional pipe has same diameter (0.3 m)?
(A) 0%
(B) 33%
(C) 41%
(D) 67%
If there is no restriction on the diameter of the
additional pipe, what would be the maximum
increase in discharge theoretically possible
from this arrangement?
(A) 0%
(B) 50%
(C) 67%
(D) 73%
Passage 26
The Darcy-Weisbach equation for head loss
fLV 2
. A
2 gD
reservoir, as shown in the figure, stores water
to a height of 8 m. The entrance from the
reservoir to the pipe (length 50 m, diameter
10 cm) is sharp, with a loss coefficient of 0.5,
and the friction factor for the pipe is 0.017.
through a pipe is given as h f
Kulkarni’s Academy
9.26
What would be the discharge through the
pipe?
(A) 0.0311 m3 /s
0.0322 m3 /s
(C) 0.0331 m3 /s
(D) 0.0341 m3 /s
(B)
133
Flow Through Pipes
A
Answer Key
9.1
B
9.2
C
9.3
C
9.4
B
9.5
C
9.6
B
9.7
D
9.8
C
9.9
C
9.10
B
9.11
B
9.12
B
9.13
D
9.14
D
9.15
D
9.16
D
9.17
A
9.18
D
9.19
A
9.20
D
9.21
C
9.22
B
9.23
A
9.24
C
9.25
D
9.26
A
E
9.1
Explanation
(B)
Re = 2500
D = 1.2 D
V = 0.6 V
VD
Re
1.2D 0.6V
VD
0.72
0.72 2500
= 1800
Hence, the correct option is (B).
9.2
(C)
9.3
(C)
9.4
(B)
D1 = 100 mm
V1 = 5 m/s
A1V1 = A2V2
V2 A1 D12
V1 A2 D22
hL
V V
1 2
2g
2
D2 = 200 mm
Fluid Mechanics
134
V12 V2
1
2 g V1
2
52 1002
1
2 9.81 2002
2
hL = 0.7167 m
hL = 0.717 m due to sudden exp.
Hence, the correct option is (B).
9.5
(C)
4A
p
Dh
4.
D
4
2
o
(C)
9.9
(C)
A = 0.01 m2
K = 2; Q = 360 m3/hr =
V
360
0.1 m3/s
3600
Q
0.1
V
10 m/s
A
0.01
KV 2 2 100
hL
10 m
2g
2 9.81
Hence, the correct option is (C).
Di2
Do Di
(B)
Total length = 25 + 150 = 175 m
A1V1 = A2V2
AV
V1 2 2
A1
(D)
A1 >> A2
V1 is negligible
Apply Bernoulli’s between 1 and 2
Hence, the correct option is (C).
9.7
9.8
9.10 (B)
Do Di = 2(Ro Ri)
9.6
Kulkarni’s Academy
1 hp pump
Discharge = Q liter per min
P = wQhL
FLQ2
P wQ.
; P Q3
12ds
P2 Q23
P1 Q13
5 Q23
1 Q3
Q2 5 3 Q
1
Hence, the correct option is (D).
P1 V12
P V2
z1 2 2 z2 hL
w 2g
w 2g
V22
hL
2g
Exit loss so neglected
hL = 25 m
25
FLQ2
hL
12d 5
0.03 175 Q 2
25
12(0.1)5
Q = 0.0239
= 23.9 103 m3/sec
= 23.9 lit/sec
Hence, the correct option is (B).
Kulkarni’s Academy
135
Flow Through Pipes
P 13 P12 P23
9.11 (B)
h
1
tan L slope
L
10
D = 0.12 m
wF
1 1
LQ 2
5
12d
2 8
wFLQ 2 5
P13
12d 5 8
V2
0.5m
2g
hL
L Q 2
L 2
F
Q
F
2
2 2
w. 5
w.
5
12d
12d
5
P13 P
8
2
FLV
2 gD
Hence, the correct option is (B).
hL
F V
L
D 2g
1
F
(0.5)
10 0.12
0.12
0.12
F
0.024
10 0.5
5
Hence, the correct option is (B).
2
9.13 (D)
L = 1500 m,
9.12 (B)
D = 0.6 m
Head difference = 20 m
hL
FLQ 2
FLQ 2
20
…. (1)
12(d )2
12(0.6)5
Apply Bernoulli’s between 1 and 2
P1 V12 P2 V22
hL V1 = V2
w 2g w 2g
P1 P2 = whL P = whL
P
wFLQ
12d 5
Q
2
(both same diameter and same length)
Q = 0.625 Q
Q
2
... (1)
20
FL(0.625Q) 2
12(d )5
Equation 1 = equation 2
D = 49.71 50 cm
Hence, the correct option is (D).
…. (2)
Fluid Mechanics
9.14
136
(D)
hL1 hL2
9.20 (D)
2
FLV
FL V 2
1 1
2 2
2 gD1
2 gD2
0.025 200 V12 0.02 100 V22
0.5
1
V
5V12 V22 2 5
V1
Hence, the correct option is (D).
9.15
9.21 (C)
9.22 (B)
F = 0.0062
F = 4F = 0.0248
I.D. = 50 mm = 50 103 m
Q = 0.001 m3/5
(D)
hL = 6 m
For max. power H = 3 hL
H = 18 m
Hence, the correct option is (D).
9.16
Kulkarni’s Academy
(D)
L = 150 + 10 = 160 m
FLQ2
hL
12d 5
0.0248 160 (0.001)2
12 (0.05) 2
hL = 1.05 m
Head generates = 10 + 1.05 = 11.05 m
Hence, the correct option is (B).
4 Ac
P
4 Hw
2 Hw
2( w H ) ( H w)
Hence, the correct option is (D).
Dh
9.17
(A)
9.18
(D)
9.19
(A)
9.23 (A)
hL1 hL2
Q12 d15 0.25 Q1 2
Q22 d 25 0.35 Q2 3
200 – 160
180 160
Q1 + Q2 = Q3
Hence, the correct option is (A).
5/2
0.3628
Hence, the correct option is (A).
9.24 (C)
High to low
FL1Q12 FL1Q22
12d15
12d 25
Kulkarni’s Academy
137
FLQ2
20 12 (0.3)5
2
Q
12d 5
0.04 3000
3
Q1 = 0.0697 m /s
Flow Through Pipes
9.26 (A)
20
H=8m
L = 50 m
D = 0.1 m
Loss coefficient = 0.5 F = 0.017
Q
0.04 2000 2
2
0.04 1000 Q2
2
20
5
5
12(0.3)
12(0.3)
2
Q2 = 0.0985 m3/s
Q Q1
0.0985 0.0697
100
100
2
Q1
0.0697
= 41.4%
Hence, the correct option is (C).
9.25
(D)
If no restriction on the diameter of addition
pipe
hL
FLQ 2
12(d )5
hL
0.017 50 Q 2
12 (0.1)5
hL
FLQ 2
12(d )5
Apply Bernoulli’s equation between 1 and 2
P1 V12
P2 V22
z1
z2 hL
w 2g
w 2g
Then hL = 8 m
Total head loss = sudden contraction + hL +
exit loss
0.5V 2 fLV 2 V 2
8
2g
2 gd 2 g
0.04 1000 Q22
20
0
12(d )5
8
Q2 = 0.12 m3/s
V = 3.96 m/s
2
FLQ
0
12d 5
Q Q1
% increase 2
100
Q1
hL
0.12 0.0697
100 = 73.23%
0.0697
Hence, the correct option is (D).
0.5 V 2 0.017 50 V 2
V2
2 9.81
2 9.81 0.1 2 9.81
Q = AV
=
4
d 2 3.96 0.0311 m3/sec
Hence, the correct option is (A).
NOTES
Boundary layer theory was proposed by Prandtl
in 1904.
10. Development of boundary layer over a
flat plate:
When a real fluid flows past the solid object
the velocity of the fluid will be same as the
velocity of object, when it comes in contact with
object. If the object is stationary the fluid will
also have zero velocity. Away from the object the
fluid velocity increases and at some distance
from the object the fluid velocity will be equal to
free stream velocity, this distance from the object
where there are velocity gradients is known as
boundary layer thickness & this region is known
as boundary layer region.
In the boundary layer region, the flow is
viscous and rotational as the flow is viscous in
the boundary layer region, Bernoulli’s equation
is not applicable in the boundary layer region.
Outside the boundary layer region as the flow
is not viscous Bernoulli’s equation can be
applied.
When a real flow fluid past a flat plate the
velocity of the fluid on the plate will be same as
that of the plate velocity if the plate is at rest the
fluid will also have zero velocity. The boundary
layer thickness grows as the distance from the
leading-edge increases upto some/certain
distance from the leading edge the flow in the
boundary layer is laminar as the laminar
boundary layer grows instability occur and flow
changes from laminar to turbulent through
transition. It is found that even in turbulent
boundary layer region close to the plate the flow
is laminar, this region is known as laminar
sublayer region.
Laminar sublayer region exists in turbulent
boundary layer region.
10.2 Boundary Condition:
Kulkarni’s Academy
139
[1] x = 0; = 0
[2] y = 0; u = 0
du
[3] y = ; u = u
[4] y = ;
0
dy
Nominal boundary layer thickness or
Boundary layer thickness []
It is the distance from the boundary to the point
in y – direction where the velocity is 99% of free
stream velocity.
Note:
For all calculation purposes at y = ; u = u.
Boundary Layer Theory
Definition:
It is the distance by which boundary should be
displaced in order to compensate for the
reduction in mass flow rate due to boundary layer
growth.
10.3 Momentum thickness ():
It is the distance by which the boundary should
be displaced in order to compensate for the
reduction in momentum due to boundary layer
growth.
u
u
1 dy
u
u
0
Displacement thickness (*)
10.4 Energy thickness (E):
It is the distance by which boundary should be
displaced in order to compensate for the
reduction in kinetic energy due to boundary layer
growth.
u u2
1 2 dy
u
u
0
E
Reduction in mass flow rate due to boundary
layer growth is
= mideal mreal
= (uu)dy
Example 1
The velocity distribution in boundary layer is
u
y
given by
[linear velocity profile]
u
Find :
(1) Displacement thickness
(2) Momentum thickness
Total mass reduction =
(u
u )dy
…. (1)
Sol.
0
(1) Displacement thickness
0
1
u
dy
u
m = u(*1)
from equation (1) and (2)
= (u u )dy u *
0
*
…. (2)
0
u
1 dy
u
y2
y
y
y3
1 dy 2
2 3 0
0
1
(u u )dy
u 0
2
2
(2) Momentum thickness ():
y
y2
* 1 dy y
2 0
0
2 3 6
>* >
Fluid Mechanics
140
Note:
*
Shape factor of boundary layer ( H )
This term is used in the analysis of flow
separation.
For linear velocity profiles, the shape factor
is 3.
Example 2
Assume that the shear stress distribution
varies linearly in laminar boundary layer
y
such that as 0 1
Find displacement thickness.
Sol.
y
0 1
du
dy
y
du 0 1 dy
y2
u 0 y
c
2
at y = 0; u = 0 , c = 0
y2
u 0 y
2
at
y = ; u = u
u 0
2
Kulkarni’s Academy
10.5 Von-Karman momentum integral
equation:
Assumptions
(1) steady flow
(2) Incompressible flow
(3) 2 – D flow
dP
(4)
0 (this condition is valid for external
dx
flows only)
From Newton second law of motion. Von –
Karman equation can be derived it is
0
d
2
u dx
For external flow
0 = shear stress on the plate surface
= momentum thickness
0
d dP
2
u dx dx int ernal / pipe flow
x = distance from the leading edge.
10.6 Significance of von-Karman equation:
…. (1)
…. (2)
[1] With the help of von-Karman equation
the boundary layer thickness can be
calculated.
[2] Shear stress on the surface of the plate
can be calculated.
[3] The drag force on the plate can be calculated.
Note:
y y2
u 2 2 y 2
1
u
2
2
0
1
u
dy
u
2
y 2
1
y
0 2 dy
y 2 2 y3
2 3
y
6 2 0
6 2
*
3
1
3
Reynolds no =
u x u x
v
Where x = distance from leading edge
For flow over flat plate in the Reynolds number
(Re) less than 5 105 then flow is taken as
laminar.
And when Re > 5 105 flow is Turbulent.
Kulkarni’s Academy
141
Boundary Layer Theory
Avg. drag coefficient [CD]
0
FD
CD
1
Au2
2
With the help of CD, drag force can be calculated.
0
1 2
u
2
Example 3
For a velocity profile for a laminar boundary
u 3 y y3
u 2 3
layer
Find:
(1) Boundary layer thickness
(2) Shear stress on the surface of the plate
(3) Drag force
(4) Avg. drag coefficient in terms of
Reynold number
U 3 y y3
U 2 23
Sol.
dU
dy
U
U
1
dy
U U
0
3 y y3 3 y y3
1
dy
2 23 2 23
0
39
280
By using Von-Karman equation
0
d
2
U dx
y 0
y 0
3U
2
3U
2
Equation (i) = equation (ii)
3U
39
d
U 2
280
dx
2
13U d
140 dx
140
d
dx
13U
… (ii)
2
140
xc
2 13U
At
x 0, 0, c 0
2
140
x
2 13U
280
x
13U
From equation (iii)
2
d 39
0 U 2
dx 280
dU
dy
0
… (i)
dU
dy
0
10.7 Local drag coefficient or skin friction
coefficient:
C fx
39
d
U 2
280
dx
280 x
13 U
280 x 2
13 Re x
2
280
13
x
Re x
….(iii)
Fluid Mechanics
4.64 x
Re x
142
Kulkarni’s Academy
Note:
As the distance from the leading-edge
increases, the shear stress decreases.
(3)
4.64x
U x
x1/2 (For any laminar boundary layer)
2
1
x2
x1
dFD 0 Bdx
L
FD 0 Bdx
0
0.323U
Re x dx
x
0
L
x1 and x2 distance from leading edge.
Note:
As the distance from the leading edge is
increasing the boundary layer thickness is
also increasing.
3U
(2) 0
2
3U
0.323U
0
Re x
4.64 x
x
2
Re x
0
0.323U
x
1
0
x
01
02
U x
FD
U L
Re L
FD 0.646 ReL BU
(4) Average drag coefficient:
CD
CD
x2
x1
FD
1
AU 2
2
0.646 Re L BU
1
( BL)U 2
2
U L
BU
( BL)U 2
1.292
CD
1.292 Re L
U L
CD
1.292 Re L
Re L
CD
1.292
Re L
Kulkarni’s Academy
143
Boundary Layer Theory
Note:
F1
F2
If F1 is drag force on first half of the plate and
F2 is the drag. Force on second half of the
plate as shear stress is more on the first half
of the plate there fore F1 F2 .
10.8 Boundary layer separation:
When fluid flows through converging
passage the velocity increases pressure
decreases i.e. the fluid flows under Negative
pressure gradient (favorable pressure
gradient). This flow is also known as
accelerating flow. The boundary layer
thickness decreases in this region due to
increase in velocity.
When the fluid flows in diverging passage
velocity decreases and pressure increases i.e.
the fluid flows under positive pressure
gradient if the angle of divergence is large the
retardation of fluid particles will be more and
at some point the fluid particles may not
support the flow and the fluid may separate
from its boundary and may reverse the flow
this is known as boundary layer separation
and this occurs at the boundary.
As the velocity gradient is zero at the separation
point, the shear stress is zero at the separation
point.
10.11 Blausius solution:
Blausius developed non – linear 3rd order
ordinary differential equation for obtaining
boundary layer solutions.
Laminar
Turbulent
5x
Re x
0.371x
1
(Re x ) 5
CFx
0.664
Re x
0.058
1
(Re x ) 5
CD
1.328
Re L
CD
0.074
1
(Re x ) 5
Fluid Mechanics
P
10.1
144
10.5
Practice Questions
Boundary layer is a thin fluid region close to
the surface of a body where
(A) Viscous forces are negligible
(B) Velocity is uniform
(C) Inertial forces can be neglected
(D) Viscous forces cannot be neglected
10.2
In the boundary layer, the flow is
(A) Viscous and rotational
(B) Inviscid and irrotational
10.6
(C) Inviscid and rotational
(D) Viscous and irroational
10.3
The hydrodynamic boundary layer thickness
is defined as the distance from the surface
where the
(A) Velocity equals to the local external
velocity
(B) Velocity equals the approach velocity
(C)
Momentum equals 99% of
momentum of the free stream
the
(D) Velocity equal 99 % of the local
external velocity
10.4
10.7
Which one of the following statements is
correct? While using boundary layer
equations, Bernoulli's equation
(A) Can be used anywhere
(B) Can be used only outside the boundary
layer
(C) Can be used only inside the boundary
layer
(D) Cannot be used either inside or outside
the boundary layer
10.8
Kulkarni’s Academy
How is the displacement thickness in
boundary layer analysis defined?
(A) The layer in which the loss of energy is
minimum
(B) The thickness upto which the velocity
approaches 99%
(C) The distance measured perpendicular to
the boundary by which the free stream
is displaced on account of formation of
boundary layer
(D) The layer which represents reduction in
momentum caused by the boundary
layer
Consider the following statements:
1.
Boundary-layer thickness in laminar
flow is greater than that of turbulent
flow
2.
Boundary-layer thickness in turbulent
flow is greater than that of laminar flow
3.
Velocity distributes uniformly in a
turbulent boundary layer
4.
Velocity has a gradual variation in a
laminar boundary-layer
Which of the statements given above are
correct?
(A) 1, 3 and 4 only (B) 1, 2, 3 and 4
(C) 1 and 2 only
(D) 2, 3, and 4 only
List I give the different items related to a
boundary layer while List II gives the
mathematical expression. Match List I with
List II and select the correct answer
(symbols have their usual meaning).
The displacement thickness at a section, for
an air stream ( 1.2 kg/m 3 ) moving with a
velocity of 10m/s over flat plate is 0.5mm .
What is the loss of mass flow rate of air due
to boundary layer formation in kg per meter
width of plate per second?
(A) 6 103
(B)
6 105
3103
(D)
2 103
(C)
Kulkarni’s Academy
10.9
145
Given that
boundary layer thickness,
* displacement thickness
e energy thickness
Boundary Layer Theory
10.14
q momentum thickness
The development of boundary layer zones
labelled P,Q,R and S over a flat plate is
shown in the given figure. Based on this
figure, match List I (Boundary layer zones)
with list II (types of boundary layer) and
select the correct answer
The shape factor H of boundary layer is
(A) H e
(B) H e
*
(C) H
(D) H
e
10.10 The laminar boundary layer thickness over a
flat aligned with the flow varies as
(A) x1/2
Column I
(A) P
(B) Q
(B) x4/5
(C) x1/2
(D) x 2
10.11 The turbulent boundary- layer thickness
varies as
(A)
x4/5
(B)
x1/5
(C)
(D)
(A)
(B)
(C)
(D)
(C) x1/2
(D) x1/7
10.12 In the laminar boundary layer flow over a flat
plate, the ratio varies as
x
(A) Re
(B)
Re
1
(D) Re1/2
Re
10.13 A flat plate is kept in an infinite fluid
medium. The fluid has a uniform freestream
velocity parallel to the plate. For the laminar
boundary layer formed on the plate, pick the
correct option matching Columns I and II
Column I
(P) Boundary layer thickness
(Q) Shear stress at the plate
(R) Pressure gradient along the plate
Column II
1.
Decreases in the flow direction
2. Increases in the flow direction
3. Remains unchanged
(A) P–1, Q–2, R–3 (B) P–2, Q–2, R–2
(C) P–1, Q–1, R–1 (D) P–2, Q–1, R–3
(C)
10.15
Column II
(1) Transitional
(2) Laminar
Viscous sub
- layer
R
(3) Laminar
S
(4) Turbulent
P – 3, Q – 1, R – 2, S – 4
P – 3, Q – 2, R – 1, S – 4
P – 4, Q – 2, R – 1, S – 3
P – 4, Q – 1, R – 2, S – 3
Air (kinematic viscosity 15 106 m2 /s ) with
a free stream velocity of 10 m/s flows over a
smooth two-dimensional flat plate. If the
critical Reynolds number is 5 105 , what is
the maximum distance from the leading edge
upto which laminar boundary layer exists?
(A) 30 cm
(B) 75 cm
(C) 150 cm
(D) 300 cm
10.16 Velocity distribution in a boundary layer
flow over a plate is given by u / u 1.5
y
; y is the distance measured
normal to the plate; is the boundary layer
thickness; and u is the maximum velocity at
where,
y if the shear stress , acting on the plate
Fluid Mechanics
146
mu
where, is the
dynamic viscosity of the fluid, K takes the
value of
(A) 0
(B) 1
is given by K
(C) 1.5
10.17
10.18
density and viscosity of the fluid respectively
are 1.23 kg/m3 and 1.79 105 Ns/m 2 .
Transition occurs at a distance xcr 0.1m
(D) None of these
The thickness of the laminar boundary layer
on a flat plate at a point A is 2 cm and at a
point B, 1 m downstream of A, is 3 cm. What
is the distance of A from the leading edge of
the plate?
(A) 0.50 m
(B) 0.80 m
(C) 1.00 m
(D) 1.25 m
from the leading edge. If the free stream
velocity is changed to u 120 m/s, X cr
becomes
(A) 0.2 m
(C) 0.05 m
10.22
The critical value of Reynolds number for
transition from laminar to turbulent boundary
layer in external flows is taken as
(A) 2300
(C)
10.19
Kulkarni’s Academy
10.21 Consider a constant pressure boundary layer
over a flat plate of length L = 3m. The free
stream velocity is u 60 m/s and the
the leading edge of the plate?
(D) 3 106
Which one of the following is the correct
relation between the boundary layer
thickness ¸ displacement thickness * and
the momentum thickness ?
(A)
*
(B)
*
(C)
*
(D)
*
10.20 For air flow over a flat plate, velocity (U) and
boundary layer thickness () can be
expressed respectively, as
3
U 3y 1 y
4.64 x
;
. If the free
U 2 2
Re x
stream velocity is 2 m/s, and air has
kinematic viscosity of 1.5 105 m2 /s and
density of 1.23 kg/m 3 , then wall shear stress
In a laminar boundary layer over a flat plate,
what would be the ratio of wall shear stress
1 and 2 at the two sections which lie at
distances x1 30 cm and x2 90 cm from
(B) 4000
5 105
(B) 0.1 m
(D) 0.005 m
(A)
1
3.0
2
(C)
1
1
32
2
(B)
1 1
2 3
(D)
1
1
33
2
10.23
Air flow in a square duct of side 10 cm. At
the entrance, the velocity is uniform at 10 m/s
and the boundary layer thickness is
negligible. At the exit, the displacement
thickness is 5 mm (on each wall). The
velocity outside the boundary layer at the exit
is
(A) 12.35 m/s
(B) 11.08 m/s
(C) 10 m/s
(D) 9 m/s
10.24 The boundary layer flow separates from the
surface if
(A)
du
dp
0 and
0
dy
dx
(B)
du
dp
0 and
0
dy
dx
du
dp
0 and
0
dy
dx
at x = 1m, is
(A)
2.3102 N/m2
(B)
43.6 103 N/m2
(C)
4.36 103 N/m2
(C)
(D)
2.18 103 N/m2
(D) The boundary layer thickness is zero
Kulkarni’s Academy
10.25
147
At the point of separation
(A) Velocity is negative
(B) Shear stress is zero
(C) Pressure gradient is negative
(D) Shear stress is maximum
10.25
Flow separation is caused by
(A) Reduction of pressure to local vapour
pressure
(B) A negative pressure gradient
(C) A positive pressure gradient
(D) Thinning of boundary layer thickness to
zero
10.27
10.28
For a laminar boundary layer with constant
dp
free stream velocity (i.e.
0 ), the
dx
u
variation of
with distance from the wall
y
is given by
Boundary Layer Theory
10.31 The laminar boundary layer over a large flat
plate held parallel to the flow is 7.2 mm thick
at a point 0.33 m downstream of the leading
edge. If the free stream speed is increased by
50%, then the new boundary layer thickness
at this location will be approximately
(A) 1.8 mm
(B) 8.8 mm
(C) 5.9 mm
(D) 4.8 mm
10.32 For the control volume shown in the figure
below, the velocities are measured both at the
upstream and the downstream ends. The flow
of density is incompressible, two
dimensional and steady.
The pressure is P0 over the entire surface of
the control volume. The drag on the airfoil is
given by
Which one among the following boundary
layer flows is LEAST susceptible to flow
separation?
(A) Turbulent boundary layer in a favorable
pressure gradient
(B) Laminar boundary layer in a favorable
pressure gradient
(C) Turbulent boundary layer in adverse
pressure gradient
(D) Laminar boundary layer in adverse
pressure gradient
10.29 In a two-dimensional, steady, fully
developed, laminar boundary layer over a flat
plate, if x is the stream wise coordinate, y is
the wall normal coordinate and u is the
streamwise velocity component, which of the
following is true?
10.30 The maximum thickness of boundary layer
in a pipe of radius 'R' is
(A) 0.1 R
(B) 0.22 R
(C) 0.5 R
(D) R
u2 h
(A)
(B) Zero
3
u2 h
(C)
(D) 2u2 h
6
Linked Answer Question 10.33 to 10.34
The boundary layer formation over a flat
plate is shown in the figure below. The
variation of horizontal velocity (u) with y and
x along the plate in the boundary layer is
approximated as: u P sin(Qy) R
Fluid Mechanics
10.33 The most acceptable boundary conditions
are
(A) At y 0, u 0 ; at y , u U ; at
y 0,
du
0
dy
(B) At y 0, u U ; at y , u U ; at
y 0,
du
0
dy
(C) At y 0, u 0 ; at y , u U ; at
y ,
du
0
dy
(D) At y 0, u U ; at y , u U ; at
du
y ,
0
dy
10.34 Expression for P, Q and R are
(A)
P 0, Q 0, R 0
(B)
P U , Q 0, R 0
(C)
P 0, Q
(D)
P U , Q
, R U
2
, R0
2
Linked Answer Question 10.35 to 10.36
A smooth flat plate with a sharp leading edge
is placed along a gas stream flowing at U =
10 m/s. The thickness of the boundary layer
at section r - s is 10 mm, the breadth of the
plate is 1 m (into the paper) and the density
of the gas 1.0 kg/m 3 . Assume that the
boundary layer is thin, two dimensional, and
follows a linear velocity distribution,
u U ( y / ) , at the section r-s, where y is
the height from plate.
148
Kulkarni’s Academy
10.35 The mass flow rate (in kg/s) across the
section q-r is
(A) Zero
(B) 0.05
(C) 0.10
(D) 0.15
10.36 The integrated drag force (in N) on the plate
between p-s, is
(A) 0.67
(B) 0.33
(C) 0.17
(D) Zero
Linked Answer Question 10.37 to 10.38
An automobile with projected area 2.6 m2 is
running on a road with speed of 120 km per
hour. The mass density and the kinematic
viscosity of air are 1.2 kg/m3 and
1.5 105 m2 /s , respectively. The drag
coefficient is 0.30.
10.37 The drag force on the automobile is
(A) 620 N
(B) 600 N
(C) 580 N
(D) 520 N
10.38 The metric horse power required to overcome
the drag force is
(A) 33.23
(B) 31.23'
(C) 23.23
(D) 20.23
Linked Answer Question 10.39 to 10.40
Consider a steady incompressible flow
through a channel as shown below.
The velocity profile is uniform with a value
of u0 at the inlet section A. The velocity
profile at section B downstream is
y
0 y
Vm ,
u Vm ,
y' H
Hy
Vm
, H y H
Kulkarni’s Academy
10.39 The ratio
149
Vm
is
u0
1
(B) 1
1 2( / H )
1
1
(C)
(D)
1 ( / H )
1 ( / H )
p pB
10.40 The ratio A
(where PA and PB are the
1 2
u 0
2
pressure at section A and B, respectively, and
is the density of the fluid) is
1
1
(A)
1 (B)
2
2
1 ( / H )
1 ( / H )
(A)
(C)
1
1 (2 / H )
2
1 (D)
1
1 ( / H )
10.41 Consider a laminar flow over a flat plate of
length L = 1m. The boundary layer thickness
at the end of the plate is w for water, a for
air for the same free stream velocity. If the
kinematic viscosities of water and air are
and
1106 m2 /s
1.6 105 m2 /s ,
respectively, the numerical value of the ratio,
w
______.
a
10.42 A fluid of constant density flows steadily
past a porous plate with a uniform free stream
velocity u as shown in the figure.
Fluid is sucked through the porous section
with a velocity of 0.1u . Velocity
distribution at section CD is given by
3
u 3 y 1 y
.
u 2 2
Boundary Layer Theory
Mass flow rate per unit width of the plate,
perpendicular to the plane of the figure across
the section BC is
3
(A) u
8
3
(B) u 0.1L
8
3
(C) u 0.1L
2
3
(D) u 0.1L
8
10.43 A thin flat plate of dimensions of
100cm 200cm is completely immersed in
an oil stream with velocity 6 m/s. The density
and dynamic viscosity of oil may be taken as
890 kg/m 3 and 0.29kg/m.s Assume a drag
coefficient given by CD 1.328Re0.5
L , where
Re L is the Reynolds number based on the
plate length. The total frictional force, if the
fluid steam is along the longer side of the
plate, is numerically closest to
(A) 4.435 N
(B) 44.35 N
(C) 443.5 N
(D) 4435 N
10.44 A flat plate is exposed to a steady, constant
density fluid flow with a free stream parallel
to the axis of the plate (case 1). In another
case, this plate is replaced by a plate which is
half the length of the previous plate (case 2),
all other conditions remaining unaltered. In
both the cases, flow over the entire length of
the plate is laminar. What is the ratio of the
drag coefficients for these two cases (Given:
the local boundary layer thickness scales
as
Rex 1/2 , where Re x is the local
x
Reynolds number at an axial coordinate x)?
cD.1
cD.1
0.500
0.666
(A)
(B)
cD.2
cD.2
cD.1
cD.1
0.707
1.000
(C)
(D)
cD.2
cD.2
Fluid Mechanics
A
150
10.9 (C)
Answer Key
10.1
10.4
10.7
10.10
10.13
10.16
10.19
10.22
10.25
10.28
10.31
10.34
10.37
10.40
10.43
D
B
A
C
D
C
A
C
B
A
C
D
D
A
C
10.2
A
10.5
C
10.8
A
10.11
A
10.14
A
10.17
B
10.20
C
10.23
A
10.26
C
10.29
B
10.32
A
10.35
B
10.38
C
10.41 0.25
10.44
C
Kulkarni’s Academy
10.3
10.6
10.9
10.12
10.15
10.18
10.21
10.24
10.27
10.30
10.33
10.36
10.39
10.42
D
D
C
D
B
C
C
B
*
D
C
C
C
D
10.10
(C)
For laminar flow
x
10.1
10.11
(A)
10.12
(D)
Explanation
4.64 x
Re
4.64
Re
x
10.4
(B)
10.5
(C)
10.6
(D)
10.7
(A)
10.8
(A)
Re
1
2
Hence, the correct option is (D).
10.2 (A)
(D)
2
(D)
10.3
1
5x
u x
Hence, the correct option is (C).
x
E
5x
Re x
10.13
(D)
10.14
(A)
10.15
(B)
Given data :
15 106 m2/s
u = 10 m/s
Given data :
* = 0.5 mm
= 1.2 kg/m3
u = 10 m/s
m u *
1.2 10 0.5
6 103 kg/s
Hence, the correct option is (A).
Re = 5 105
Re
u x u x
5 105
10 x
15 106
x = 0.75 x
= 75 cm
Hence, the correct option is (B).
Kulkarni’s Academy
10.16
151
(C)
u
1.5 y
1.5
u
Boundary Layer Theory
10.18
(C)
10.19
(A)
10.20
(C)
3
u 3y 1 y
4.64 x
;
u 2 2 6
Re x
y
y = 0; = 0
u = 2 m/s
y
= 1.5 105 m2/s
= 1.23 kg/m3
du
0
dy
k ( u )
0
du
dy 0
y ,
u
. u 1.5
=
At y = 0 i.e., on the wall
= 0
u
k ( u )
1.5
3u
2
3 1.23 1.5 10 5 2
4.64 1
2
2 1
1.5 105
0 = 4.35 103 N/m2
k = 1.5
Hence, the correct option is (C).
10.17
|x1m 0
1.5
Hence, the correct option is (C).
(B)
10.21
(C)
L=3m
3 cm
2 cm
= 1.79 105 pq.s.
A
x1
B
1m
x
1
x
1
2
x2
u = 60 m/s ; = 1.23 kg/m3 ;
3
x1
x1
4
2
1 x1 1 x1 9
4 + 4x1 = 9x1
5x1 = 4
x1 = 0.8 m
Hence, the correct option is (B).
Transition occurs. xcr = 0.1 from leading edge
x=?
Re x
u x
1.23 60 0.1
412290.50
1.79 105
=
1.23 120 x
1.79 105
x = 0.05 m
Hence, the correct option is (C).
Fluid Mechanics
10.22
152
(C)
Kulkarni’s Academy
Y
1
x
1
x
2
2
x1
du
dy
1
90
3
2
30
1
When y = ;
2
y 0;
Hence, the correct option is (C).
10.23
(*)
10.27
(A)
r
du
0
dy
du
max
dy
(A)
In laminar flow due to law velocities,
separation is happing so turbulent is least
susceptible to flow separation.
Hence, the correct option is (A).
10.29 (B)
10.28
Boundary layer thickness is negligible
A2 81 cm 2
* 5 mm
A1 100 cm 2
y
10 cm
U 1 10 m/s
x
u
u
y
x
Hence, the correct option is (B).
at exit
9
10
10.30
(D)
Fully developed
R max
A 81 cm 2
Apply continuity
A1V1 = A2V2
10 100 = 81 V2
V2 = 12.35 m/s
Hence, the correct option is (A).
10.24
(B)
10.25
(B)
10.26
(C)
Developing
flow
Hence, the correct option is (D).
10.31 (C)
4.64 x
Re x
= 7.2 mm
X = 0.33
u2 1.5u1
4.64x
u x
R
Kulkarni’s Academy
153
x
u x
x
u
Boundary Layer Theory
10.33
(C)
y = 0; u = 0,
x1 = x2
1
2
x
u1
x
u 2
y = ; u = u,
x1 u2
x1 u1
y = ;
1.5u1
Hence, the correct option is (C).
1
1.5
2
7.2mm
2
5.87 mm
1.5
10.34
(D)
y = 0; u = 0
0 = P(0) + R
R=0
Hence, the correct option is (C).
u = P sin Q(y)
(A)
10.32
du
0
dy
y ;
u u
h
u = PsinQ
0
U
du
0
dy
du
P cos(Qy ) Q
dy
y
h
0 = PQ.cos(Q)
PQ 0 b/c if p = 0, Q = 0 u = 0 irrespective
of y which is not possible.
dx
dF0 = 0(dx 1)
FD 0dx
Then cos(Q) = 0
u
y
u h
Q
h
u
u
y y
1 dy 1 dr
u
u
h h
0
0
Q
h
6
Von-Karman equation
0
d
d
0 u2
2
u dx
dx
FD 0dx
d
FD u
.dx u2
dx
h
FD u2 . (one side of aerofoil)
6
u 2 h
Total drag = FD
3
Hence, the correct option is (A).
2
2
2
u P sin
.
2
P = u
Hence, the correct option is (D).
10.35
(B)
u = 10 m/s
u = u(y/)
= 10 mm
w=1m
= 1 kg/m3
Fluid Mechanics
154
q
Kulkarni’s Academy
r
u
y2
y
1
dy
y
0 2 2 2
0
u
10
5mm
2
=
p
mreduction * 1 u
s
q
= 1 (5 1) 10
= 50 103 = 0.05
Hence, the correct option is (B).
r
m pq
m rs
p
s
1m
q
10 m/s
10.36
(C)
Drag force: It is the force exerted by the fluid
on the plate in a direction parallel to the
relative motion. When the angle of incidence
of the plate is zero the drag is due to shear
only.
10 103 m
0.05 kg/s
q
r
p
m pq ( 1) u 1 10 103 10
0.1 kg/s
= 0.1 kg/sec
1
p
s
r
u
dy
1
m rs (dy 1)u
0
mrs u
0
u
2
Ppq = 0.1 10 = 1N
r
u u
mv
Pqr = 0.05 10 = 0.5 N
s
Momentum entering pq
y
dy
U 10 m/s
10 10 3 m
Momentum through rs
110 10 103
2
= 0.05 kg/s
1
r
dy
mqr m pq mrs
= 0.1 – 0.05 – 0.05 kg/sec
u
y
u
s
m pq (dy 1) u
Kulkarni’s Academy
155
Momentum = (dy 1) u2
Boundary Layer Theory
(C)
10.39
= u 2 dy
u0
A
4 B
0
Prs
u2 y 2
0
2
3
dy
A
B
H
Prs = 0.33 N
2
r
q
p
0.5
1
1
0.33
Fdrag = 1 – (0.5 + 0.33) = 0.17 N
Hence, the correct option is (C).
10.37
Vm y
0 y
u
Vm y H
Vm ( H y )
(D)
A = 2.6 m
5
V 120 33.33 m/s
18
Mass flow rate at section A
mA A v
( H 1) v0
CD = 0.3
= Hu0
= 1.5 105 m2/s
= 1.2 kg/m3
CD
1 2
u A
2
1
0.3 1.2 33.332 2.6
2
Hence, the correct option is (D).
dy
519.89 33.33 J/S
P = 17328.13 W
17328.13
23.23HP
746
Hence, the correct option is (C).
m12 (dy 1) u
0
m12
0
m12
P=FV
vm y
H
y
(C)
OR
…. (1)
FD
FD = 519.89 N
10.38
m B m12 m23 m34
1
FD CD u2 A
2
H y H
2
Vm y
dy
Vm
2
m34
m 23 Vm ( H 2 )
Total mass flow at exit (or section B)
Vm
V
Vm ( H 2 ) m
2
2
m B Vm Vm H 2 Vm
Fluid Mechanics
Vm(H )
156
…. (2)
Kulkarni’s Academy
10.41
Given data : L = 1 m, w = 1 106 m2/s
a = 1.6 105 m2/s
For steady flow
m A mB
w
?
a
Hu0 Vm ( H )
Vm
H
u0 H
For laminar flow
1
1
H
(A)
Hence, the correct answer is 0.25.
PA PB
?
1 2
Pu0
2
10.42
(D)
U B
C U
Apply Bernoulli’s equation between A and
B (outside the boundary layer)
A
B
U0
PA
Vm
PB
PA u02 PB Vm2
w 2g w 2g
PA PB V u
g
2g
2
m
PA PB
Vm
u0
5x
; v
VL
v
w
v
1106
w
0.25
a
va
1.6 105
Hence, the correct option is (C).
10.40
0.25
2
0
u02 Vm2
2 v02 1
PA PB Vm2
2 1
1 2
u0
u0
2
1
1
H
1
2
1
H
Hence, the correct option is (A).
A
D
0.1U
m AB m BC mCD m AD
m AB AV ( 1)u u
m AD L 0.1u 0.1 Lu
3 y y3
mCD u 3 dy
2 2
0
3 y 2
1 y3
u 3
2 2 0 2 4 0
5
u
8
m BC m AB mCD m AD
5
= u u 0.1 Lu
8
5
0.1L
u
8
3
u 0.1L
8
Hence, the correct option is (D).
Kulkarni’s Academy
10.43
157
(C)
l=2m
= 890 kg/m3
w= 1 m
= 0.29 kg/m-s
u = 6 m/s
CD 1.328 Re
CD
1
2
FD
1
A.u2
2
1
FD CD Au2
2
1
= 0.0068 890 (2 1) 6 2
2
= 221.71 N
Total drag force = 2 221.71
= 443.43 N
Hence, the correct option is (C).
10.44
(C)
cD
cD1
cD2
1
L
L
2
L
1
0.707
2
Hence, the correct option is (C).
Boundary Layer Theory
NOTES
Chapter-11
VORTEX MOTION
11.1 Introduction:
The motion of fluid along a curve path is known
vortex motion.
Vortex motion is of two types –
(1) Forced vortex
(2) Free vortex
(1) Forced vortex motion:
Motion of a fluid in a curved path under the
influence of external agency (Torque) is known
as forced vortex motion. As there is a continuous
expenditure of energy in forced vortex motion.
therefore, Bernoulli’s equation is not applicable.
For forced vortex motion the equation
v = r
then v r
is applicable for forced vortex motion.
Example:
(1) Liquid in a container when rotated
(2) Motion of fluid in the impeller of a
centrifugal pump.
1
r
This equation is applicable for free vortex
equation.
Example:
(1) Motion of fluid in the diffuser of the
centrifugal pump.
(2) Flow of fluid in pipe bend.
(3) Whirl pool
(4) Flow of liquid wash basin.
v
Note:
Free vortex is an irrotational flow.
Note:
Forced vortex motion is a rotation flow.
(2) Free vortex motion:
In free vortex motion the fluid moves in curved
path due to internal fluid action but not due to
external torque. As there is no expenditure of
energy, therefore Bernoulli’s equation is
applicable for free vortex motion.
d
mvr 0
dt
mvr = c
c
vr c
m
vr = c
11.2 Generalised equation for vortex
motion:
Kulkarni’s Academy
159
Vortex Motion
c
v
r
mv 2
CF
r
Volume = dA. dr
P2 P1
m
volume
m = .volume
= .dAdr
PdA
P
v22
P2
2
v22
2
v12
2
gz2 P1
gz2 gz1
v12
2
gz1
Divide with g
dAdrv 2
P
P
dr dA
r
r
v 2
P
dr P
dr
r
r
P v
r
r
This equation gives the variation of pressure in
radial direction.
P1 v12
P
v2
z1 2 2 z 2
g 2
g 2
Bernoulli’s equation is applicable for free vortex.
11.4 Forced vortex motion equation:
2
dP
P2
r2
P1
r1
dP
P2 P1
dP
P2 P1
v 2
11.3 Free vortex motion equation:
For free – vortex motion
p2
r2
p1
r1
dp
c
r
c2
r3
P2 P1
r 2 2
r3
z2
dr gdz
z1
2 2 2 2
.r2
.r1 gz2 gz1
2
2
r22 r12
P2 P1 g ( z2 z1 ) _____(1)
2 2
P
P
dr
dz
r
z
vr = c; v
dr gdz
2
dr gdz
r
This equation is valid for both free and forced
vortex.
dP
r
For forted vortex v = r
We know that from hydrostatic law
P
w g
z
P = f (r, z)
v2
v22 v12
gz2 gz1
2
2
v22
v12
P2
gz2 P1
gz1
2
2
P2 v22
P v2
z2 1 1 z1
g 2 g
g 2 g
Bernoulli’s equation is not applicable for forced
vortex.
11.5 Observation:
z2
dr gdz
z1
c 2
2r22
c 2
2r12
gz2 gz1
From equation 1.
r2 r2
P2 P1 2 2 1 g ( z2 z1 )
2 2
Used when P1 P2
Fluid Mechanics
160
Kulkarni’s Academy
Let us select two point 1 & 2 on the surface P1 =
11.6 Isobars in forced vortex:
P2
r2 r2
0 2 2 1 g ( z2 z1 )
2 2
11.7 Volume of paraboloid:
Fig. (a)
Fig. (b)
Volume =
1
R2 H
2
r2 r2
2 2 1 g ( z2 z1 )
2 2
2 2 2
r2 r1 ( z2 z1 )
2g
From above equation we can say that the variation
between r & z is parabolic.
Example 1
Show that in a forced vortex motion the rise
of liquid at ends is equal fall of the liquid at
the centre when no water spills over. (i.e., x
= y)
Sol.
Fig. (c)
If point 1 is taken on the surface r1 = 0
2 2
r2 0 ( z2 z1 ) z
2g
2 r22
z
2g
1
2
R 2 ( H y ) R 2 ( H y x) R 2 ( x y )
1
1
R 2 x R 2 x R 2 y
2
2
2 R 2
H
2g
1 2
1
R x R2 y
2
2
x=y
Kulkarni’s Academy
P
11.1
Practice Questions
A cylindrical container is filled with a liquid
up to half of its height. The container is
mounted on the centre of a turn-table and is
held fixed using a spindle. The turn-table is
now rotated about its central axis with a
certain angular velocity. After some time
interval, the fluid attains rigid body rotation.
Which of the following profiles best
represents the constant pressure surfaces in
the container?
161
Vortex Motion
11.2
Which one of the following is an irrotational
flow?
(A) Free vortex flow
(B) Forced vortex flow
(C) Couette flow
(D) Wake flow
11.3
A right circular cylinder is filled with a liquid
upto its top level. It is rotated about its
vertical axis at such a speed that half the
liquid spills out, then the pressure at the point
of intersection of the axis and bottom surface
is
(A) Same as before rotation
(B) Half of the value before rotation
(C) quarter of the value before rotation
(A)
(D) Equal to the atmospheric pressure
11.4
(B)
Which combination of the following
statements about steady incompressible
forced vortex flow is correct?
P.
Shear stress is zero at all points in the
flow.
Q.
Vorticity is zero at all points in the flow.
R.
Velocity is directly proportional to the
radius from the centre of the vortex
S. Toal energy per unit mass is constant in the
entire flow field
(C)
11.5
(A) P and Q
(B) R and S
(C) P and R
(D) P and S
Forced vortex flow is similar to solid body
rotation. For this case
(A) The shear strain rate is zero but the local
angular velocity is non-zero
(B) The shear strain rate is non-zero but the
local angular velocity is zero
(D)
(C) Both the shear strain rate and the local
angular velocity are zero
(D) Both the shear strain rate and the local
angular velocity are non-zero
Fluid Mechanics
162
Kulkarni’s Academy
11.6 Choose the correct combination of true
11.10 A leaf is caught is a whirlpool. At a given
statements from the following:
instant, the leaf is at a distance of 120 m from
P. In a free vortex, the total pressure varies
the centre of the whirlpool. The whirlpool
from streamline to streamline
can be described by the following velocity
Q. In a forced vortex, the total pressure
60 103
distribution; Vr
varies from streamline to streamline
m/s and
2r
R. In a free vortex, the static pressure
increases with radial distance from the
300 103
V
m/s ,
centre at the same elevation
2r
S. In a forced vortex, the static pressure
Where r (in metres) is the distance from the
decrease with radial distance from the
centre of the whirlpool. What will be the
centre at the same elevation
distance of the leaf from the centre when it
(A) P,Q,R
(B) R, S
has moved through half a revolution?
(C) P,Q,R,S
(D) Q,R
11.7 A cylindrical vessel open at the top is filled
(A) 48 m
(B) 64 m
with water and rotated at a constant angular
(C) 120 m
(D) 142 m
velocity about its vertical axis such that the
bottom of the vessel is just exposed at the
11.11 The U tube arrangement shown rotates about
axis. The volume of water spilled as a
axis BB at 60 / r.p.m. Initially (before
fraction of the volume of the cylinder is
rotation) the level in the arms of the U tube is
(A) 1/3
(B) 2/5
60 cm. The steady state difference in the
(C) 1/2
(D) 2/3
levels of the two limbs is
11.8 An open circular tank of 1m height and 0.3 m
diameter contains 0.8 m of water. If the tank
is rotated about the vertical axis such that
there is no spillage of water, the maximum
angular velocity of the tank is, nearly
(A) 18.65 rad/s
(B) 18.65 rad/minute
(C) 1.865 rad/s
(D) 1.865 rad/minute
11.9 A closed cylinder having a radius R and
(A) 12.5 cm
(B) 25 cm
height H is filled with oil of density . If the
cylinder is rotated about its axis at an angular
(C) 20 cm
(D) 10 cm
velocity of , then thrust at the bottom of the
11.12 The constant angular velocity at which a
cylinder is
liquid rotates in a cylinder about a vertical
(A) R 2gH
axis such that the pressure at a point on the
2 2
R
axis is the same as at a point 2m higher at a
(B) R2
4
radius 2m is
2
2 2
(C) R ( R gH )
(A) 2 rad/s
(B) 1 rad/s
2 2
2 R
(D) R
gH
(C) rad/s
(D) 2 rad/s
4
Kulkarni’s Academy
163
11.13 Both free vortex and forced vortex can be
expressed mathematically in terms of
tangential velocity V at the corresponding
radius r. Choose the correct combination
Vortex Motion
A
Answer Key
11.1
A
11.2
A
11.3
D
11.4
B
11.5
A
11.6
D
Free Vortex
Forced Vortex
11.7
C
11.8
A
11.9
D
(A) V = r const
Vr = const
11.10
B
11.11
D
11.12
C
(B) V 2 r const
V = r const
11.13
D
11.14
A
(C) Vr = const
V 2 r const
(D) Vr = const
V r const
11.14 A U-tube of a very small bore, with its limbs
in a vertical plane and filled with a liquid of
density , up to a height of h, is rotated about
a vertical axis, with an angular velocity of
, as shown in the figure.
The radius of each limb from the axis of
rotation is R. Let Pa be the atmospheric
pressure and g, the gravitational acceleration.
The angular velocity at which the pressure at
the point O becomes half of the atmospheric
pressure is given by
(A)
Pa 2 gph
R 2
(C)
Pa 2gh
2R 2
(B)
2( Pa gh)
R 2
(D)
Pa gh
2R 2
E
Explanation
11.1
(A)
11.2
(A)
11.3
(D)
11.4
(B)
As fluid is flow, shear stress will be
present
Vorticity = 2 rotation
V = rw; v r
Hence, the correct option is (B).
11.5
(A)
11.6
(D)
11.7
(C)
11.8
(A)
z
r 2 w2
2g
(0.15)2 w2
2 9.81
W = 18.6761 rad/s
Hence, the correct option is (A).
0.4
Fluid Mechanics
11.9
164
(D)
Kulkarni’s Academy
11.10
Ftotal = Force due to weight of fluid (F1) +
Force due to rotation (F2)
F1 = gV { V volume
(B)
Whirl pool is an example of free vortex
motion and it is irrotational flow.
Vr
= gR2H
F1 = gH.R2
dr
dt
V r
When fluid is rotating, pressure varies with
respect to radius
d
dt
dP V 2
dr
r
In force vortex v = rw
Vr
60 103
2r
dP
rw2
dr
V
300 103
2r
P
w2 r 2
2
Vr
dr
1
V rd
5
w2 r 2
2rdr
2
o
R
F2
F2
w2 R 4
4
w2 R 2
Ft R
gH
4
1
dr
d
5
r
e
2
Hence, the correct option is (D).
Note:
If the force on the top surface is to be
calculated then it is only due to rotation, there
is no weight force on the top surface of the
container.
r
2
1
dr
d
50
r
120
5
r2
120
r2 120 e
5
= 64 m
Hence, the correct option is (B).
Kulkarni’s Academy
11.11
Vortex Motion
(D)
2N 2 60
w
2rad / s
60
60
w2 2 2
r2 r1 z2 z1
2g
4
0.752 0.252 z2 z1
2 9.81
z2 – z1 = 0.1019 m
10 cm
Hence, the correct option is (D).
11.12
165
(C)
z
r 2 w2
2g
w 9.81 3.14 rad/s
Hence, the correct option is (C).
(D)
11.14
(A)
We know that
P2 P1
Pa
2
P2 = Pa
r1 = 0
r2 = R
V = Rw
P1
2
g ( z2 z1 )
Pa w2 R 2
gh
2
2
Pa
Pa
w2 R2
gh
2
2
Pa 2gh w2 R2
w2
Pa 2gh
R 2
w
Pa 2gh
R 2
Hence, the correct option is (A).
22 w2
2
2 9.81
11.13
P2 P1
w2 r22 r12
V 2
dr gdz
r
