Department of Aeronautical engineering
School of Mechanical engineering
Vel Tech Dr RR & SR Technical University
Course Material
U4MEA10- Fluid Mechanics
1
U4MEA10
FLUID MECHANICS
Unit - I Basic Concepts And Properties
9
Fluid – definition, distinction between solid and fluid - UNIT Is and dimensions - Properties of fluids
- density, specific weight, specific volume, specific gravity, temperature, viscosity, compressibility,
vapour pressure, capillary and surface tension - Fluid statics: concept of fluid static pressure, absolute
and gauge pressures - pressure measurements by manometers and pressure gauges.
Unit – II
Bernoulli’s Equation And Boundary Layer Concepts
9
Fluid Kinematics - Flow visualization - lines of flow - types of flow - continuity equation (one
dimensional differential forms)- fluid dynamics - equations of motion - Euler’s equation along a
streamline - Bernoulli’s equation – applications - Venturi meter, Orifice meter, Pitot tube - Boundary
layer flows, boundary layer thickness, boundary layer separation - drag and lift coefficients.
Unit - III
Flow Through Pipes
9
Viscous flow - Navier - Stoke’s equation (Statement only) - Shear stress, pressure gradient
relationship - laminar flow between parallel plates - Laminar flow through circular tubes (Hagen
poiseulle’s) - Hydraulic and energy gradient - flow through pipes - Darcy -weisback’s equation - pipe
roughness -friction factor-minor losses - flow through pipes in series and in parallel - power
transmission.
Unit - IV
Dimensional Analysis And Hydraulic Turbines
9
Dimensional analysis - Buckingham’s p theorem- applications - similarity laws and models.Hydro
turbines: definition and classifications - Pelton turbine – Francis turbine - Kaplan turbine - working
principles - velocity triangles - work done - specific speed - efficiencies -performance curve for
turbines.
Unit - V
Pumps
9
Pumps: definition and classifications - Centrifugal pump: classifications, working principles, velocity
triangles, specific speed, efficiency and performance curves - Reciprocating pump: classification,
working principles, indicator diagram, work saved by air vessels and performance curves cavitations in pumps -priming- slip- rotary pumps: working principles of gear, jet and vane pump.
Text Books
1. 1.Streeter, V.L., and Wylie, E.B., “Fluid Mechanics”, McGraw-Hill, 1983.
2. Kumar, K.L., “Engineering Fluid Mechanics”, Eurasia Publishing House (P) Ltd., New Delhi
(7th edition), 1995.
Bansal, R.K., “Fluid Mechanics and Hydraulics Machines”, (5 th edition), Laxmi publications (P)
Ltd., New Delhi, 1995.
3.
Reference Books
1.
2.
3.
White, F.M., “Fluid Mechanics”, Tata McGraw-Hill, 5th Edition, New Delhi, 2003.
Ramamirtham, S., “Fluid Mechanics and Hydraulics and Fluid Machines”, Dhanpat Rai and
Sons, Delhi, 1998.
Som, S.K., and Biswas, G., “Introduction to fluid mechanics and fluid machines”, Tata
McGraw-Hill, 2nd edition, 2004.
2
UNIT – I
Fluid – definition, distinction between solid and fluid
Units and dimensions
Properties of fluids - density, specific weight, specific volume,
specific gravity, temperature, viscosity, compressibility, vapour
pressure, capillary and surface tension
Fluid statics: concept of fluid static pressure, absolute and gauge
pressures
Pressure measurements by manometers and pressure gauges.
3
BASIC CONCEPTS AND PROPERTIES
PART – A
1. Define the following properties.
Density, weight density, specific volume and specific gravity of fluid
(i)Density (or) specific mass (or) Mass Density:
The mass density of a fluid is the mass which is possesses per unit volume
Mass density ( )=
Mass of fluid
m
 ( K g / m3 )
Volume of the fluid v
(ii) Weight density (or) specific weight (w)
The weight density or specific weight of a fluid is the weight it posses per unit
volume.
Weight density(w) =
Weight of fluid
mass  Acceleration due to gravity

(N/m3 )
Volume of fluid
Volume of fluid
w=  g
(iii) Specific volume (v)
Specific volume is the reciprocal of specific density. The specific volume of a
fluid is the volume occupied by the unit mass of the fluid.
1
Volume of fluid Mass of fluid 1
Specific volume (v)=


Mass of fluid
Volume

2. Differentiate between i) ideal fluid & Real Fluid.
(ii) Specific weight and specific volume of a fluid
Ideal Fluid:
A fluid which is incompressible and is having no viscosity is known as ideal fluid.
4
Ideal fluid is only an imaginary fluid as all fluids, have some viscosity.
Real Fluid:
A fluid which possess viscosity is known as real fluid
Specific Weight:
The Specific weight of a fluid is the weight it possesses per unit volume.
Weight density
Weight of Fluid (w)
(or)
Specific weight
Volume of Fluid (v)
w = g
Specific volume:
Specific Volume is the reciprocal of specific density. The specific volume of a
fluid is the volume occupied by the unit mass of the fluid.
3. State the Newton’s law of Viscosity.
It states that the shear stress () on a fluid element layer is directly proportional
to the rate of shear strain. The constant of proportionality is called the coefficient of
viscosity.
Mathematically, it is expressed by equation   
du
dy
Fluids which obey the above relation are known as Newtonian fluids
Fluids which do not obey the above relation are called Non-Newtonian Fluid.
4. Distinguish Between Surface Tension and capillarity.
SURFACE TENSION
Surface tension is defined as the tensile
force acting on the surface of a Liquid in
contact with a gas or on the surface
between two immiscible Liquids such
that contact surface behaves like a
membrane under tension.
Surface tension is expressed in N/m (or)
5
CAPILLARITY
Capillarity
is
defined
as
a
phenomenon of rise or fall of a liquid
surface in a small tube relative to the
adjacent general level of Liquid when
the tube is held vertically in the fluid.
The rise of Liquid surface is known as
dyne/Cm
capillary raise while the fall of the
Liquid surface is known as capillary
depression capillary rise or fall, h=
4cos / gd
5. Define Kinematics Viscosity and drive its unit.
It is defined as the ratio between the dynamic viscosity and density of fluid. It is
denoted by the Greek symbol () (Nu)
 = Dynamic Viscosity
Density

 2
[m / s]

6. What is the difference between dynamic viscosity and kinematic viscosity?
State their units
Dynamic Viscosity: ()
Dynamic viscosity (or) coefficient of dynamic viscosity is defined as the ratio
between to the shear stress and rate of shear deformation.


du
dy

Ns
m2
Kinematic Viscosity:
It is defined as the ratio between the dynamic viscosity and density of fluid. It is
denoted by the Greek symbol () (Nu).

2
 = Dynamic Viscosity Density v   m / s
7. Define Newtonian and Non-Newtonian fluids.
Newtonian Fluid:
A fluid which the shear stress is directly proportional to the rate of shear stain [or
viscosity gradient] is known as a Newtonian fluid.
Non-Newtonian Fluid:
6
A real fluid in which is not proportional to the rate of shear strain [(or) velocity
gradient] is known as a Non-Newtonian fluid.
8. Define the term Viscosity.
Viscosity: ()
Viscosity is defined as the property of a fluid which offers resistance to the
movement of one layer of fluid over another adjacent Layer of the fluid.


du
dy
This property is due to cohesion and interaction between molecules of the fluid.
9. State Newton’s law of viscosity.
Newton’s Law of viscosity:
It states that the shear stress on a fluid element layer is directly proportional to
the rate of shear strain. The constant of proportionality is called the coefficient of
viscosity.
du

  . ;  
dy
 du 
 dy 
 
The fluid viscosity is due to cohesion and interaction between molecules of
the fluid.
For example
High viscosity Fluid Tar and caster op
Low viscosity fluid  Kerosene, Petrol and water
10. What is the effect of temperature on viscosity of water and that of air?
When the viscosity of the liquid decreases with increase in temperature since the
molecules present in the Liquid is less.
7
When the viscosity of the air increases with increases in temperature.
11. Explain the importance of compressibility in fluid flow.
Compressibility is the reciprocal of the bulk modulus of Elasticity, K which is
defined as the ratio of compressive stress to volumetric strain.
Increase of Pressure
Volumetric strain
-dpV
=
dv
Bulk Modulus K=
Compressibility is given by =1/K
12. Explain the phenomenon of capillarity obtain an expression for capillary rise
of a Liquid.
Capillarity is defined as a phenomenon of rise or fall of of a Liquid surface in a
small tube relative to the adjacent general level of liquid when the tube is held
vertically in the Liquid.
The rise of liquid surface is known as capillary rise. While the fall of liquid
surface is known as capillary depression. It is expressed in terms of cm (or ) mm of
Liquid.
Expression for Capillary rise:
Consider a glass tube of small diameter ’d’ opened at both ends and is inserted
in a liquid say water, the Liquid will rise in the tube above the level of liquid.
Let h-height of the Liquid in the tube. Under a state of equilibrium, the weight of
Liquid height h is balanced by the force at the surface of the liquid in the tube. But
the force at the surface of the Liquid in the tube is due to surface tension.
Let  = surface tension of liquid,  =Angle of contact between liquid glass tube.
The weight of liquid of height ‘h’ in the tube =(Area of tube x h)  x g.
8
=

4
d2 h   g
 (1)
Where,
 = Density of liquid
Vertical component of surface tensile force
= ( x Circumference) x Cos 
= ( x d x cos)
2
Equating the equation (1) & (2) we get,
/4d2 h  g =  x d x cos
   d  cos
 / 4d 2    g
4 cos 
h
 gd
h
when = 0
Capillary Rise of Liquid h=
4 cos 
 gd
13. What are the types of the fluids?
Types of the fluids:
1.
2.
3.
4.
5.
Ideal Fluid
Real Fluid
Newtonian Fluid
Non-Newtonian Fluid
Ideal Plastic Fluid.
14. Define Pascal’s Law.
It states that the pressure or intensity of pressure at a point in a static fluid is
equal is equal in all directions.
Px = Py = P2
15. Define Hydraulic Law.
It states that rate of increase of pressure in a vertical direction is equal to
weight density of the fluid at that point.
9
P = gZ.
16. Define Manometer.
It is defined as the devices used for measuring the pressure at a point in a
fluid by balancing the column of fluid by the same or another column at the fluid.
17. Define absolute pressure.
It is defined as the pressure which is measured with reference to absolute
vacuum pressure.
18. Define Gauge pressure.
It is defined as the pressure which is measured with the help of a pressure
measuring instrument, in which the atmospheric pressure is taken as datum. The
atmospheric pressure on the scale is marked as zero.
19. Define vacuum pressure.
It is defined as the pressure below the atmospheric pressure.
20. What are the different types of mechanical gauges?
There are different types of mechanical gauges.
Mechanical gauges are best suitable for measuring very high fluid pressure.
Incase of steam boilers where manometer can not be used, a mechanical gauge can
be conveniently used.
1. Bourdon tube pressure gauge
2. Diaphragm pressure gauge
3. Dead weight pressure gauge.
21. What are units and dimension?
S.No
Quantity
Unit
DIMENSIONS
10
1.
2.
3.
4.
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Geometric
Length
Area
Volume
Slope
Kinematic
Time
Velocity
(linear)
Velocity
(angular)
Acceleration
(linear)
Acceleration
(angular)
Discharge
Gravitationa
l
acceleration
Kinematic
velocity
Dynamic
Mass
Force
Weight
Mass
density
Specific
weight
Dynamic
viscosity
Surface
tension
Elastic
modulus
Pressure
Shear
intensity
generally
adopted
MLT SYSTEM
FLT SYSTEM
M
M2
M3
L
L2
L3
L
L2
L3
Sec
M/sec
T
LT-1
T
LT-1
Rad / sec2
T-1
T-1
M/sec
LT-2
LT-2
Rad /sec2
T2
T2
Cum /sec
M/sec2
L3 T-1
LT-2
L3T-1
LT-2
M/sec2
L2 T-1
L2T-1
Kg
Newton
Newton
Kg /cum
M
MLT-2
MLT-2
ML-3
FL-1 T2
F
F
FL-4T2
Newton/cu
m
Newton
/cum
Newton/m
ML-2T2
FL-3
ML-1T-1
FL-3T
MT-2
FL-1
Newton/m2
ML-1T2
FL-2
Newton/m2
Newton/m2
ML-1T2
ML-1T2
FL-2
FL-2
11
23
24
25
26
Work,
energy
Impulse
momentum
Torque
Power
Newton m
ML2T2
FL
Newton sec
MLT-1
FT
Newton m
Newton /sec
ML2T-2
ML2T3
FL
FLT-1
22. Differentiate between fundamental units and derived units.
The fundamental or primary units are the simplest in their form possessing a
single dimension. When the units of measurements of the primary quantities are
defined, the measurements of all other quantities can be easily obtained.
Example: Length (L), Time (T), Mass (M), Temperature ()
The derived secondary quantities possess more than one dimension, and are
expressed by a combination of dimensions.
Example: Velocity (LT-1), linear acceleration (LT-2), force (MLT2) etc.
12
PART - B
1. Calculate the specific weight, density and specific gravity of one litre of a
liquid which weights 7 N.
Solution: Given:
Volume
1
1
m3 1 litre 
m3 or 1 litre  1000cm3 
1000 
1000

= 7 N.
= 1 litre 
Weight
(i) Specific weight (w)
(ii) Density ()
Weight
7N

Volume  1  m3


 1000 
w 7000

kg / m3  713.5kg / m3 .
=
g 9.81

(iii) Specific gravity

Density of liquid 713.5

 Density of water  1000 kg / m3 

Density of water 1000
= 0.7135.
2. Calculate the density, specific weight and weight of one litre of petrol of
specific gravity = 0.7.
Solution:
Given:
Volume
= 1 litre
= 1x 1000 cm3 
Sp. Gravity,
1000 3
m  0.001m3
6
10
= S = 0.7
(i) Density ()
Using equation (1.1A),
Density ()
= S x 1000 kg/m3 = 0.7 x 1000 = 700 kg/m3.
13
(ii) Specific weight (w)
Using equation (1.1)
w = xg
= 700x9.81 N/m3 = 6867 N/m3.
We know that specific weight 
w
or

Weight
Volume
W
W
or 6867 
0.001
0.001
W  6867  0.001  6.867 N.
3. A flat plate area 1.5x106 mm2 is pulled with a speed of 0.4 m/s relative to another
plate located at a distance of 0.15 mm from it. Find the force and power required
to maintain this speed, if the fluid separating them is having viscosity as 1 poise.
Solution:
Given:
Area of the plate,
A = 1.5 x 106 mm2 = 1.5m2
Speed of plate relative to another plate, du = 0.4 m/s
Distance between the plates,
dy = 0.15 mm = 0.15 x 10-3m
1 Ns
.
Viscosity
 = 1 poise
10 m 2
Using equation (1.2), we have
du 1
0.4
N
 
 
 266.66 2
3
dy 10 015  10
m
(i)  Shear force,
F=  x area =266.66 x 1.5 = 400 N.
(ii) Power* required to move he plate at the speed 0.4 m/sec
= F x u = 400 x 0.4 = 160 W.
4. Calculate the dynamic viscosity of an oil, which is used for lubrication between
a square plate of size 0.8 m x0.8 m and an inclined plane with angle of inclination
30 as shown in Fig. The weight of the square plate is 300 N and it slides down
the inclined plane with a uniform velocity of 0.3 m/s. The thickness of oil film is
1.5 mm.
Solution:
Given:
14
Area of plate,
Angle of plane,
Weight of plate,
Velocity of plate,
Thickness of oil film,
A = 0.8 x 0.8 = 0.64 m2
 = 30
W = 300 N
u = dy
t = dy
= 1.5 mm = 1.5 x 10-3m
Let viscosity of fluid between plate and inclined plane is . Component of
weight W, along the plane = W cos 60 =150 N
Thus the shear force, F, on the bottom surface of the plate = 150 N
F
150

N / m2
Area 0.64
Now using equation (1.2), we have
du
 
dy
And shears stress,

where du = Change of velocity = u-0=u=0.3 m/s
dy = t = 1.5 x 10-3 m
150
0.3


0.64
1.5  10 3
150  1.5  103


 1.17 Ns / m2  1.17  10  11.7 poise.
0.64  0.3
5. The space between two square flat parallel plates is filled with oil. Each side of
the plate is 60 cm. The thickness of the oil film is 12.5 mm. The upper plate,
which moves at 2.5 meter per see requires a force of 98.1 N to maintain the speed.
Determine:
(i)
(ii)
the dynamic viscosity f the oil in poise, and
the kinematic viscosity of the oil in stokes if the specific gravity of the
oil is 0.95. (AMIE, Winter 1977)
Solution:
Given:
Each side of a square plate = 60 cm = 0.60 m
15
 Area,
A = 0.6 x 0.6 = 0.36 m2
Thickness of oil film,
dy = 12.5 mm = 12.5 x10-3 m
Velocity of upper plate,
u = 2.5 m/sec
 Change of velocity between plates, du = 2.5 m/sec
Force required on upper plate,
F = 98.1 N
 Shear stress,

(i)
Let  = Dynamic viscosity of oil
Using equation (1.2),

Force F 98.1N
 
Area A 0.36m2
 
N
98.1 12.5  10 3

 1.3635 2s
0.36
2.5
m
du
dy
or
98.1
2.5

0.36
12.5  10 3
 1Ns  10 poise 


 m2

 1.3635  10  13.635 poise.
(ii)
Sp. gr. of oil, S = 0.95
Let  =kinematic viscosity of oil
Using equation (1.1 A),
Mass density of oil,
 = S x1000 = 0.95 x1000 = 950 kg/m3
N
1.3635  2s 

m 
Using the relation,   , we get  

950
= .001435 m2/sec = .001435 x 104 cm2/s
= 14.35 stokes.
( cm2 /s = stoke)
6. The dynamic viscosity of an oil, used for lubrication between a shaft and sleeve
is 6 poise. The shaft is diameter 0.4 m and rotates at 190 r.p.m. Calculate the
power lost in the bearing for a sleeve length of 90 mm. The thickness of the oil
16
film is 1.5 mm.
Solution:
Given:
Viscosity
 = 6 poise
=
Dia. of shaft,
Speed of shaft,
Sleeve length,
Thickness of oil film,
N
6 Ns
 0.6 2s
2
10 m
m
D = 0.4 m
N = 190 r.p.m.
L = 90 mm = 90 x 10-3m
t = 1.5 mm = 1.5 x 10-3 m
Tangential velocity of shaft,
Using the relation
fig
u
 DN
60
du
 
dy

  0.4  190
60
 3.98m / s
Where du = Change of velocity = u – 0 = u = 3.98 m/s
dy = Change of distance = t = 1.5 x 10-3 m
3.98
  10 
 1592 N / m2
3
1.5  10
This is shear stress on shaft
 Shear force on the shaft,
F = Shear stress x Area
= 1952 x D x L = 1592 x  x .4 x 10-3 = 180.05N
D
T  Force 
Torque on the shaft,
2
0.4
 180.05 
 36.01Nm
2
2 NT 2  190  36.01


 716.48W .
 Power lost
60
60
7. A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 N
s/m2 and specific gravity 0.9. A metallic plate 1.2 m x 0.2 cm is to be lifted up with
a constant velocity of 0.15 m/sec, through the gap. If the plate is in the middle of
17
the gap, find the force required. The weight of the plate is 40 N.
Solution:
Given:
Width of gap
= 2.2 cm, viscosity,  = 2.0 N s/m2
Sq.gr. of fluid
= 0.9
 Weight density of fluid = 0.9 x 1000 = 900 kgf/m3 = 900 x 9.81 N/m3
( 1 kgf = 9.81 N)
Volume
= 1.2 m x 1.2 m x 0.2 cm
= 1.2 x 1.2 x .002 m3 = .00288 m3
Thickness of plate
= 0.2 cm
Velocity of plate
= 0.15 m /sec
Weight of plate
= 40 N.
When plate is in the middle of the gap, the distance of the plate from vertical
surface, of the gap
 Width of gap  thickness of plate   2.2  0.2 

 1cm  .01m.

2
2


Now the shear force on the left side of the metallic plate,
F1 = Shear stress x Area
 du 
0.15 
2
     Area  2.0  
  1.2  1.2 N  Area  1.2  1.2m 
 .01 
 dy 1
 43.2 N.
Similarly, the shear force on the right side of the metallic plate,
F2 = Shear stress x Area
.15 
 2.0  
  1.2  1.2  43.2N
 .01 
 Total shear force = F1 + F2 = 43.2 + 43.2 = 86.4 N.
In this case the weight of plate (which is acting vertically downward) and
upward thrust is also to be taken into account.
The upward thrust = Weight of fluid displaced
= (Weight density of fluid) x Volume of fluid displaced
18
= 9.81 x 900 x .00288 N
Volume of fluid displaced  Volume of plate  .00288 
= 25.43 N.
The net force acting in the downward direction due to weight of the plate and
upward thrust
= Weight of plate - Upward thrust
= 40-25.43 = 14.57 N
 Total force required to lift the plate up
= Total shear force + 14.57 = 86.4+14.57 = 100.97 N.
8. The surface tension of water in contact with air at 20C is 0.0725 N/m. The
pressure inside a droplet of water is to be 0.02 N/cm2 greater than the outside
pressure. Calculate the diameter of the droplet of water.
Solution:
Given:
Surface tension,
 = 0.0725 N/m
Pressure intensity, p in excess of outside pressure is
p  0.02 N / cm2  0.02  10 4
N
m2
Let
d = dia. of the droplet
4
4  0.0725
or 0.02  10 4 
Using equation (1.14), we get p 
d
d
4  0.0725
d
4  .00145m  .00145  1000  1.45 mm.
0.02   10 
9. Find the surface tension in a soap bubble of 40 mm diameter when the inside
pressure is 2.5 N/m2 above atmospheric pressure.
Solution:
Given:
Dia. of bubble,
d = 40 mm = 40 x 10-3 m
Pressure in excess of outside,
p = 2.5 N/m2
For a soap bubble, using equation (1.15), we get
8
8 
p
or 2.5 
d
40  10 3
19
2.5  40  103

N / m  0.0125N / m.
8
10. The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm2
(atmospheric pressure). Calculate the pressure within the droplet if surface
tension is given as 0.0725 N/m of water.
Solution:
Given:
Dia. of droplet,
Pressure outside the droplet
Surface tension,
d = 0.04 mm = .04 x 10-3 m
= 10.32 N/cm2
= 10.32 x 104 N/m2
 = 0.0725 N/m
The pressure inside the droplet, inn excess of outside pressure is given by
equation
or
 Pressure inside the droplet
4 4  0.0725

 7250 N / m2
d
.04  10 3
7250 N
 4 2  0.725N / cm2
10 cm
= p + Pressure outside the droplet
p
= 0.725 + 10.32 = 11.045 N/cm2.
11. Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed
vertically in (a) water and (b) mercury. Take surface tension  = 0.0725 N/m for
water and  =0.52 N/m for mercury in contact with air. The specific gravity for
mercury is given as 13.6 and angle of contact = 130.
Solution:
Given:
Dia. of tube,
Surface tension,  for water
 for mercury
Sp. gr. of mercury
d = 2.5 mm = 2.5 x 10-3 m
= 0.0725 N/m
= 0.52 N/m
= 13.6
20
 Density
(a) Capillary rise for water ( = 0)
= 13.6 x 1000 kg/m3.
h
Using equation (1.20), we get
4
4  0.0725

  g  d 1000  9.81 2.5  103
= .0118 m = 1.18 cm.
(b) For mercury
Angle of constant between mercury and glass tube,  = 130
4 cos
4  0.52  cos130

Using equation (1.21), we get h 
  g  d 13.6  1000  9.81  2.5  10 3
= - .004 m = -0.4 cm.
The negative sign indicates the capillary depression.
12. Calculate the capillary effect in millimeters in a glass tube of 4 mm diameter,
when immersed in (i) water, (ii) mercury. The temperature of the liquid is 20C
and the values of the surface tension of water and mercury at 20C in contact with
air are 0.073575 N/m respectively. The angle of contact for water is zero that for
mercury. 1.30. Take density of water at 20C as equal to 998 kg/m3. (U.P.S.C.
Engg. Exam., 1974)
Solution:
Given:
Dia of tube, d = 4 mm = 4 x 10-3m
The capillary effect (i.e., capillary rise or depression) is given by equation
(1.20) as
h
4 cos
p gd
where  = surface tension in kgf/m
 = angle of contact, and  = density
(i) Capillary effect for water
 = 0.073575 N/m,  = 0
 = 998 kg/m3 at 20C
21
4  0.073575  cos 0
13600  9.81  103
= - 2.45 x 10-3 m = - 2.46 mm.
h

The negative sign indicates the capillary depression.
13. Find out the minimum size of glass tube that can be used to measure water
level if the capillary rise in the tube is to be restricted to 2 mm. Consider surface
tension of water in contact with air as 0.073575 N/m. (Converted to SI Units,
A.M.I.E., Summer 1985)
Solution:
Given:
Capillary rise,
h = 2.0 mm = 2.0 x 10-3 m
Surface tension,
 = 0.073575 N/m
Let dia. of tube
=d
The angle  for water
=0
The density for water,
 = 1000 kg/m3
Using equation (1.20), we get
4
4  0.073575
or 2.0  10 3 
  gd
1000  9.81  d
4  0.073575
d
 0.015m  1.5cm.
1000  9.81  2  10 3
h

Thus minimum diameter of the tube should be 1.5 cm.
14. A hydraulic press has a ram of 30 cm diameter and a plunger of 4.5 cm
diameter. Find the weight lifted by the hydraulic press when the force applied at
the plunger is 500 N.
Solution:
Given:
22
Dia. of ram,
Dia. of plunger,
Force on plunger,
Find weight lifted
D = 30 cm 0.3 m
d = 4.5 cm = 0.045 m
F = 500 N
=W
Area of ram,
A
a
Area of plunger,

4

4
D2 
d 
2


4
4
 0.3 2  0.07068m2
 0.045 2  .00159m2
Pressure intensity due to plunger

Force on plunger F
500
 
N / m2 .
Area of plunger a .00159
Due to Pascal’s law, the intensity of pressure will be equally transmitted in all
directions. Hence the pressure intensity at the ram

500
 314465.4 N / m2
.00159
But pressure intensity at ram


 Weight
Weight
W
W


N / m2
Area of ram A .07068
W
 314465.4
.07068
= 314465.4 x .07068 = 22222 N = 22.222 kN.
15. The diameters of a small piston and a large piston of a hydraulic jack at 3 ate 3
cm and 10 cm respectively. A force of 80 N is applied on the small piston. Find
the load lifted by the large piston when:
(a) The piston is 40 cm above the large piston.
(b) Small piston is 40 cm above the large piston.
The density of the liquid in the jack is given as 1000 kg/cm3.
23
Solution:
Given:
Dia. of small piston,
d = 3 cm
 Area of small piston,
a
Dia. of large piston,
 Area of larger piston,
Force on small piston,
Let the load lifted

d2 

  3   7.068cm2
2
4
4
D = 10 cm
p
2
A    10   78.54cm2
4
F = 80 N
= W.
(a) When the piston are at the same level.
Pressure intensity on small piston
F
80

N / cm2
a 7.068
This is transmitted equally on the large piston.
 Pressure intensity on the large piston
80

7.068
 Force on the large piston = Pressure x Area
80

 78.54 N  888.96 N.
7.068
(b) When the small piston is 40 cm above the large piston.
Pressure intensity on the small piston
F
80 N
 
a 7.068 cm2
 Pressure intensity at section A – A
F
  Pr essure int ensity due to height of 40cm of liquid.
a
But pressure intensity due to 40 cm of liquid
=  x g x h = 1000 x 9.81 x 0.4 N/m2
1000  9.81  .40

N / cm2  0.3924 N / cm2 .
104
 Pressure intensity at section
80
A A 
 0.3924
7.068
24
= 11.32 + 0.3924 = 11.71 N/cm2
 Pressure intensity transmitted to the large piston = 11.71 N/cm2
 Force on the large piston
= Pressure x Area of the large piston
= 11.71 x A = 11.71 x 78.54 = 919.7 N.
16. What are the gauge pressure and absolute pressure at a point 3 m below the
free surface of a liquid having a density of 1.53 x 103 kg/m3 if the atmospheric
pressure is equivalent to 750 mm of mercury? The specific gravity of mercury is
13.6 and density of water = 1000 kg/m3. (A.M.I.E., Summer 1986)
Solution:
Depth of liquid,
Density of liquid,
Atmospheric pressure head,
Z1 = 3 m
1 = 1.53 x 103 kg/m3
Z0 = 750 mm of Hg
750

 0.75m of Hg
1000
 Atmospheric pressure, patm = 0 x g x Z0
where 0 = density of Hg = 13.6 x 1000 kg/m3
and Z0 = Pressure head in teams of mercury.

patm = (13.6 x 1000) x 9.81 x 0.75 N/m2 ( Z0 = 0.75)
= 100062 N/m2
Pressure at a point, which is at a depth of 3 m from the free surface of the
liquid is given by,
p = 1 x g x Z1
= (1.53 x 1000) x 9.81 x 3 = 45028 N/m2
 Gauge pressure, p = 45028 N/m2.
Now absolute pressure
= Gauge pressure + Atmospheric pressure
= 45028 + 100062 = 145090 N/m2.
17. The right limb of a simple U-tube manometer containing mercury is open to
the atmosphere while left limb is connected to a pipe in which a fluid of sp. Gr.
0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the
right limb. Find the pressure of fluid in the pipe if the difference of mercury level
in the two limbs is 20 cm.
Solution:
Given:
25
Sp.gr. of fluid,
S1 = 0.9
 Density of fluid,
1 = S1 x 1000 = 0.9 x 1000 = 900 kg/m3
Sp.gr. of mercury,
S2 = 13.6
 Density of mercury,
2 = 13.6 x 1000 kg/m3
Difference of mercury level
h2 = 20 cm = 0.2 m
Height of fluid from A-A,
h1 = 20 – 12 = 8 cm = 0.08 m
Let p = Pressure of fluid in pipe
Equating the pressure above A-A, we get
p + 1gh1 = 2gh2
or p + 900 x 9.81 x 0.08 = 13.6 x 1000 x 9.81 x .2
p = 13.6 x 1000 x 9.81 x .2 – 9.81 x 0.08
= 26683-706 = 25977 N/m2 = 2.597 N/cm2.
18. A simple U-tube manometer containing mercury is connected to a pipe in
which a fluid of sp. gr. 0.8 and having vacuum pressure is flowing. The other end
of the manometer is open to atmosphere. Find the vacuum pressure in pipe, if the
difference of mercury level in the two limbs is 40 cm and the height of fluid in the
left from the centre of pipe is 15 cm below.
Solution:
Given:
Sp.gr. of fluid,
Sp.gr. of mercury,
Density of fluid,
Density of mercury,
S1 = 0.8,
S2 = 13.6
1 = 800
2 = 13.6 x 1000
Difference of mercury level, h2 = 40 cm = 0.4 m. Height of liquid in left limb,
h1 = 15 cm = 0.15 m. Left the pressure in pipe = p. Equating pressure above datum
line A-A, we get

2gh2 + 1gh1 + p = 0
p = -[2gh2 + 1gh1]
= - [13.6 x 1000 x 9.81 x 0.4 + 800 x 9.81 x 0.15]
26
= - [53366.4 + 1177.2] = - 54543.6 N/m2 = - 5.454 N/cm2.
19. A single column manometer is connected to a pipe containing a liquid of sp.gr.
0.9 as shown in Fig. Find the pressure in the pipe if the area of the reservoir is 100
times the area of the tube for the manometer reading shown in Fig. The specific
gravity of mercury is 13.6.
Solution:
Given:
Sp. gr. of liquid in pipe,
 Density
Sp. gr. of heavy liquid,
Density,
S1 = 0.9
1 = 900 kg/m3
S2 = 13.6
2 = 13.6 x 1000
Area of reservoir
A
  100
Area of right lim b a
Height of liquid,
h1 = 20 cm = 0.2 m
Rise of mercury in right limb
h2 = 40 cm = 0.4 m
Let
PA = Pressure in pipe
Using equation (2.9), we get
PA 
a
h2   2 g  1 g   h2  2 g  h1 1 g
A

1
 0.4 13.6  1000  9.81  900  9.81  0.4  13.6  1000  9.81  0.2  900  9.81
100

0.4
133416  8829   53366.4  1765.8
100
 533.664  53366.4  1765.8 N / m2  52134 N / m2  5.21N / cm2 .
20. A differential manometer is connected at the two points A and B of two pipes
as shown in Fig. The pipe A contains a liquid of sp. gr. = 1.5 while pipe B contains
a liquid of sp.gr. = 0.9. The pressures at A and B are 1 kgf/cm2 and 1.80 kgf/cm2
27
respectively. Find the difference in mercury level in the differential manometer.
Solution:
Given:
Sp. gr. of liquid at A,
Sp. gr. of liquid at B,
Pressure at A,
S1 = 1.5
 1 = 1500
S2 = 0.9
 2 = 900
pA = 1 kgf/cm2 = 1 x 104 kgf/m2
= 1.8 x 9.81 N/m2
( 1 kgf = 9.81 N)
Pressure at B, pB = 1.8 kgf/cm2
= 1.8 x 104 kgf/cm2
= 1.8 x 104 x 9.81 N/m2
( 1 kgf = 9.81 N )
Density of mercury = 13.6 x 1000 kg/m3
Taking X-X as datum line.
Pressure above X-X in the left limb
= 13.6 x 1000 x 9.81 x h + 1500 x 9.81 x (2+3) pA
= 13.6 x 1000 x 9.81 x h + 7500 x 9.81 + 9.81 x 104
Pressure above X-X in the right limb
= 900 x 9.81 x (h+2) + pB
= 900 x 9.81 x (h+2) + 1.8 x 104 x 9.81
Equating the two pressures, we get
13.6 x 1000 x 9.81h + 7500 x 9.81 + 9.81 x 104
= 900 x 9.81 x (h+2) +1.8 x 104 x 9.81
Diving by 1000 x 9.81, we get
13.6h + 7.5 + 10 = (h+2.0) x .9 + 18
or
13.6h + 17.5 = 0.9h + 1.8 + 18 = .9h + 19.8
or
(13.6-0.9)h=19.8-17.5 or
12.7h = 2.3
2.3
h
 0.181m  18.1cm.

12.7
20. A differential manometer is connected at the two points A and B as shown in
Fig. At B air pressure is 9.81 N/cm2 (abs), find the absolute pressure at A.
Solution:
Air pressure at B
or
Density of oil
Density of mercury
= 9.81 N/cm2
pB = 9.81 x 104 N/m2
= 0.9 x 1000 = 900 kg/m3
= 13.6 x 1000 kg/m3
28
Let the pressure at A is pA
Taking datum line at X-X
Pressure above X-X in the left limb
= 1000 x 9.81 x 0.6 + pB
= 5886 + 98100 = 103986
Pressure above X-X in the left limb
= 13.6 x 1000 x 9.81 x 0.1 + 900 x 9.81 x 0.2 + pA
= 13341.6 + 1765.8 + pA
Equating the two pressure head
103986 = 13341.6 + 1765.8 + pA


pA = 103986-15107.4 = 88876.8
88876.8 N
N
 8.887 2 .
pA = 88876.8 N/m2 =
2
10000cm
cm
2
Absolute pressure at A = 8.887 N/cm .


22. Find out the differential reading ‘h’ of an inverted U-tube manometer
containing oil of specific gravity 0.7 as the manometric fluid when connected
across pipes A and B as shown in Fig. below, conveying liquids of specific
gravities 1.2 and 1.0 and immiscible with manometric fluid. Pipes A and B are
located at the same level and assume the pressures at A and B to be equal.
(A.M.I.E., Winter 1985)
Solution:
Given:
Fig. shows the arrangement. Taking X-X as datum line.
Let
PA = Pressure at A
PA = Pressure at B
Density of liquid in pipe A
= Sp. gr. x 1000
= 1.2 x 1000
= 1200 kg/m2
29
Density of liquid in pipe B
= 1 x 1000 = 1000 kg/m3
Density of oil
= 0.7 x 1000 = 700 kg/m3
Now pressure below X-X in the left limb.
= pA – 1200 x 9.81 x 0.3 – 700 x 9.81 x h
Pressure below X-X in the right limb
pA – 1200 x 9.81 x 0.3 – 700 x 9.81 x h = pB – 1000 x 9.81 (h+0.3)
But
pA = pB (given)

-1200 x 9.81 x 0.3 – 700 x 9.81 x h = -1000 x 9.81 (h+0.3)
Dividing by 1000 x 9.81,
1.2 x 0.3-0.7h =-(h+0.3)
or
0.3 x 1.2 + 0.7h = h+0.3 or 0.36-0.3 = h-0.7h = 0.3h
h


0.36  0.30 0.06

m
0.30
0.30
1
1
m   100  20cm.
5
5
30
UNIT – II
Fluid Kinematics
Flow visualization
Lines of flow
Types of flow
Continuity equation (one dimensional differential forms)Fluid dynamics
Equations of motion
Euler’s equation along a streamline
Bernoulli’s equation-applications
Venturi meter, Orifice meter, Pitot tube - Boundary layer flows,
Boundary layer thickness,
Boundary layer separation
Drag and lift coefficients.
31
BERNOULLI’S EQUATION AND BOUNDARY LAYER CONCEPTS
PART – A
1. Define Kinematics of flow.
It is defined as that branch of science which deals with motion of particles
without considering the forces causing the motion.
2. What are the methods of describing fluid flow?
The fluid motion is described by two methods they are
(i)
Lagrangian method and
(ii)
Eulerian method
In the Langrangian Method, a single fluid particle is followed during its motion
and its velocity, acceleration , density etc are described . In case of Eulerian method
the velocity, acceleration, density pressure and density etc. are described at a point
in flow field. The eulerian method is commonly used in fluid mechanics.
3. Distinguish between; steady flow and Un steady flow
Steady flow is defined as that type flow in which the fluid characteristics like
velocity , pressure , density etc at a point do not change with time. Thus for steady
flow, mathematically.
 v 
 0,
 
 t  x0 , y0 , z0
 p 
0



t



 x0 , y0 , z0
Unsteady flow is that type of flow in which the velocity pressure and density
at a point changes with respect to time
 v 
 p 
 
0,  
0

 t  x0 , y0 , z0
  t   x0 , y0 , z0
4. Distinguish between uniform Non uniform flows.
Uniform flow is defined as that type of flow in which the velocity at any given time
does not change with respect to space [i.e. Length of direction of the flow]
32
 v 
=0
 
 s  t=constant
v= Changes of velocity
s= length of flow in the direction s
Non-Uniform Flow:
Non-uniform flow is that type of flow in which the velocity at any given time
changes with respect to space. Thus, mathematically for non- uniform flow.
 v 
0
 
 s  t=constant
5. Distinguish between Laminar and Turbulent flow
Laminar flows is defined as that type of flow in which the fluid particles
move along well defined path or stream line and all the stream lines are straight
and parallel. Thus, the particles move in laminas or layers gliding smoothly over the
adjacent layer. This type of flow is called stream line flow or viscous flow.
Turbulent flow is that type of flow in which the fluid particles move in a ZigZag way. Due to the movement of fluid particles in a Zig-Zag way, the eddies
formation takes place which are responsible for high energy loss.
6. Distinguish between compressible and in compressible flow.
Compressible flow is that type of flow in which the density of the fluid
changes from point to point ie. density is not constant for the fluid. Thus
mathematically, for compressible flow.
  constant
In compressible flow is that type of flow in which the density is constant for
the fluid flow. Liquids are generally incompressible. Mathematically for
compressible flow.
  constant
7. Distinguish between rotational and in rotational flow
Rotational flow is that type of flow in which the fluid particles while flowing
along stream lines, also rotate about their own axis . And if the fluid particles while
33
flowing along stream Lines, do not rotate about their own axis, that type of flow is
called irrotational flow.
8. Define the equation of continuity obtain an expression for continuity equation
for three dimensional flow.
According to Law of Conservation of mass. Rate of flow at section 1-1 =rate of flow
at section 2.2
1A1V1=2A2V2
The above equation is applicable to compressible as incompressible fluids and
is called continuity equation. If the fluid is incompressible then 1 = 2 and continuity
equation reduces to.
A1V1 = A2V2
9. Explain the term local Acceleration and convective Acceleration.
Local acceleration: is designed as the rate of increase of velocity with respect
to time at a given point in a flow field.
u v w
, ,
is known as local acceleration
t t t
Convective Acceleration:
It is defined as the rate of change of velocity due to the change of position of fluid
particles in a fluid flow.
10. Type of flow line
Path line
Stream line
Streak line or filament lines
Potential lines or Equi-potential lines
Flow net
11. Explain the terms:
(i)
(ii)
Path Line
Stream Line
34
Path Line:
A path line time is defined as , the path or line traced by a single particle of
fluid during a period of time. Path line shows the direction of velocity of the same
fluid.
Stream Line:
This is an imaginary curve drawn through a flowing fluid in such a way that
the tangent of which at any point . the pattern of flow of fluid may be represented by
a series of stream Lines obtained by drawing a series of curves into the following
fluid such that the velocity vector at any point is tangential to the curves.
12. Define Equipotential line.
A line along which the velocity potential is constant, is called equipotential line.
13. Define flow net.
A grid obtained by drawing a series of equipotential lines and stream lines is
called Flow net.
14. What is the Euler’s equation of motion? How will you obtain Bernoulli’s
equation from if equation of motion.
According to Newton’s second law of motion, the net force Fx acting on a
fluid element in the direction of x is equal to mass m of the fluid element multiplied
by the acceleration ax in the x – direction
Fx = M.ax
(i)
(ii)
(iii)
(iv)
(v)
Fg – Gravity force
Fp – The pressure force
Fv – Force due to viscosity
Ft – force due to turbulence
Fc – force due to compressibility
Thus in equation, the net force,
35
Fx = (Fg)x + (Fp)x + (Fv)x + (Ft)x + (Fc)x
(i) If the force due to compressibility, Fc is negligible, the resulting net force F x = (Fg)x
+ (Fp)x + (Fv)x + (Ft)x and equation of motions are called Reynold’s equation of
motion.
(ii) For flow, where Ft is negligible the resulting equations of motion are known as
Navier – Stoles equation
(iii) If the flow is assumed to be ideal, viscous force (Fp) is zero and equation of
motions are known as Euler’s equation of motion.
15. State the Bernoulli’s theorem?
It states that in a steady ideal flow of an incompressible fluid, the total energy
at any point of the fluid is constant. The total energy consists of pressure energy,
kinetic energy and potential energy or datum energy. These energies per unit weight
of the fluid are;
Pressure energy = p  p
w
g
Kinetic energy =
v2
2g
Datum energy = z
So,
p V2

+Z = constant
ω 2g
16. What is a venturi meter?
A venture meter is a device used for measuring the rate of flow of a fluid flowing
through a pipe. It consists of three parts;
(i)
A short converging part
(ii)
Throat and
(iii) Diverging part
17. What is a orifice meter?
It is a device used for measuring the rate of flow of fluid through a pipe. It is
a cheaper device as compared to venturi meter. It is also works on the same
principle as that of venture meter. It consists of a flat circular plate which has a
circular sharp edged hole called orifice which is concentric with the pipe.
36
18. What is pitot tube?
It is a device used for measuring the velocity of flow at any point in a pipe or
a channel. It is based on the principle that, if the velocity of flow at appoint becomes
zero, the pressure there is increased due to the conversion of kinetic energy into
pressure energy. In its simplest form, the pitot – tube consists of a glass tube, bent at
a right angles.
19. What is the Euler’s equation of motion? How will you obtain Bernoulli’s
equation from if equation of motion.
According to Newton’s second law of motion, the net force Fx acting on a
fluid element in the direction of x is equal to mass m of the fluid element multiplied
by the acceleration ax in the x – direction.
Fx = M.ax
(vi)
(vii)
(viii)
(ix)
(x)
Fg – Gravity force
Fp – The pressure force
Fv – Force due to viscosity
Ft – force due to turbulence
Fc – force due to compressibility
Thus in equation, the net force,
Fx = (Fg)x + (Fp)x + (Fv)x + (Ft)x + (Fc)x
1. If the force due to compressibility, Fc is negligible, the resulting net force F x =
(Fg)x + (Fp)x + (Fv)x + (Ft)x and equation of motions are called Reynold’s
equation of motion.
2. For flow, where Ft is negligible the resulting equations of motion are known
as Navier – Stoles equation
3. If the flow is assumed to be ideal, viscous force (Fp) is zero and equation of
motions are known as Euler’s equation of motion.
20. Types of flows
Uniform flow, non uniform flow, stream line flow, turbulent flow, steady
flow, unsteady flow, compressible flow, incompressible flow, rotational flow,
irrigational flow, one dimensional flow, two dimensional flow, three dimensional
flow etc
21. Define Drag.
37
The component of the total Force (FR) in the direction of motion is called drag.
It I denoted by FD.
22. Define Lift.
The component of the total force (FR) in the direction perpendicular to the
direction of motion is known as lift. It is denoted by FL.
PART - B
1. Derive the Bernoulli’s equation from Euler’s equation.
EULER’S EQUATION OF MOTION
This is equation of motion in which the forces due to gravity and pressure are
taken into consideration. This is derived by considering the motion of a fluid
element along a stream-line as:
Consider a stream-line in which flow is taking place in S-direction. Consider
a cylindrical element of cross-section dA and length dS. The forces acting on the
cylindrical element are:
Pressure force pdA in the direction of flow.
p 

1. Pressure force  p 
ds dA opposite to the direction of flow.
s 

2. Weight of element gdAds.
Let  is the angle between the direction of flow and the line of action of the
weight of element.
The resultant force on the fluid element in the direction of S must be equal to
the mass of fluid element  acceleration in the direction S.
p 

 pdA -  p +
ds dA  gdAds cos 
s 

= pdAds  as
where as is the acceleration in the direction of S.
38
..... (1)
Now as =

dv
, where v is a function of s and t.
dt
v ds v vv v



s dt t
s
t
If the flow is steady,




ds

 v
dt

v
0
t
vv
as 
s
Substituting the value of as in equation (1) and simplifying the equation, we get
p
vv
dsdA  g dAds cos  =  dAds 
s
s
p
vv

 g cos  =
Dividing y dsdA,
s
s
p
vv
 g cos  + v
0
or
s
s

we have
cos  =
dz
ds
1 p
dz vv
g

0
 s
ds s
p
 gdz  vdv  0


or
or
p
 gdz  vdv  0

....(2)
Equation (2) is known as Euler’s equation of motion.
Bernoulli’s equation is obtained by integrating the Euler’s equation of motion
(2) as

dp
 gdz   vdv = constant
 
If flow is incompressible,  is constant and
39
p
v2
 gz +
 constant

2
p
v2
z
 cons tant
g
2g
p v2

 z  cons tant
g 2g

or
or
.....(3)
Equation (3) is a Bernoulli’s equation in which
p
 pressure energy per unit weight of fluid or pressure Head.
g
V2/2g = Kinetic energy per unit weight or kinetic Head.
z
= potential energy per unit weight or potential Head.
2. The water is flowing through a pipe having diameters 20 cm and 10 cm at
sections 1 and 2 respectively. The rate of flow through pipe is 35 litres/s. The
section 1 is 6m above datum and section 2 is 4 m above datum. If the pressure at
section 1 is 39.24 N/cm2, find the intensity of pressure at section 2.
Solution:- Given
At section 1, D1 = 20 cm = 0.2 m

2
.2   0.0314m2
4
p1  39.24 N/cm2
A1 
= 39.24  10 4N / m2
z1  6.0 m
At section 2, D2 = 0.10 m

2
A 2   0.1  .00785 m2
4
z2  4 m
p2  ?
Rate of flow,
Q  35lit / s 
35
 0.035 m3 / s.
1000
40
Now
Q = A1V1 = A2V2

V1 
Q 0.035

 1.114 m/s
A1 .0314
Q
0.035
V2 

 4.456 m/s
A 2 .00785
and
Applying Bernoulli’s equation at sections 1 and 2, we get
p1 v12
p2 v 2 2

 z1 

 z2
g 2g
g 2g
 4.456   4.0
p2
39.24  104 1.114 

 6.0 

1000  9.81 2  9.81
1000  9.81 2  9.81
2
or
or
40  0.063  6.0 
or
46.063 


2
p2
 1.012  4.0
9810
p2
 5.012
9810
p2
 46.063  5.012  41.051
9810
p2 = 41.051  9810 N/m2

41.051 9810
N / cm2  40.27 N/cm2 .
4
10
3. Water is flowing through a pipe having diameter 300 mm and 200 mm at the
bottom and upper end respectively. The intensity of pressure at the bottom end is
24.525 N/cm2 and the pressure at the upper end is 9.81 N/cm2. Determine the
difference in datum head if the rate of flow through pipe is 40 lit/s.
Solution:
Given:
Section 1,
D1 = 300 mm = 0.3 m
p1 = 24.525 N/cm2 = 24.525  104 N/m2
Section 2, D2 = 200 mm = 0.2 m
p2 = 9.81 N/cm2 = 9.81  104 N/m2
41
Rate of flow
= 40 lit/s.
40
 0.04 m3 / s.
1000
or
Q
Now
A1V1 = A2V2 = rate of flow = 0.04
V1 
.04
.04
0.04


 0.5658 m / s.
A1  D 2  0.3 2
 
4 1
4
0.566 m/s.
V2 
.04
.04
0.04


 1.274 m/s
A 2  D 2  .2 2
 
 
4 2
4

Applying Bernoulli’s equation at (1) and (2), we get
p1 v12
p2 v 2 2

 z1 

 z2
g 2g
g 2g
24.525  104 .566  .566
9.81 104 1.274 

 z1 

 z2
1000  9.81
2  9.81
1000  9.81 2  9.81
2
or
or
or
25 + .32 + z1 = 10 + 1.623 + z2
25.32 + z1 = 11.623+ z2

z2 – z1 = 25.32 – 11.623 = 13.697 = 13.70 m
 Difference in datum head = z2 – z1 = 13.70 m.
4. Explain the working principle of venturimeter.
Venturimeter. A venturimeter is a device used for measuring the rate of a flow of a
fluid flowing through a pipe. It consists of three parts:
(i) A short converging part, (ii) Throat, and (iii) Diverging part. It is based on
the Principle of Bernoulli’s equation.
Expression for Rate of Flow through Venturimeter
42
Consider a venturimeter fitted in a horizontal pipe through which a fluid is
flowing (say water), as shown in figure.
Let
d1 = diameter at inlet or at section (1),
p1 = pressure at section (1)
v1 = velocity of fluid at section (1),
 2
d1
4
d2, p2, v2, a2 are corresponding values at section (2).
a = area at section (1) =
and
Applying Bernoulli’s equation at sections (1) and (2), we get
p1 v12
p
v 2

 z1  2  2  z2
g 2g
g 2g
As pipe is horizontal, hence
z 1 = z2
2
p1 v1
p2 v 2 2
p1  p2 v 22 v12




or


g 2g g 2g
g
2g 2g
p  p2
But 1
is the difference of pressure heads at sections 1 and 2 and it is equal to h
g
p  p2
or 1
=h
g
Substituting this value of
p1  p 2
in the above equation, we get
g
v 22 v12

...(1)
2g 2g
Now applying continuity equation at sections 1 and 2
h
a1v1  a2 v 2 or v1 
a2 v 2
a1
Substituting this value of v1 in equation (1)
2
or
 a2 v 2 


v 2 2  a1 
v 2  a 2  v 2 a 2  a 2 
h

 2 1  22   2  1 2 2 
2g
2g
2g  a1  2g  a1

2
a
v 22  2gh 2 1 2
a1  a2
43

v 2  2gh
 Discharge,
a12
a1

2
2
2
a1  a2
a1  a22
2gh
Q = a2v2
 a2
a1
a12  a22
 2gh 
a1a2
a12  a22
 2gh …(2)
Equation (2) gives the discharge under ideal conditions and is called
theoretical discharge. Actual discharge will be less than theoretical discharge.

Qact  Cd 
a1a2
a12  a22
 2gh
....(3)
where Cd = Co-efficient of venturimeter and its value is less than 1.
Value of ‘h’ given by differential U-tube manometer
Case I. Let the differential manometer contains a liquid which is heavier than the
liquid flowing through the pipe. Let
Sh = sp. Gravity of the heavier liquid
So = sp. Gravity of the liquid flowing through pipe
x = difference of the heavier liquid column in U-tube
Then
S

h  x  h  1
 So 
…(4)
Case II. If the differential manometer contains a liquid which lighter than the liquid
flowing through the pipe, the value of h is given by
 S 
h  x 1  h 
 So 
where S1 = sp. gr. Of lighter liquid in U-tube
So = sp. Gr. Of fluid flowing through pipe
x = difference of the lighter liquid columns in U-tube.
….(5)
Case III. Inclined Venturimeter with Differential U-tube manometer.
The above two cases are given for a horizontal venturimeter. This case is
related to included venturimeter having differential U-tube manometer. Let the
44
differential manometer contains heavier liquid then h is given as
S

p
 p

h   1  z1    2  z2   x  h  1
 g
  g

 So 
….(6)
Case IV
Similarly, for inclined venturimeter in which differential manometer contains
a liquid which is lighter than the liquid flowing through the pipe, the value of h is
given as
 S 
p
 p

h   1  z1    2  z2   x 1  l 
…. (7)
 g
  g

 So 
5. A horizontal venturimeter with inlet and throat diameters 30 cm and 15 cm
respectively is used to measure the flow of water. The reading of differential
manometer connected to the inlet and the throat is 20 cm of mercury. Determine
the rate of flow. Take Cd = 0.98.
Solution, Given:
Dia. at inlet,
 Area at inlet,
Dia, at throat,

d1 = 30 cm


2
a1  d12   30   706.85 cm2
4
4
d2 = 15 cm

a2   152  176.7 cm2
4
Cd = 0.98
Reading of differential manometer = x = 20 cm of mercury.
 Difference of pressure head is given by
or
S

h  x  h  1
 So 
where Sh = sp. gravity of mercury = 13.6, So = sp. gravity of water = 1
13.6 
 20 
 1  20  12.6 cm  252.0 cm of water
 1

45
The discharge through venturimeter is given by
Q  Cd
a1a2
a12  a22
= 0.98 
 2gh
706.85  176.7
 706.85 
 176.7 
 2  981 252
2
86067593.36
684.4
499636.9  31222.9
125756
 125756 cm3 / s 
lit / s  125.756 lit / s.
1000

86067593.36
2

6. A horizontal venturimeter with inlet diameter 20 cm and throat diameter 10 cm
is used to measure the flow of water. The pressure at inlet is 17.658 N/cm2 and the
vacuum pressure at the throat is 30 cm of mercury. Find the discharge of water
through venturimeter. Take Cd = 0.98.
Solution Given:
Dia. at inlet,

Dia. at throat,

d1 = 20 cm

2
a1    20   314.16 cm2
4
d2 = 10 cm

a2   102  78.74 cm2
4
p1  17.658 N/cm2  17.658  10 4 N / m2
 for water
 1000
kg
m3

and
p1 17.658  104

 18 m of water
g 9.81 1000
p2
 30 cm of mercury
g
= -0.30 m of mercury = - 0.30  13.6 = -4.08 m of water
p
p
 Differential head
 h  1  2  18   4.08 
g g
= 18 + 4.08 = 22.08 m of water = 2208 cm of water
The discharge Q is given by
46

Q  Cd
a1a2
a12  a22
 0.98 

 2gh
314.16  78.54
 314.16    78.74 
2
2
 2  981 2208
50328837.21
 165555 cm3 / s  165.555 lit/s.
304
7. Explain the Working Principle of Orifice meter.
Orifice Meter or Orifice Plate
It is a device used for measuring the rate of flow of a fluid through a pipe. It is
a cheaper device as compared to venturimeter. It also works on the same principle as
that of venturimeter. It consists of a flat circular plate which has a circular sharp
edged hole called orifice, which is concentric with the pipe. The orifice diameter is
kept generally 0.5 times the diameter of the pipe, though it may vary from 0.4 to 0.8
times the pipe diameter.
A differential manometer is connected at section (1), which is at a distance of
about 1.5 to 2.0 times the pipe diameter upstream from the orifice plate, and at
section (2), which is at a distance of about half the diameter of the orifice on the
down stream side from the orifice plate.
Let
p1 = pressure at section (1),
v1 = velocity at section (1),
a1 = area of pipe at section 1), and
p2, v2, a2 are corresponding values at section (2). Applying Bernoulli’s
equation at sections (1) and (2) we get
p1 v12
p
v 2

 z1  2  2  z2
g 2g
g 2g
 p1
  p2
 v 22 v12

  z1     z2  
 g
  g
 2g 2g
or
But
 p1
  p2

  z1     z2   h  Differential head
 g
  g

47
v 22 v12
h

or 2gh = v 22  v12
2g 2g

v 2  2gh  v12
or
…..(i)
Now section (2) is at the vena contracts and a2 represents the area at the vena
contracts. If a0 is the area of orifice then, we have
Cc 
a2
ao
where Cc = Co-efficient of contraction

a2  ao  Cc
….(ii)
By continuity equation, we have
a1v1  a2 v 2 or v1 
aC
a2
v2  0 c v2
a1
a1
….(iii)
Substituting the value of v1 in equation (i), we get
v 2  2gh 
or
v2

The discharge
2
a 0 2 Cc 2 v 2 2
a12
  a 2

 a0  2 2
2
 2gh    Cc v 2 or v 2  1   0  Cc 2   2hg
  a1 

 a1 
v2 
2gh
2
a 
1   0  Cc 2
 a1 
Q  v 2  a2  v 2  a0Cc

a0Cc 2gh

a2  a0Cc from (ii)
....(iv)
2
a 
1   0  Cc 2
 a1 
The above expression is simplified by using
48
a 
1  0 
 a1 
Cd  Cc
2
2
a 
1   0  Cc 2
 a1 
2

a 
1   0  Cc 2
 a1 
Cc  Cd
a 
1  0 
 a1 
2
Substituting this value of Cc in equation (iv), we get
2
Q  a0  Cd
=
a 
1   0  Cc 2
 a1 
a 
1  0 
 a1 
Cda0 2gh
a 
1  0 
 a1 
2

2

2gh
a 
1   0  Cc 2
 a1 
Cda0a1 2gh
....(1)
a12  a0 2
where Cd = Co-efficient of discharge for orifice meter.
The co-efficient of discharge for orifice meter is much smaller than that for a
venturimeter.
8. An orifice meter with orifice diameter 15 cm is inserted in a pipe of 30 cm
diameter. The pressure difference measured by a mercury oil differential
manometer on the two sides of the orifice meter gives a reading of 50 cm of
mercury. Find the rate of flow of oil of sp. gr. 0.9 when the co-efficient of
discharge of the meter = 0.64.
Sol. Given:
Dia. Of orifice,
 Area,
Dia. of pipe,
d0 = 15 cm

2
15   176.7 cm2 a
4
d1 = 30 cm
ao 
49

2
 30   706.85 cm2
4
S0 = 0.9
a1 
 Area,
Sp. gr. of oil
Reading of Differential manometer, x = 50 cm of mercury
S

13.6 
h  x  g  1  50 
 1 cm of oil
 0.9

 So 
 50 14.11  705.5 cm of oil
 Differential head,
Cd = 0.64
 The rate of the flow, Q is given equation
Q  Cd .
a0a1
a12  a0 2
= 0.64 
 2gh
176.7  706.85
 706.85   176.7 
2
2
 2  981  705.5
94046317.78
 137414.25 cm3 / s  137.414 Litres/s.
684.4
9. A sub-marine moves horizontally in sea and has its axis 15m below the surface
of water. A pitot-tube properly placed just in front of the sub-marine and along its
axis is connected to the two limbs of a U- tube containing mercury. The difference
of mercury level is found to be 170mm. Find the speed of the sub-marine knowing
that the sp. Gr. Of mercury is 13.6 and that of sea – water is 1.026 with respect of
fresh water.
(A.M.I.E., Winter, 1975)
=
Solution. Given:
Diff. of mercury level,
Sp. Gr. Of mercury,
Sp. Gr. Of sea – water,
x=170 mm=0.17m
Sg=13.6
So=1.026
50

 Sg

13.6
h=x   1  0.17 
 1  2.0834m
S
1.026


 o


V= 2gh  2  9.81  2.0834  6.393m / s
=
6.393  60  60
km / hr  23.01km / hr.
1000
10. A pitot-tube is inserted in a pipe of 300mm diameter. The static pressure in
pipe is 100mm of mercury (vacuum). The stagnation pressure at the centre of the
pipe, recorded by the pitot-tube is 0.981 N/cm2. Calculate the rate of flow of water
through pipe, if the mean velocity of flow is 0.85 times the central velocity. Take
Cv=0.98.
(Converted to S.I. Units, A.M.I.E., Summer, 1987)
Solution. Given:
Dia. of pipe,
 Area,
d=300mm=0.30m


a= d2  (.3)2  0.07068m 2
4
4
Static pressure head
=100mm of mercury (vacuum)
100

 13.6  1.36m of water
1000
Stagnation pressure
=.981 N/cm2=.981  104 N/m2
 Stagnation Pressure head =

.981  104
.981  104

 1m
g
1000  9.81
h=Stagnation pressure head – Static pressure head
=1.0-(-1.36)=1.0+1.36=2.36m of water
 Velocity at centre
=C  2gh
=0.98  2  9.81  2.36  6.668m / s
Mean velocity,
V  0.85  6.668  5.6678m / s
 Rate of flow of water = V  area of pipe
=5.6678  .07068m 3 / s  0.4006m 3 / s.
51
11. A 30 cm diameter pipe, conveying water, branches into two pipes of diameters
20cm and 15cm respectively. If the average velocity in the 30cm diameter pipe is
2.5 m/s, find the discharge in this pipe. Also determine the velocity in 1.5 cm pipe
if the average velocity in 20cm diameter pipe is 2m/s.
Sol. Given:
D1=30cm=0.30m



A1 = D12  .3 2  0.07068m 2
4
4
V1=2.5m/s
D2=20cm=0.20m


2

A2 =  .2   .4  0.0314m 2 ,
4
4
V2=2m/s
D3=15cm=0.15m


2

A 3 = .15    0.225  0.01767m 2
4
4
Find (i) Discharge is pipe 1 or Q1
(ii) Velocity in pipe of dia. cm or V3
Let Q1,Q2 and Q3 are discharges in pipe , and respectively
Then according to continuity equation
Q1=Q2+Q3
(i)
(ii)
……(1)
The discharge Q1 in pipe 1 is given by
Value of V3
Q2 = A2V2=.0314 2.0=.0628 m3/s
Substituting the values of Q1 and Q2 in equation (1)

But

0.1767-0.0628=0.1139m3/s
Q3=.1767-0.0628=0.1139m3/s
Q3=A3 V3=.01767  V3 or .1139=.01767 V3
.1139
V3 
 6.44m / s.
.01767
12. Water flows through a pipe AB 1.2m diameter at 3m/s and then passes through
52
a pipe BC 1.5m diameter. At C, the pipe branches. Branch CD is 0.8m in diameter
and carries one-third of flow in AB. The flow velocity in branch CE is 2.5 m/s.
Find the volume rate of flow in AB, the velocity in BC, the velocity in CD and the
diameter of CE.
(AMIE, Winter 1982; Osmania University, 1990)
Sol. Given:
Diameter of pipe AB,
Velocity of flow through AB,
Dia. of pipe BC,
Dia. of branched pipe CD,
Velocity of flow in pipe CE,
Let the flow rate in pipe
Velocity of flow in pipe
Velocity of flow in pipe
Diameter of pipe
Then flow rate through
DAB=1.2m
VAB=3.0 m/s
DBC=1.5m
VCD=0.8m
VCE=2.5 m/s
AB=Q m3/s
BC=VBCm/s
CD=VCDm/s
CE= DCE
CD=Q/3
And flow rate through
CE=Q-Q/3=
2Q
3
i) Now volume flow rate through AB=Q=VAB  Area of AB


2
2
 3.0   DAB   3.0  1.2 
4
4
3
=3.393 m /s.
20. Explain Boundary Layer thickness.
Boundary Layer Thickness (): It is defined as the distance from the boundary of the
solid body measured in the y – direction to the point, where the velocity of the fluid
is approximately equal to 0.99 times the free steam (U) velocity of the fluid. It is
denoted by the symbol. For laminar and turbulent zone it is denoted as:
1. lam = Thickness of laminar boundary layer,
2. tur = Thickness of turbulent boundary layer, and
3. ’ =Thickness of laminar sub – layer
Displacement Thickness (*): It is defined as the distance, measured perpendicular
to the boundary of the solid body, by which the boundary should be displaced to
compensate for the reduction in flow rate on account of boundary layer formation. It
is denoted by *. It is also defined as:
53
“The distance, perpendicular to the boundary, by which the free stream is
displaced due to the formation of boundary layer”
Expression for *:
Consider the flow of a fluid having free – stream velocity equal to U over a
thin smooth plate as shown in figure. At a distance x from the leading edge. The
velocity of fluid at B is zero and at C, which lies on the boundary layer, is U. Thus
velocity varies from zero at B to U at C, where BC is equal to the thickness of
boundary layer i.e.,
Distance BC = 
Let
y = distance of elemental strip from the plate,
dy = thickness of the elemental strip,
u = velocity of fluid at the elemental strip
b = width of plate.
Then area of elemental strip, dA = b x dy
Mass of fluid per second flowing through elemental strip
=  x Velocity x Area of elemental strip
= u x dA = u x b x dv
 (i)
If there had been no plate, then the fluid would have been flowing with a
constant velocity equal to free – stream velocity (U). Then mass of fluid per second
flowing through elemental strip would have been
=  x Velocity x Area = x U x b x dy
 (ii)
As U is more than, hence due to the presence of the plate and consequently
due to the formation of the boundary layer, there will be a reduction in mass flowing
per second through the elemental strip.
This reduction in mass / sec flowing through elemental strip
= mass / sec given by equation (ii) – mass / sec given by equation (i)
= Ubdy - ubdy = b(U-u) dy
 Total reduction in mass of fluid /s flowing through BC due to plate
54


0
0
   b U  u  dy   b  U  u  dy
  iii 
if fluid is incompressible
Let the plate is displaced by a distance * and velocity of flow for the distance
* is equal to the free-stream velocity (i.e., U). Loss of the mass of the fluid / sec
flowing through the distance *
=  x Velocity x Area
=  x U x * x b
{Area = * x b}

(iv) Equating equation (iii) and (iv), we get

 b  U  u  dy    U   * b
0
Canceling b to both sides, we have

 U  u dy  U   * Or
0

 U is constant and can



 be taken inside the integral 

U  u  dy
1
 *   U  u  dy  
U0
U
0

 *    1 
0

u
 dy
U
22. A plate of 600 mm length and 400 mm wide is immersed in a fluid of sp.gr. 0.9
and kinematic viscosity (v=1) 10-4 m2/s. The fluid is moving with a velocity of 6
m/s. Determine (i) boundary layer thickness, (ii) shear stress at the end of the
plate, and (iii) drag force on one side of the plate.
Solution:
As no velocity profile is given in the above problem, hence Blasius’s solution
will be used.
Given: Length of plate,
Width of plate,
Sp.gr.of fluid,
 Density,
Velocity of fluid,
Kinematic viscosity,
L = 600 mm = 0.60 m
b = 400 mm = 0.40 m
S = 0.9
 = 0.9 x 1000 = 900 kg / m3
U = 6 m/s
v = 10-4 m2/s
55
ReL 
Reynold number
U  L 6  0.6

 3.6  10 4
v
10 4
As ReL is less than 5 x 105, hence boundary layer is laminar over the entire
length of the plate.
(i) Thickness of boundary layer at the end of the plate from Blasius’s solutions is


4.91x
,
Rex
4.91  0.6
3.6  10 4
where x =0.6 m and R ex  3.6  10 4
 0.0155m  15.5mm
(ii) Shear stress at the end of the plate is
U 2 0.332  900  6 2
 0  0.332

 56.6 N / m2
4
ReL
3.6  10
(iii) Drag force (FD) on one side of the plate is given by
FD 
1
 AU 2  CD
2
Where CD from Blasius’s solution is
1.328
1.328
CD 

 0.00699
ReL
3.6  10 4
FD 
1
 AU 2  C D
2
1
  900  0.6  0.4  6 2  .00699
2
 A=L  b=0.6  0.4
 26.78 N
23. A flat plate 1.5m x 1.5m moves at 50 km/hour in stationary air of density 1.15
kg/m3. if the co-efficient of drag and life are 0.15 and 0.75 respectively,
determine:
i.
ii.
iii.
iv.
The lift force,
The drag force
The resultant force, and
The power required to keep the plate in motion (A.M.I.E, Winter 1997)
56
Solution :
Given
Area of the plate,
A = 1.5 x 1.5 = 2.25 m2
Velocity of the plane,
U = 50 km/hr =
Density of air
Co-efficient of drag
Co-efficient of lift
 = 1.15 kg/m3
CD = 0.15
CL = 0.75
i)
Lift Force (FL) Using equation.
FL  CL A 
ii)
U 2
1.15  13.892
 0.75  2.25 
N=187.20N Ans.
2
2
Drag Force (FD) using equation
FD  CD  A 
iii)
50  1000
m / s = 13.89m/s
60  60
U 2
1.15  13.892
 0.15  2.25 
N=37.44N Ans.
2
2
Resultant Force (FR) Using equation
FR  FD2  FL2  37.442  187.202 N
= 1400+35025  190.85 N
iv)
Power Required to keep the Plate in Motion
P=

Force in the direction of motion  Velocity
kW
1000
FD  U 37.425  13.89

kW  0.519 kW. Ans
1000
1000
24. A man weighting 90 kgf descends to the ground from an aeroplane with the
help of a parachute against the resistance of air. The velocity with which the
parachute, which is hemispherical in shape, comes down is 20 m/s. finds the
diameter of the parachute. Assume CD = 0.5 and density of air =1.25 kg/m3.
Solution, Given:
Weight of man
Velocity of parachute
Co-efficient of drag
Density of air
Let the dia, of parachute
W = 90 kgf = 90 x 9.81 N = 882.9 N ( 1 kgf = 9.81 N)
U = 20 m/s
CD = 0.5
 = 1.25 kg/m3
=D
 Area
A

4
D2m2
57
When the parachute with the man comes down with a uniform velocity, U=20
m/s, the drag resistance will be equal to the weight of man, neglecting the weight
parachute. And projected are of the hemispherical parachute will be equal to
 Drag,
4
d2.
FD = 90 kgf = 90 x 9.81 = 882.9 N (using equation)
FD  CD  A 
U 2
 882.9=0.5 
 D2 

2

4
D4 
1.25  202
2
882.9  4  2.0
 8.9946 m2
0.5    1.25  20  20
D  8.9946  2.999 m. Ans
25. A kite 0.8 m x 0.8 m weighing 0.4 kgf (3.924 N) assumes an angle, of 12 to the
horizontal. The string attached to the kite makes an angle of 45 to the horizontal.
The pull on the string is 2.5 kgf (24.525 N) when the wind is flowing at a speed of
30 km/hour. Find the corresponding co-efficient of drag and lift. Density of air is
given as 1.25 kg/m3
Solution,
Given:
Projected area of kite,
weight of kite,
Angle made by kite with horizontal,
Pull on the string
A = 0.8 x 0.8 = 0.64 m2
W = 0.4 kgf = 0.4 x 9.81 = 3.924 N
1 = 45
P = 2.5 kgf = 2.5 x 9.81 = 24.525 N
Speed of wind,
U = 30 km/hr =
Density of air,
Drag force,
of motion
 = 1.25 kg/m3
FD = Force exerted by wind in the direction
30  1000
m / s = 8.333 m/s
60  60
(i.e. in the X-X direction)
= Component of pull, P along X-X
58
= P cos 45 = 24.525 cos 45 = 17.34 N
And Lift Force, FL = force excerted by wind on the kite perpendicular to the
direction of motion (i.e, along r-Y direction)
= Component of P in vertically downward
direction + Weight of kite (W)
= P sin 45 + W = 24.525 sin 45 + 3.924 N
= 17.34 + 3.924 = 21.264 N.
i) Drag co-efficient (CD). using equation, we have
FD  CD  A 
CD 
U 2
2
2  FD
2  17.34

 0.624. Ans.
2
A U
0.64  1.25  8.3332
ii) Lift co-efficient (CL). using equation, we have
FL  CL  A 
CL 
U 2
2
2  FL
2  21.264

 0.765. Ans.
2
A   U
0.64  1.25  8.3332
26. A jet plane which weights 29.43 kN and having a wing area of 20m 3 flies at a
velocity of 950 km/hour, when the entire delivers 7357.5 kW power. 656% of the
power is used to overcome the drag resistance of the wing. Calculate the coefficients of lift and drag for the wing. The density of the atmospheric air is 1.21
kg/m3.
Solution, given :
Weight of plane,
Wing area,
W = 29.43 kN = 29.43 x 1000 N = 29430 N
A = 20 m3.
Speed of plane
U = 950 km/hr =
Engine power,
Power used to overcome drag
P = 7357.5 kW
Resistance
= 65% of 7357.5 =
 Density of air,
 = 1.21 kg/m3
59
950  1000
 263.88 m / s
60  60
65
 7357.5  4782.375 kW
100
Let
Now power used in kW to
over come drag resistance
CD = Coefficient of drag and CL = coefficient of lift.
=
FD  U
F  263.88
or 4782.375= D
1000
1000
 FD 
But from equation, we have
4782.375  1000
263.88
FD  CD .A.

U 2
2
4782.375  1000
263.88 2
 CD  20  1.21
263.88
2
 CD 
4782.375  1000  2
 0.0215. Ans.
20  1.21 263.883
The lift force should be equal to weight of the plane
FL  CL .A.
But
CL 
U 2
2
FL = W = 29430 N
or 29430 = CL  20  1.21
29430  2
 0.0349. Ans.
20  1.21 263.88 2
60
263.882
2
UNIT – III
Viscous flow
Navier - Stoke’s equation (Statement only)
Shear stress, pressure gradient relationship
Laminar flow between parallel plates
Laminar flow through circular tubes (Hagen poiseulle’s)
Hydraulic and energy gradient
Flow through pipes
Darcy -weisback’s equation –
Pipe roughness –
Friction factorMinor losses –
Flow through pipes in series and in parallel
Power transmission.
FLOW THROUGH PIPES
61
PART - A
1. What do you understand by the terms major energy loss and minor energy
losses in pipes?
The loss energy in pipe is classified as major energy loss and minor energy
lossed. Major energy loss is due to friction while minor energy losses are due to
sudden expansion of pipe, sudden contraction of pipe, bend in pipe and an
obstruction in pipe.
2. How will you determine the loss of head due to friction in pipes by using (i)
Darcy formula and (ii) chezys formula?
4fLV 2
Energy loss due to friction is given by darcy formula, hfhf =
2gd
The head loss due to friction in pipe can also be calculated by Chezy’s
formula V = c mi
Where,
C = chezys constant
m = hydraulic mean depth = d/4
V = velocity of flow
h
i  f  loss of head per unit length
L
3. Derive an expression for the loss of due to
(i) Sudden enlargement (ii) Sudden contraction of a pipe
(V - V2 ) 2
(i) Loss of head due to sudden expansion of pipe, hc = 1
2g
V1 = velocity in small pipe
V2 = velocity in large pipe
(ii) Loss of head due to sudden contraction of pipe,
 1 -1 
hc = 

 Cc 
2
V22
2g
Ce = Coefficient of contraction
V2
= 0.375 2 [for Cc = 0.62]
2g
62
V22
= 0.5
[if value of Cc is not given]
2g
4. Define the terms:
(i) Hydraulic gradient line
(ii) Total energy line
(i) Hydraulic gradient line:
The line representing the sum of pressure head and datum head of a flowing
fluid in a pipe with respect to some reference line is called hydraulic gradient line
(H.G.L)
(ii) Total Energy line:
The line representing the sum of pressure head, datum head and velocity
head of a following fluid in a pipe with respect to some reference line is known as
total energy line [T.E.L]
5. What is a siphon? On what principle it works.
Siphon is a long bent pipe used to transfer liquid from a reservoir at a higher
level to another reservoir at a lower level, when the two reservoirs are separated by a
high level ground.
6. What is compound pipe? What will be loss of head when pipes are connected in
series?
When pipes of different lengths and different diameter are connected end to
end, (in series) it is called as compound pipe. The rate of flow through each pipe
connected is series is same.
7. Explain the terms (i) pipes in parallel (ii) equivalent pipe (iii) Equivalent size of
the pipe?
(i) Pipes in parallel:
When the pipes are connected in parallel, the loss of head in each pipe is
63
same. The rate of flow in main pipe is equal to sum of the rate of flow in each pipe,
connected in parallel.
(ii) Equivalent size of the pipe:
The diameter of equivalent pipe is called equivalent size of the pipe.
The equivalent size of the pipe is obtained from
L L1 L 2 L3
 

d 5 d15 d 52 d 53
L = Equivalent length of pipe = L1 + L2+L3 and d1, d2, d3 = are diameters of pipe
connected in series.
Equivalent size of the pipes = d
(iii) Equivalent pipe:
A single pipe of uniform diameter, having same discharge and same loss of
head as compound pipe consisting of several pipes of different lengths and
diameters is known as equivalent pipe
8. Explain the phenomenon of Water hammer?
When a liquid is flowing through a long pipe fitted with a valve at the end of
the pipe and the valve is closed suddenly a pressure wave of high intensity is
produced behind the valve. This pressure wave of high intensity is transmitted
along the pipe with sonic velocity. This pressure wave of high intensity is having
the effect of hammering action on the walls of the pipe. The phenomenon is known
as water hammer
9. Find the expression for the power transmission through pipe. What is the
condition for maximum transmission of power and corresponding efficiency of
transmission?
Let H = total head available at inlet of pipe
hf = Loss of head due to friction
64
Efficiency of power transmission through pipes  =
H - hf
H
Condition for maximum transmission of power through pipe hf =
efficiency = 66.67%
65
H
and maximum
3
PART – B
1. Explain the flow of viscous fluid through circular pipe and derive Hagen
Poiseuilles.
FLOW OF VISCOUS FLUID THROUGH CIRCULAR PIPE
For the flow of viscous fluid through circular pipe, the velocity distribution
across a section, the ratio of maximum velocity to average velocity, the shear stress
distribution and drop of pressure for a given length is to be determined. The flow
through the circular pipe will be viscous or laminar, if the Reynolds number (R e*) is
less than 2000. The expression for Reynold number is given by
Re 
where
VD

 = Density of fluid flowing through pipe
V = Average velocity of fluid
D = Diameter of pipe and
 = Viscosity of fluid.
Consider a horizontal pipe of radius R. The viscous fluid is flowing from left
to right in the pipe. Consider a fluid element of radius r, sliding in a cylindrical fluid
element of radius (r + dr). Let the length of fluid element be x. If ‘p’ is the intensity
of pressure on the face AB, then the intensity of pressure on face CD will be
p 

 p  x x  . Then the forces acting on the fluid element are:


1. The pressure force, p  r2 on face AB.
p  2

2. The pressure force,  p 
x  r on face CD.
x


3. The shear force,   2rx on the surface of fluid element. As there is no
acceleration, hence the summation of all forces in the direction of flow must
be zero i.e.
or
p  2

pr 2   p 
x  r    2r  x  0
x


p
 xr 2    2r  x  0
x
66

or

p
.r  2  0
x
p r

x 2
....(1)
p
across a section is
x
constant. Hence shear stress distribution across a section is linear.
The shear stress  across a section varies with ‘r’ as
(i) Velocity Distribution. To obtain the velocity distribution across a section, the
du
y is substituted in equation (1).
value of shear stress   
dy
du
y , is measured from the pipe wall. Hence
dy
y = R – r and dy = -dr
du
du

 
dr
dr
But in the relation  = 

Substituting this value in (1), we get

du
p r
du 1 p

or

r
dr
x 2
dr 2 x
Integrating this above equation w.r.t., ‘r’, we get
u
1 p 2
r C
4 x
….(2)
where C is the constant of integration and its value is obtained from the boundary
condition that at r = R, u = O.


1 p 2
R C
4 x
1 p 2
C
R
4 x
O
Substituting this value of C in equation (2), we get
67
1 p 2 1 p 2
r 
R
4 x
4 x
1 p 2 2
R  r 
=4 x 
u
....(3)
p
and R are constant, which means the velocity,
x
u varies with the square of r. Thus equation (3) is a equation of parabola. This shows
that the velocity distribution across the section of a pipe is parabolic. This velocity
distribution is shown in figure (b).
In equation (3), values of ,
(ii) Ratio of Maximum Velocity to Average Velocity
The velocity is maximum, when r = 0 in equation (3). Thus maximum velocity, Umax
is obtained as
Umax  
1 p 2
R
4 x
….(4)
The average velocity, u , is obtained by dividing the discharge of the fluid
across the section by the area of the pipe (R2). The discharge (Q) across the section is
obtained by considering the flow through a circular ring element of radius r and
thickness dr as shown in figure (b). The fluid flowing per second through this
elementary ring
dQ = velocity at a radius r  area of ring element
= u  2r dr
1 p 2 2
R  r   2r dr

4 x 

R
R
0
0
Q   dQ   
1 p 2 2
R  r  2r dr
4 x


68

R
1  p 
 2  R2  r 2 rdr


0
4  x 

R
1  p 
 2  R2r  r 3 dr


0
4  x 




R
 R 2r 2 r 4 
R4 R4 
1  p 
1  p 


2




2






 x 
4  x 
2
4
4

2
4 



0


1  p 
R4
  p  4

2



R


4  x 
4 8  x 
 Average velocity,
or
  p  4
R
Q
8  x 
u

Area
R2
  p  2
u
R
8  x 
Dividing equation (4) by equation (5),
1 p 2
R
Umax
4 x

 2.0
1  p  2
u

R
8  x 
 Ratio of maximum velocity to average velocity = 2.0.
(iii) Drop of pressure for a given Length (L) of a pipe
From equation (5), we have
u
1  p  2
R
8  x 
 -p  8u
or 
 2
 x  R
Integrating the above equation w.r.t. x, we get
69
...(5)
8u
dx
R2
8u
8u
- p1  p2   2  x1  x 2  or p1  p2   2  x 2  x1 
R
R
8u
 2 L  x 2  x1  L from figure
R
8uL
D


 R= 
2
2

D / 2 
1
1
2
2
  dp  

or

32uL
, where p1  p2 is the drop of pressure.
D2
p  p2
Loss of pressure head  1
g
p1  p2  
p1  p2
32uL
 hf 
g
gD2

....(6)
Equation (6) is called Hagen Poiseuille Formula.
2. A crude oil of viscosity 0.97 poise and relative density 0.9 is flowing through a
horizontal circular pipe of diameter 100 mm and of length 10 m. Calculate the
difference of pressure at the two ends of the pipe, if 100 kg of the oil is collected
in a tank in 30 seconds.
0.97
 0.097 Ns/m2
10
Sol. Given:
  0.97 poise =
Relative Density
 0, or Density
Dia. Of pipe,
= 0.9
= 0.9  1000  900kg/ m3
D = 100 mm = 0.1 m
L = 10 m
M = 100 kg
t = 30 seconds
Mass of oil collected,
in time,
Calculate difference of pressure or (p1 – p2).
The difference of pressure (p1 – p2) for viscous or laminar flow is given by
32uL
D2
100

kg / s
30
p1  p2 
Now, mass of oil/sec
70
where u  averagevelocity 
Q
Area
 0  Q  900  Q

0  900
100
 900  Q
30
100
1
Q

 0.0037 m3 / s.
30 900
Q
.0037 .0037
u


 0.471 m/s.
2
 2

Area
D
.1
 
4
4



For laminar or viscous flow, the Reynolds number (Re) is less than 2000. Let us
calculate the Reynolds number for this problem.
Re  
Reynolds number,
VD

where   0  900,
V = u  0.471, D  0.1 m, = 0.097
.471 0.1
Re  900 
 436.91

0.097
As Reynolds number is less than 2000, the flow is laminar.
32uL 32  0.097  .471 10

p1  p2 

N / m2
2
D2
.1
= 1462.28 N/m2 = 1462.28  10-4 N/cm2 = 0.1462 N/cm2.
3. A fluid of viscosity 0.7 Ns/m2 and specific gravity 1.3 is flowing through a
circular pipe of diameter 100 mm. The maximum shear stress at the pipe wall is
given as 196.2 N/m2, find (a) the pressure gradient (b) the average velocity and (c)
Reynold number of the flow.
Sol. Given:
  0.7
Ns
m2
Sp. gr. = 1.3
 Density
= 1.3  1000  1300 kg/m3
Dia. Of pipe,
D = 100 mm = 0.1 m
Shear stress,
0 = 196.2 N/m2
Find
(i)
(ii)
(iii)
dp
dx
Average velocity, u
Reynold number, Re
Pressure gradient,
71
dp
dx
The maximum shear stress (0) is given by
(i) Pressure gradient,
p R
p D
p 0.1
or 196.2 = -    
x 2
x 4
x 4
p
196.2  4

 7848 N/m2 per m
x
0.1
0  

 Pressure Gradient
= -7848N/m2 per m.
(ii) Average velocity, u
1
1  1 p 2 
u  Umax   
R
2
2  4 x 
1  p  2


R
8  x 
1
2

  7848   .05 
8  0.7
 3.50 m/s.



Umax  
1 p 2 
R 
8 x 



D 1

  .05
2 2

R=
(iii) Reynold number, Re
uD uD uD


v
/

3.50  0.1
= 1300 
 650.00.
0.7
Re 
4. Calculate: (a) the pressure gradient along flow, (b) the average velocity, and (c)
the discharge for an oil of viscosity 0.02 Ns/m2 flowing between two stationary
parallel plates 1 m wide maintained 10 mm apart. The velocity midway between
the plates is 2 m/s.
(Delhi University, 1982)
Sol. Given:
  .02 Ns/m2
Viscosity,
Width,
b=1m
Distance between plates, t = 10 mm = .01 m
Velocity midway between the plates, Umax = 2 m/s.
72
 dp 
(i) Pressure gradient 

 dx 
1 dp 2
1  dp 
2
t or 2.0 = .01


8 dx
8  .02  dx 
dp
2.0  8  .02

 3200 N/m2 per m.
dx
.01 .01
Umax  

(ii) Average velocity ( u )
Umax 3

2
u
2 Umax 2  2
u

 1.33 m/s.
3
3
Using equation

(iii) Discharge (Q)
= Area of flow  u  b  t  u  1 .01 1.33  .0133 m3 / s.
5. Derive Darcy-Weisbach equation.
Expression for loss of head due to Friction in pipes
Consider a uniform horizontal pipe, having steady flow as shown in figure.
Let 1-1 and 2-2 are two sections of pipe.
Let
and
p1 = Pressure intensity at section 1-1,
V1 = Velocity of flow at section 1-1,
L = length of the pipe between sections 1-1 and 2-2
d = diameter of pipe,
f’ = frictional resistance per unit wetted area per unit velocity,
hf = loss of head due to friction
p2, V2 = are values of pressure intensity and velocity at section 2-2.
Applying Bernoulli’s equations between sections 1-1 and 2-2,
Total head at 1-1 = Total head at 2-2 + loss of head due to friction between 1-1
and 2-2
73
p1 V12
p2 V22

 z1 

 z2  hf
g 2g
g 2g
or
But

z1 = z2 pipe is horizontal
V1 = V2 as dia. Of pipe is same at 1-1 and 2-2
p1 p2

 hf
g g
or hf 
p1 p2

g g
....(i)
But hf is the head lost due to friction and hence intensity of pressure will be
reduced in the direction of flow by frictional resistance.
Now frictional resistance = frictional resistance per unit wetted area per unit
velocity  wetted area  velocity2
or
F1 = f’  dL  V2
= f’  P  L  V2
[ wetted area = d  L velocity = V = V1 = V2]
[ d = Perimeter = P] ….(ii)
The forces acting on the fluid between section 1-1 and 2-2 are:
1. Pressure force at section 1-1 = p1  A
where A = Area of pipe
2. Pressure force at section 2-2 = p2  A
3. Frictional force F1 as shown in figure.
Resolving all forces in the horizontal direction, we have
p1A – p2A – F1 = 0
…(1)
or
p1  p2  A  F1  f ' P  L  V2
or
p1  p2 
f ' P  L  V 2
A
But from equation (i),
p1 – p2 = ghf
Equation the value of (p1 – p2), we get
ghf 
f ' P  L  V 2
A
74
[
From (ii), F1  f 'PLV2 ]
hf 
or
In equation (iii),
f' P
  L  V2
g A
...(iii)
P Wetted perimeter
d
4



 2 d
A
Area
d
4

f' 4
f ' 4LV 2
2
hf 
 L  V 

g d
g
d
Putting
f' f
 ,
g 2
Equation (iv), becomes as
...(iv)
where f is known as co-efficient of friction.
hf 
4.f LV 2 4f. L. V 2
.

2g
d
d  2g
...(2)
Equation (2) is known as Darcy-Weisbach equation. This equation is commonly used
for finding loss of head due to friction in pipes.
Sometimes equation (2) is written as
hf 
f. L. V 2
d  2g
….(2A)
Then f is known as friction factor.
6. Find the diameter of a pipe of length 2000 m when the rate of flow of water
through the pipe is 200 litres/s and the head lost due to friction is 4 m. Take the
value of C = 50 in Chezy’s formulae.
Sol. Given:
Length of pipe,
Discharge,
Head lost due to friction
Value of Chezy’s constant,
Let the diameter of pipe
L = 2000 m
Q = 200 litre/s = 0.2 m3/s
hf = 4 m
C = 50
=d
Velocity of flow,
V=
Hydraulic mean depth,
Discharge
Q
0.2 0.2  4



 2  2
Area
d2
d
d
4
4
d
m=
4
75
Loss of head per unit length,
i
hf
4

 .002
L 2000
Chezy’s formula is given by equation as V = C mi
Substituting the values of V, m, I and C, we get
0.2  4
d
 50
 .002 or
2
d
4
d
0.2  4 .00509
 .002  2

4
d  50
d2
Squaring both sides,
d
.005092 .0000259
4  .0000259
 .002 

or d5 
 0.0518
4
4
4
d
d
.002

d  5 0.0518  .0518 
1/ 5
 0.553 m  553 mm.
7. An oil of sp.gr. 0.7 is flowing through a pipe of diameter 300 mm at the rate of
500 litres/s. Find the head lost due to friction and power required to maintain the
flow for a length of 1000 m. Take v = .29 stokes.
Sol. Given :
Sp.gr. of oil,
Dia. of pipe,
Discharge,
Length of pipe,
Velocity,
 Reynolds number,
 Co-efficient of friction,
S = 0.7
d = 300 mm = 0.3 m
Q = 500 litres/s = 0.5 m3/s.
L = 1000 m
Q
0.5
0.5  4
V


 7.073 m/s
Area  d2   0.32
4
V  d 7.073  0.3
4
Re 

 7.316  10 
4
v
0.29  10
f
.079
Re
1
4

Head lost due to friction,
76
0.79
 7.316  104 
1
4
 .0048
hf 
4  f  L  V 2 4  .0048  1000  7.0732

 163.18 m
d  2g
0.3  2  9.81

Power required
g.Qhf
kW
1000
where  = density of oil = 0.7  1000  700 kg/m3

 Power required
700  9.81 0.5  163.18
 560.28 kW.
1000
8. Find the loss of head when a pipe of diameter 200 mm is suddenly enlarged to a
diameter of 400 mm. The rate of flow of water through the pipe is 250 litres/s.
Sol. Given:
Dia. of smaller pipe,
 Area,
Dia. of large pipe,
 Area,
Discharge,
Velocity,
Velocity,
D1 = 200 mm = 0.20 m


2
A1  D12  .2   0.03141 m2
4
4
D2 = 400 mm = 0.4 m

2
A 2    0.4   0.12564 m2
4
Q = 250 litres/s = 0.25 m3/s
Q
0.25
V1 

 7.96 m / s
A1 .03141
Q
0.25
V2 

 1.99 m / s
A1 .12564
Loss of head due to enlargement is given by equation as
he
 V  V2 
 1
2g
2
7.96  1.99 

2g
2
 1.816 m of water.
9. At a sudden enlargement of a water main from 240 mm to 480 mm to 480 mm
diameter, the hydraulic gradient rises by 10 mm. Estimate the rate of flow.
(A.M.I.E., Summer, 1977)
Sol. Given:
77
Dia. of smaller pipe,
D1 = 240 mm = 0.24 m


2
A1  D12  .24 
4
4
D2 = 480 mm = 0.48 m

2
A 2  .48 
4
 Area,
Dia. of large pipe,
 Area,
Rise of hydraulic gradient*, i.e.,


p2   p1
10
1

m
 z2      z1   10mm 
g   g
1000 100


Let the rate of flow = Q
Applying Bernoulli’s equation to both sections, i.e., smaller pipe section, and
large pipe section.
p1 V12
p
V2

 z1  2  2  z2  Head loss due to enlargement
g 2g
g 2g
But head loss due to enlargement,
he
 V  V2 
 1
2
2g
From continuity equation, we have A1V1 = A2V2

 2
2
A 2 V2 4 D2  V2  D2 
 .48 
V1 

    V2  
 V2  2V22  4V2


A1
 .24 
 D1 
D12
4
Substituting this value in (ii), we get
he
 4V2  V2 

2g
2
 3V2 

2g
2

9V22
2g
Now substituting the value of he and V1 in equation (i),
p1  4V2 
p
V2
9V 2

 z1  2  2  z2  2
g
2g
g 2g
2g
2
78
...(i)
 p

16V22 V22 9V22  p2


   z2    1  z1 
2g
2g
2g
 g
  g

or
p
 p

1
But hydraulic gradient rise =  2  z2    1  z1  
 g
  g
 100
16V22 V22 9V22
6V22
1
1



or

2g
2g
2g
100
2g
100

2  9.81
 0.1808  0.181 m/s
6  100

V2 
 Discharge,
Q = A2  V2


2
 D22  V1  .48   .181  0.03275 m3 / s.
4
4
= 32.75 litres/s.
10. A horizontal pipe of diameter 500 mm is suddenly contracted to a diameter of
250 mm. The pressure intensities in the large and smaller pipe is given as 13.374
N/cm2 and 11.772 N/cm2 respectively. Find the loss of head due to contraction if
Ce = 0.62. Also determine the rate of flow of water.
Solution:
Given:
Dia. of large pipe,
D1 = 500 mm = 0.5 m
Area,
A1 

 0.5 2  0.1963m2
Dia. of smaller pipe,
4
D2 = 250 mm = 0.25 m
 Area,
A2 

.25 2  0.04908m2
4
Pressure in large pipe,
p1 = 13.734 N/cm2 = 13.734 x 104 N/m2
Pressure in smaller pipe, p2 = 11.772 N/cm2 = 11.772 x 104 N/m2
Cc = 0.62
Head lost due to contraction
79
2
2
2
1

V22  1
  0.375 V2

1.0


1.0



2 g  0.62
2g
 Cc

From continuity equation, we have A1V1 = A2V2
or
V2
 2
2g

2
2
D 2  V2 
A2 V2 4 2
D2 
0.25 
V2

2
V1 


 V2 
 V  

A1
4
 0.50 
 D1 
D12
4
Applying Bernoulli’s equation before and after contraction,
p1 V12
p
V2

 z  2  2  z2  hc
 g 2g
 g 2g
But
z1 = z2
(pipe is horizontal)

But
p1 V12 p2 V22



 hc
 g 2g  g 2g
V2
V
hc  0.375 2 and V1  2
2g
4
Substituting these values in the above equation, we get
13.734  104  V2 / 4  11.772  10 4 V22
V2



 0.375 2
9.81  1000
2g
1000  9.81 2 g
2g
V22
V2
14.0 
 12.0  1.375 2
16  2 g
2g
2
2
V
1 V2
V2
14  12  1.375 2 
 1.3125 2
2 g 16 2 g
2g
2
or
or
or
2.0  1.3125 
V22
2.0  2  9.81
or V2 
 5.467 m / s.
2g
1.3125
(i) Loss of head due to contraction,
V22 0.375   5.467 
hc  0.375

 0.571m.
2g
2  9.81
2
(ii) Rate of flow of water, Q = A2V2 = 0.04908 x 5.467 = 0.2683 m3/s = 268.3 lit/s.
11. Water is flowing through a horizontal pipe of diameter 200 mm at a velocity of
3 m/s. A circular solid plate of diameter 150 mm is placed in the pipe to obstruct
the flow. Find the loss of head due to obstruction in the pipe if Cc = 0.62.
Solution:
Given:
80
Dia, of pipe,
Velocity,
D = 200 mm = 0.20 m
V = 30. m/s
A
Area of pipe,

4
D2 

 0.2 2  0.03141m2
Dia. of obstruction
4
d = 150 mm = 0.15 m
 Area of obstruction,
a

.15 2  0.01767 m2
4
Cc = 0.62
The head lost due to obstruction is given by equation (11.10) as

V 2 2g 
A

 10 

 Cc  A  a 


3 3 
.03141

 1.0 

2  9.81  0.62 .03141  .01767 


2
2
9
 3.687  1.02  3.311m.
2  9.81
12. A horizontal pipe line 40 m long is connected to a water tank at one end and
discharges freely into the atmosphere at the other end. For the first 25 m of its
length from the tank, the pipe is 150 mm diameter and its diameter is suddenly
enlarged to 300 mm. The height of water level in the tank is 8 m above the centre
of the pipe. Considering all losses of head which occur, determine the rate of
flow. Take f = .01 for both sections of the pipe. (Osmania University, 1992;
A.M.I.E., Summer, 1978)
Solution:
Given:
Total length of pipe,
Length of 1st pipe,
Dia. of 1st pipe,
Length of 2nd pipe,
Dia. of 2nd pipe,
Height of water,
Co-efficient of friction,
L = 40 m
L1 = 25 m
d1 = 150 mm = 0.15 m
L2 = 40 – 25 = 15 m
d2 = 300 mm = 0.3 m
H=8m
f = 0.01
81
Applying Bernoulli’s theorem to the free surface of water in the tank and
outlet of pipe as shown in Fig. and taking reference line passing through the centre
of pipe.
or
where
p2 V22
008 

 0  all losses
 g 2g
V2
8.0  0  2  hi  h f 1  he  h f 2
2g
V2
hi = loss of head at entrance = 0.5 1
2g
......  i 
4  f  L1  V12
hf1 = head lost due to friction in pipe 1 
d1  2 g
he = loss head due to sudden enlargement 
 V1  V2 
2
2g
4  f  l2  V22
hf2 = Head lost due to friction in pipe 2 
d2  2 g
But from continuity equation, we have A1V1=A2V2


2
d 2  V2   2
A2 V2 4 2
d2
0.3 

V1 

    V2  
  V2  4 V2
 2
A1
 .15 
 d1 
d1
4
Substituting the value of V1 in different head losses, we have
82
0.5V12 0.5   4V2 
8V 2
hi 

 2
2g
2g
2g
2
hf1 

4  0.01  25   4V22 
0.15  2  g
s
4  .01  25  16 V22
V2

 106.67 2
0.15
2g
2g
he 
 V1  V2 
hf 2 
2

2g
 4 V2  V2 
2g
2

9V22
2g
4  .01  15  V22 4  .01  15 V22
V2


 2.0  2
0.  2 g
0.3
2g
2g
Substituting the values of these losses in equation (i), we get
8.0 


V2 
V22 8V22
V 2 9V 2
V2

 106.67 2  2  2  2
2g 2g
2g 2g
2g
V22
V2
1  8  106.67  9  2  126.67 2
2g
2
8.0  2  g
8.0  2  9.81

 1.2391  1.113m / s
126.67
126.67
 Rate of flow,
Q  A2  V2 

4
 0.3 2  1.113  0.07867 m3 / s  78.67 litres / s.
13. The difference in water surface levels in two tanks, which are connected by
three pipes in series of lengths 300m, 170m and 210m and of diameters 300mm,
200mm and 400 mm respectively, is 12m. Determine the rate of flow of water if coefficient of friction are. 005,0052 and .0048 respectively, considering: (i) minor
losses also (ii) neglecting minor losses.
(Delhi University, 1987).
Sol. Given:
83
Difference of water level,
Length of pipe 1,
Length of pipe 2,
Length of pipe 3,
H=12m
L1=300m and dia., d1 = 300 mm = 0.3 m
L2=170m and dia., d2 = 200 mm = 0.2 m
L3=210m and dia., d3 = 400 mm = 0.4 m
Also,
f1=.0025, f2=.0052 and f3 = .0048
(i) Considering Minor Losses, Let V1, V2 and V3 are the velocities in the 1st, 2nd and
3rd pipe respectively.
From continuity, we have A1V1=A2V2=A3V3

 2
2
d
A1V1 4 1
d2
0.3 
V2 

V1  12 V1  
  V1  2.25V1
 2
A2
d
.2


2
d2
4
and
A V d2
0.3 
V3  1 1  12 V1  
 V1  0.5625V1
A3
d3
 0.4 
2
Now using equation, we have
0.5V12 4f1L1V12 0.5V22 4f2 L2 V22  V2  V3  4f3L3 V32 V32
H






2g
d1  2g
2g
d2  2g
2g
d3  2g 2g
2
0.5V12 4  .005  300  V12 0.5   2.25V1


2g
0.3  2g
2g
2
Substituting V2 and V3 , 12.0 
+4  .0052  170 
 2.25V1 
0.2  2g
2


 2.25V1  .562V1 
2g
4  .0048  210   .5625V1  .5625V1 
+

0.4  2g
2g
2
or
V12
12.0=
0.5  20.0  2.53  89.505  2.847  3.189  0.316 
2g
84
2
2
V12

118.887 
2g

V1 
 Rate of flow,
12  2  9.81
 1.407 m/s
118.887
Q=Area  velocity=A 1  V1


2
d 1   V1  .3 2  1.407  0.09945 m 3 / s
4
4
= 99.45 lires/s.
=
H
4f1L1V12 4f2 L2 V22 4f3L3 V32


d1  2g d2  2g d3  2g
or
=

2
2
V12  4  .005  300 4  .0052  170   2.25  4  .0048  210  .5625  




2g 
0.3
0.2
0.4

12.0=
V12
V2
 20.0  89.505  3.189   1  112.694
2g
2g
V1 
2  9.81  12.0
 1.445m / s
112.694

2
 Discharge, Q=V1  A1  1.445   .3   0.1021 m 3 / s  102.1litres / s.
4
14. Three pipes of 400mm, 200mm and 300mm diameters have lengths of 400m,
200m and 300m respectively. They are connected in series to make a compound
pipe. The ends of this compound pipe are connected with tanks whose difference
of water levels is 16m. If co-efficient of friction for these pipes is same and equal
to 0.005, determine the discharge through the compound pipe neglecting first the
minor losses and then including them.
Sol. Given:
Difference of water levels,
Length and dia. of pipe 1,
Length and dia. of pipe 2,
Length and dia. of pipe 3,
H=16m
L1=400m and d1 = 400 mm = 0.4m
L2=200m and d2 = 200 mm = 0.2m
L3=400m and d3 = 300 mm = 0.3m
Also,
f1=f2=f3 = 0.005
85
(i) Discharge through the compound pipe first neglecting minor losses.
Let V1, V2 and V3 are the velocities in the 1st, 2nd and 3rd pipe respectively.
From continuity, we have A1V1=A2V2=A3V3

 2
2
d
A1V1 4 1
d12
0.4 

V2 

V 
V 
 V1  4V1
 2 1 d 22 1  0.2 
A2
d2
4
and
 2
2
d
A1V1 4 1
d12
0.4 

V3 

 V1  2 V1  
 V1  1.77V1
 2
A3
d3
 0.2 
d3
4
Now using equation, we have
H
4f1L1V12 4f2 L2 V22 4f3L3 V32


d1  2g d2  2g d3  2g
4  0.005  400  V12 4  0.005  200   4V1  4  0.005  300
2
16 


 1.77V1 
0.4  2  9.81
0.2  2  9.81
0.3  2  9.81
2
or
=
V12  4  0.005  400 4  0.005  200  16 4  0.005  300  3.157 




2  9.81 
0.4
0.2
0.3

16=

V1 
V12
V12
20

320

63.14

 403.14


2  9.81
2  9.81
16  2  9.81
 0.882m / s
403.14

2
 Discharge, Q=A1  V1   0.4   0.882  0.1108 m 3 / s.
4
(ii) Discharge through the compound pipe considering minor losses also.
Minor losses are:
(a) At inlet,
hi 
0.5V12
2g
86
(b) Between 1st pipe and 2nd pipe, due to contraction,
0.5V22 0.5  4V1

2g
2g
2
hc 
=


V2  4V1 
0.5  16  V12
V2
 8 1
2g
2g
(c) Between 2nd pipe and 3rd pipe, due to sudden enlargement,
he
 V  V3 
 2
2g
=  2.23  
2
2

 4V1  1.77V1 
2g
2

V3  1.77 V1 
V12
V2
 4.973 1
2g
2g
V2 1.77V1 
V2
V2
 1.77 2 1  3.1329 1
(d) At the outlet of 3 pipe, ho  3 
2g
2g
2g
2g
2
rd
The major losses are

4f1  L1  V12 4f2  L2  V22 4f3  L3  V32


d1  2g
d2  2g
2g
4  0.005  400  V12 4  0.005  200   4V1  4  0.005  300  1.77V1 



0.4  2  9.81
0.2  2  9.81
0.3  2  9.81
2
V12
2  9.81
 Sum of minor losses and major losses
=403.14 
 0.5V12
V2
V2
V2 
V2
V2

 8  1  4.973 1  3.1329 1   403.14 1  419.746 1
2g
2g
2g 
2g
2g
 2g
87
2
V12
419.746 
 16
2g


V1 
16  2  9.81
 0.864m / s
419.746
 Discharge, Q=A1V1 

2
 0.4   0.864  0.1085 m 3 / s.
4
15. Three pipes of lengths 800m, 500m and 400m and of diameters 500mm,400mm
and 300mm respectively are connected in series. These pipes are to be replaced by
a single pipe of length 1700m. Find the diameter of the single pipe.
Sol. Given:
Length of pipe 1,
Length of pipe 2,
Length of pipe 3,
L1=800m and dia., d1 = 500 mm = 0.5m
L2=500m and dia., d2 = 400 mm = 0.4m
L3=400m and dia., d3 = 300 mm = 0.3m
Length of single pipe, L=1700m
Let the diameter of equivalent single pipe = d
Applying equation,
L L1 L 2 L 3



d 5 d15 d15 d15
or
1700 800 500 400
 5  5  5  25600  48828.125  164609  239037
d5
.5
.4
.3

d5 

d= .007188 
1700
 .007118
239037
0.2
 0.3718  371.8
16. A main pipe divides into two parallel pipes which again forms one pipe as
shown in fig. The length and diameter for the first parallel pipe are 2000 m and
1.0m respectively, while the length and diameter of 2nd parallel pipe are 2000m
and 0.8m. Find the rate of flow each parallel pipe, if total flow in the main is 3.0
m3/s. The co-efficient of friction for each parallel pipe is same and equal to .005.
88
Sol. Given:
Length of pipe 1,
Dia. of pipe 1,
Length of pipe 2,
Dia. of pipe 2,
Total flow,
L1=2000 m
d1=1.0 m
L2=2000m
d2=0.8m
Q=3.0m3/s
Let
From equation,
Using equation, we have


f1=f2=f=.005
Q1=discharge in pipe 1
Q2=discharge in pipe 2
Q=Q1+Q3=3.0
4f1L1V12 4f2 L2 V22

d1  2g
d2  2g
4  0.005  2000  V1 4  .005  2000  V22

1.0  2  9.81
0.8  2  9.81
or
V12 V22
V2

or V12  2
1.0 0.8
0.8

V1 
V2
V
 2 .......(ii)
0.8 .894
Q1 
Now
 2

V
d1  V1  (1) 2  2
4
4
.894


V1 
V2 
.894 
 2


d 2  V2  (.8) 2  V2   .64  V2
4
4
4
Substituting the value of Q1 and Q2 in equation (i), we get
Q2 
and
 V2


  .64 V2  3.0 or 0.8785 V2  0.5026 V2  3.0
4 0.894 4
or
V2 .8785  .5026  3.0 or V=
Substituting this value in equation (ii)
89
3.0
 2.17m / s
1.3811
V1 
V2
2.17

 2.427 m/s
.894 0.894
 2

d1  V1  12  2.427  1.906m 3 / s.
4
4
Hence
Q1 

Q1  Q  Q1  3.0  1.906  1.094m 3 / s.
17. A pipe line of 0.6 m diameter is 1.5 km long. To increase the discharge, another
line of the same diameter is introduced parallel to the first in the second half of
the length. Neglecting minor losses, find the increase in discharge if 4f=0.04. The
head at inlet is 300mm.
(A.M.I.E., December, 1975)
Sol. Given:
Dia. of pipe line,
Length of pipe line,
D=0.6m
L=1.5 km=1.5  1000 = 1500 m
4f=0.04 or f=.01
Head at inlet,
h=300mm=0.3m
Head at outlet,
= atmospheric head =0
 Head lost,
hf=0.3m
1500
 750m
Length of another parallel pipe, L1 
2
Dia. of another parallel pipe,
d1=0.6m
Fig shows the arrangement of pipe system.
Ist Case. Discharge for a single pipe of length 1500m and dia. =0.6m.
4fLV * 2
This head lost due to friction in single pipe is hf 
d  2g
where V* = velocity of flow for single pipe
or

0.3=
V* 
4  .01  1500  V* 2
0.6  2g
0.3  0.6  2  9.81
 0.2426m / s
4  01  1500
 Discharge, Q*=V*  Area=0.2456 

2
 .6   0.0685m 3 / s....(i)
4
2nd Case. When an addition pipe of length 750m and diameter 0.6 m is connected in
90
parallel with the last half length of the pipe.
Let
Q1= discharge in 1st parallel pipe
Q2=discharge in 2nd parallel pipe
Q=Q1+Q2

where Q= discharge in main pipe when pipes are parallel.
But as the length and diameters of each parallel pipe is same

Q=Q2=Q/2
Consider the flow through pipe ABC or ABD
Head loss through ABC
= Head lost through AB + head lost through BC
…..(ii)
But head lost due to friction through ABC = 0.3 m given
Head lost due to friction through AB=
4  f  750  V2
where V= velocity of flow
0.6  2  9.81
through AB

Q
Q
40


Area   0.6 2  .36
4
 Head lost due to friction through AB

4  0.1  750  4Q 
2

  31.87 Q
0.6  2  9.81   .36 
2
Head lost due to friction through BC

4  f  L1  V12
d1  2g



4  0.1  750 
Q




0.6  2  9.81  2   .6 2 

4





Disch arg e

Q

 V1  


2
2   .6  
.6 2


4
4

4  0.1  750
16

Q2  7.969 Q2
2
2
0.6  2  9.81 4    .36
Substituting these values in equation (ii), we get
91
0.3=31.87 Q2 + 7.969 Q2=39.839 Q2
0.3
Q=
 0.0867m 3 / s
39.839

 Increase in discharge = Q-Q*=0.0867-.0685=0.0182 m3/s.
18. Two sharp ended pipes of diameters 50 mm and 100 mm respectively, each of
length 100 m are connected in parallel between two reservoirs which have a
difference of level of 10 m. If the co-efficient of friction for each pipe is (4f) 0.32,
calculate the rate of flow for each pipe and also the diameter of a single pipe 100
m long which would give the same discharge, if it were substituted for the
original two pipes. (A.M.I.E., Winter 1982, Summer 1987)
Solution:
Given:
Dia. of 1st pipe,
Length of 1st pipe,
Dia. of 2nd pipe,
Length of 2nd pipe,
Difference in level in reservoirs,
Co-efficient of friction,
Let
d1 = 50 mm = 0.05 m
L1 = 100 m
d2 =100 mm = 0.10 m
L2 = 100 m
H = 10 m
4f = 0.32
V1 = velocity of flow in pipe 1, and
V2 = velocity of flow in pipe 2.
When the pipes are connected in parallel, the loss of head will be same in both
the pipes. For the first pipe, loss of head is given as
92
4 f  L1  V12 0.32  100  V12
H

d1  2 g
0.05  2  9.81
 4f=.32 
10  32.619V12
or

V1 
10
 0.5535m / s
32.619
 Rate of flow 1st pipe, Q1  V1  A1  0.5536 
 .5536 

4

4
 d1 
2
 0.5 2  .001087 m3 / s  1.087 litres / s.
For the 2nd pipe, loss of head is given by,
10  H 

V2 
4 f  L2  V22 0.32  100  V22

d2  2 g
0.10  2  9.81
10  .10  2  9.81
 0.783 m / s
.32  100
 Rate of flow in 2nd pipe, Q2  A2  V2 

Let

4

4
d22  V2
.12  .783  .00615m3 / s  6.15 litres / s.
D = diameter of a single pipe which is substitute for the two original
pipes
L = length of single pipe = 100 m
V = velocity through pipe
The discharge through single pipe,
Q  Q1  Q2  1.087  6.15  7.237 litres / s  .007237 m3 / s

V
Q
.007237 4  .007237 .009214



m/ s
 2
Area
 D2
D2
D
4
93
Loss of head through single pipe is
.009214 
0.32  100  
2

4 f  LV
D2 

H

D  2g
D  2  9.81
or
10.0 
or
D5 

2
.32  100  .0092142 .0001384

2  9.81  D5
D5
.0001384
 .00001384
10
D  .00001384 
1/ 5
 0.1067 m  106.7 mm.
19. A pipe of diameter 300 mm and length 3500 m is used for the transmission of
power by water. The total head at the inlet of the pipe is 500 m. Find the
maximum power available at the outlet of the pipe, if the value of f = .006.
Solution:
Given:
Diameter of the pipe,
Length of the pipe,
Total head at inlet,
Co – efficient of friction
d = 300 mm = 0.30 m
L = 3500 m
H = 500 m
f = 0.006
For maximum power transmission, using equation
hf 
Now
hf 
H 500

 1.667 m
3
3
4  f  L  V 2 4  .006  3500  V 2

 14.27 V 2
d  2g
0.3  2  9.81
Equating the two values of hf we get
166.7 = 14.27 V2 or V 
94
166.7
 3.417 m / s
14.27
Discharge,
Q = V x Area
 3.417 
  2

2
d  3.417  .3   0.2415m3 / s
4
4
Head available at the end of the pipe
 H  hf  H 
H 2 H 2  500


 333.33m
3
3
3
 Maximum power available


 g  Q  head at the end of pipe
1000
kW
1000  9.81  .2415  333.33
kW  689.7 kW
1000
UNIT - IV
Dimensional analysis
Buckingham’s p theorem
Applications
95
Similarity laws and models.
Hydro turbines: definition and classifications
Pelton turbine – Francis turbine - Kaplan turbine - working principles
- velocity triangles - work done - specific speed - efficiencies performance curve for turbines.
DIMENSION ANALYSIS AND HYDRAULIC TURBINES
PART – A
1. What are units and dimension?
S.No
Quantity
Unit
DIMENSIONS
96
1.
2.
3.
4.
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
Geometric
Length
Area
Volume
Slope
Kinematic
Time
Velocity
(linear)
Velocity
(angular)
Acceleration
(linear)
Acceleration
(angular)
Discharge
Gravitationa
l
acceleration
Kinematic
velocity
Dynamic
Mass
Force
Weight
Mass density
Specific
weight
Dynamic
viscosity
Surface
tension
Elastic
modulus
Pressure
Shear
intensity
Work, energy
generally
adopted
MLT SYSTEM
FLT SYSTEM
M
M2
M3
L
L2
L3
L
L2
L3
Sec
M/sec
T
LT-1
T
LT-1
Rad / sec2
T-1
T-1
M/sec
LT-2
LT-2
Rad /sec2
T2
T2
Cum /sec
M/sec2
L3 T-1
LT-2
L3T-1
LT-2
M/sec2
L2 T-1
L2T-1
Kg
Newton
Newton
Kg /cum
Newton/cu
m
Newton
/cum
Newton/m
M
MLT-2
MLT-2
ML-3
ML-2T2
FL-1 T2
F
F
FL-4T2
FL-3
ML-1T-1
FL-3T
MT-2
FL-1
Newton/m2
ML-1T2
FL-2
Newton/m2
Newton/m2
ML-1T2
ML-1T2
FL-2
FL-2
Newton m
ML2T2
FL
97
24
25
26
Impulse
momentum
Torque
Power
Newton sec
MLT-1
FT
Newton m
Newton /sec
ML2T-2
ML2T3
FL
FLT-1
2. Differentiate between fundamental units and derived units.
The fundamental or primary units are the simplest in their form possessing a
single dimension. When the units of measurements of the primary quantities are
defined, the measurements of all other quantities can be easily obtained.
Example: Length (L), Time (T), Mass (M), Temperature ()
The derived secondary quantities possess more than one dimension, and are
expressed by a combination of dimensions.
Example: Velocity (LT-1), linear acceleration (LT-2), force (MLT2) etc.
3. What do you mean by dimensional analysis?
The process of obtaining a relation between a number of quantities by the use
of dimension is known as dimensional analysis
Consider the equation, S = ut +1/2 at2
Performing a dimensional analysis:
[L] = [LT-1] [T] + [LT-2][T2] = [L] + [L]
The above equation is a dimensionally homogeneous equation i.e. the
dimensions of every term on each side of the equation are identical.
4. What is a dimensionally homogeneous equation? Give examples?
An equation is said to be dimensionally homogeneous if the dimensions of
every term on each side of the equation are identical. Every equation representing a
physical phenomenon derived from an analytical approach will satisfy this
condition. Such equations are independent of the system of units.
Consider the equation V2 = U2 + 2as
Where, V and u are the final initial velocities of a body moving along a straight line,
a is the acceleration and s is the corresponding displacement.
98
Performing dimensional analysis.
[LT-1]2 = [LT-1]2 + [LT2] [L]
[L2T-2] = [L2T2] + [L2T2]
the above equation shows that term in the above equation has the same
dimension. Hence the above equation is a dimensionally homogeneous equation.
5. What are the two systems adopted to express derived units? (or) What are the
methods of dimensional analysis?
In dimensional analysis, if the number of variable involved in a physical
phenomenon is known, then the relation among the variables can be determined by
the following two methods.
1. Rayleigh’s method and
2. Bucking ham’s/ theorem
Rayleigh’s method is used for determining the expression for a variable of three
(M, L, T) or maximum four (M, L, T,). If the number of independent variables
becomes more than four, then it is very difficult to find the expression for the
dependent variable. But these difficulties can be overcome by Buckingham’s 
theorem.
6. Why Buckingham’s  - theorem is considered superior over the Rayleigh’s
method for dimensional analysis?
The rayleigh’s method of dimensional analysis becomes more laborious if the
variables are more than the number of fundamental dimensions (M, L, T). This
difficulty is overcome by using Buckingham’s
 - theorem in which dimensional analysis can be done for n variables. If
there are n variables (independent and dependent variables) in a physical
phenomenon and if these variables contain m fundamental dimensions (M, L, T),
then the variables are arranged into (n – m) dimension less terms ( called as  terms)
7. What do you mean by repeating variables? How are the repeating variables
99
selected for dimensional analysis?
Repeating variables those which are present in all  terms used in
Buckingham  - theorem. Number of repeating variables should be equal to number
of fundamental units.
Rules to follow for selection of repeating variable are:
1. These are to be selected taxing one from geometric characteristic [e.g. length (i),
area (A), breadth (b), depth (d), volume (v)], one from fluid characteristic [density
(T), dynamic viscosity () etc] and one from flow characteristic [velocity (v),
acceleration (a), acceleration due to gravity (g) discharge (Q), speed (N) etc]
2) T, L, V or L, d, V, or , I, v, or , d, v are the most preferable combination.
3) Dependent variable should never be chosen as repeating variables.
4) Any two repeating variables should not have the same dimensions.
5) The repeating variables together must have the same number of fundamental
dimensions.
8. What do you mean by dimensionless numbers? Name some of it.
Dimensionless numbers are those numbers which are obtained by dividing
the inertia force by viscos force or gravity force or pressure force or surface tension
force or elastic force. As this is a ratio of two forces, it will be dimensionless number.
These dimensionless numbers are also called non – dimensional parameters.
The following are the important dimension less numbers:
1.
2.
3.
4.
5.
Reynolds numbers
Froude’s number
Euler’s number
Weber’s number
Mach’s number
9. What are the different laws on which models are designed for dynamic
similarity?
Model laws or laws of similarity are the laws on which the models are
designed for dynamic similarity. Models are designed on the basis of ratio of the
force, which is dominating in the phenomenon.
100
The following are the model laws:
1.
2.
3.
4.
5.
Reynolds model law
Froude model law
Euler model law
Weber model law
Mach model law
10. Explain the terms: distorted models and undistorted models. What is the use
of distorted models?
Undistorted Models: Undistorted models are those models which are geometrically
similar to their prototypes or in other words the scale ratio for the linear dimensions
of the model and its prototype is same. The behaviour of the prototype can be easily
predicted from the results of undistorted model.
Distorted Model:
A model is said to be distorted if it is not geometrically similar to its
prototyped model. Different scale ratios for the linear dimensions are adopted. For
example, incase to rivers, harbours, reservoirs etc. two different scale ratios, one for
horizontal dimensions and other for vertical dimensions are taken. Thus the models
of revivers, harbors and reservoirs will become distorted models.
The followings are the advantages of distorted models.
1. The vertical dimensions of the model can be measured accurately.
2. The cost of the model can be reduced.
3. Turbulent flow in the model can be maintained.
11. What do you mean by model analysis?
Model analysis is the means of asserting and eliminating certain undesirable
conditions through model experiments and research that results in improvements in
the existing works. Safe and economic design and construction of new works and
knowledge on many aspects of hydraulic engineering can be achieved.
Model analysis are made for two purposes:
101
(a) to obtain information about the likely performance of the prototype, and
(b) to help in the design and to avoid costly mistakes
12. Write the drawbacks of analytical methods.
The following are the disadvantages of analytical methods used for study and
analysis of many problems in fluid mechanics.
1.
2.
3.
4.
It involve a number of approximations and assumptions and hence
application of the analytical methods are restricted.
It involve highly complicated equations which cannot be solved.
The solutions to various complex flow patterns cannot be obtained by
analytical methods alone.
It is impossible or impracticable in some cases to make a satisfactory and
complete mathematical analysis of the problems
13. “Models are the only resources to the nearest approach to the solution of some
hydraulic problems. Justify this statement?
In a model, there is ample scope to try several alternatives designs before
adopting a final one. These trails cost less. A model study provides not only
qualitative but also quantitative indication of the characteristics of the prototype. A
hydraulic model offers itself as a powerful design tool which establishes a valid
system from the observations on which the performance of the prototype could be
inferred. Models are very useful in studying the relative merits of alternative designs
Hence the statement “Models are the only resource to the nearest approach to
the solution of some hydraulic problems “is justified.
14. What are the various fields where models have great application?
The following are some examples where model studies have been of great value.
1. Dams: The design of every dam and all its connected works like spillways,
penstocks and gates are studied in order to get detailed information on the
flow of water and its effect on the structure. A model study can be helpful in
deciding locality and site of the dam.
102
2. Rivers and Harbours : Model analysis is devoted to the dredging of rivers
straightening of channels, protection of banks and bottoms from erosion,
various forms of river control and improvement.
Model design of harbour is necessary to duplicate the natural tidal cycles and to
study the wave action of harbour.
3. Hydraulic Machines : Models tests are useful to obtain performance data for
hydraulic turbines and other turbo machines and centrifugal pumps.
4. Structures : Many structural tests like deflection tests and destructive and
non-destructive testing of structures can be performed on models.
5. Ships : Models are helpful in investigating drag forces, and wake patterns of
naval vessels.
6. Seepage problems : Model studies of seepage flow is done to find out the
uplift pressures on hydraulic structures.
16. What you mean by hydraulic similitude?
The observations made on the performance of the model are useful to predict
the performance of the prototype. Hence it is very necessary that the model should
represent the prototype in every respect i.e the model should represent the prototype
should have similar properties. The similarity between a prototype and its model is
called similitude.
For absolute similitude between a model and the prototype the following
types of similarities should exist.
a) Geometric similarity
b) Kinematics similarity and
c) Dynamic similarity
17. What are the demerits of distorted models?
The following are the demerits of distorted models:
103
i)
ii)
iii)
Due to unequal horizontal and vertical scales the pressure and velocity
distribution are not truly reproduced in the model.
The wave pattern in the model will be different from that in the prototype
due to depth distortion.
Slopes, bends and earth cuts are not truly reproduced.
18. What do you mean by scale effect?
This is a defect which occurs in certain models due to which the computed
properties of the prototype from model experiments deviate much from the actual
properties of the prototype.
For example, a model cannot match with prototype if it large depths, high
velocities, surface tension factor, flow conditions and force. Here the models do not
have exact properties with prototype. Hence, the scale effect occurs.
19. What are the different types of forces acting in moving fluid?
Types of Forces Acting in Moving Fluid :
For the fluid flow problems, the forces acting on fluid mass may be any one, or a
combination of several of the following forces :
1.
2.
3.
4.
5.
6.
Inertia Force Fi.
Viscous force Fv.
Gravity force Fg.
Pressure force Fp.
Surface tension force,
Fs Elastic force Fe.
1. Inertia Force (F1) :
It is equal to the product of mass and acceleration of the flowing fluid and
acts in the direction opposite to the direction of acceleration. It is always existing
in the fluid flow problems.
2. Viscous Force (Fv) :
104
It is equal to the product of shear stress () due to viscosity and surface area of
the flow. It is present in fluid flow problems where viscosity is having an important
role to play.
3. Gravity Force (Fg) :
It is equal to the product of mass and acceleration due to gravity of the flowing
fluid. It is present in case of open surface flow.
4. Pressure Force (Fp) :
It is equal to the product of pressure intensity and cross sectional area of the
flowing fluid. It is present in case pipe flow.
5. Surface Tension Force(Fs) :
It is equal to the product of surface tension and length of surface of the flowing fluid.
6. Elastic Force (Fe) :
It is equal to the product of elastic stress and area of the flowing fluid.
For a flowing fluid, the above – mentioned forces may not always be present.
And also the forces, which are present in a fluid flow problem, are not of equal
magnitude. There are always one or two forces which dominate the other forces.
These dominating forces govern the flow of fluid.
20. What are “Unit Quantity” and “Specific Quantity”?
The rate of flow, speed, power etc., of hydraulic machines are all functions of
working head. To facilitate correlation, comparison and use of experimental data,
these quantities are usually reduced to unit heads. Each is expressed as a function of
head and its valve corresponding to a unit value of head is determined. These
reduced quantities are known as unit quantities. Eg. Unit discharge, unit speed, unit
power and unit torque etc. Thus two similar turbines having different data can be
compared by reducing the data of both turbines under unit head.
A specific quantity is obtained by reducing any quantity to a value
corresponding to unit head and some unit size. The unit size may be the inlet
diameter of runner in case of reaction turbine or least jet diameter in pelton wheels.
105
When two different turbines are to be compared, it can be done by reducing their
data to specific quantities.
21. Define the terms: Hydraulic machines, Turbines and pumps?
Hydraulic machines are those machines which convert either hydraulic
energy into mechanical energy or mechanical energy into hydraulic energy.
Turbines are the hydraulic machines which convert hydraulic energy into
mechanical energy.
Ex: Pelton turbine, Francis turbine, Kaplan turbine.
Pumps are the hydraulic machines which convert mechanical energy into
hydraulic energy.
Ex: Centrifugal pump, Reciprocating pump.
23. How will you classify the turbines?
1. According to the type of energy at inlet:
a. Impulse turbine
b. Reaction turbine
2. According to the direction of flow through runner:
a. Radial flow turbine
b. Axial flow turbine
c. Mixed flow turbine.
3. According to the head at the inlet of turbine:
a. High head turbine
b. Medium head turbine and
c. Low head turbine
4. According to the specific speed of turbine:
a. Low specific speed turbine
b. Medium specific speed turbine and
c. High specific speed turbine
106
24. Differentiate between:
(a) The impulse and reaction turbines
(b) Radial and axial flow turbines
(c) Inward and outward turbines.
(a) If the energy available at the inlet of turbine is only kinetic energy, the
turbine is known as impulse turbine. As the water flows over the vanes, the
pressure is atmospheric from inlet to outlet of turbine.
If water possesses kinetic energy as well as pressure energy, the turbine is
known as reaction turbine. As the water flows through the runner, the water is
under pressure and the pressure energy charges to kinetic energy. The runner is
completely enclosed in an air-tight casing and the runner and casing is full of water.
(b) If the water flows in the radial direction through the runner, the turbine is
called radial flow turbine.
If the water flows through the runner along the direction parallel to the axis of
rotation of the runner the turbine is called axial flow turbine.
Ex. Kaplan turbine
(c) If the water flows from outwards to inwards radially the turbine is known as
inward radial flow turbine.
If water flows radially from inwards to outwards the turbine is known as
outward radial flow turbine
25. Define the terms: speed ratio, flow ratio and jet ratio?
Speed Ratio: It is the ratio of tangential velocity of wheel at inlet (U i) to the velocity
given by 2gH is known as speed ratio.
 Speed ratio =
U1
2g  H
where H = Head on turbine
Flow ratio:
107
It is the ratio of velocity of flow at inlet (Uf1) to the velocity given by
 Flow ratio =
Vf1
2g  H
2g  H
where H = Head on turbine
Jet ratio (m): It is defined as the ratio of the pitch diameter (D) of the pelton wheel
to the diameter of the jet (d).
m
D
( = 12 for most cases)
d
26. What is the significance of specific speed in a turbine?
The significance of specific speed in a turbine are:
a. Plays an important role for selecting the type of turbine.
b. Performance of turbine can be predicted by knowing the specific speed of
a turbine.
The type of turbine for different specific speed is given below:
S.
Specific speed (MKS)
No.
1. 10 to 35
2. 35 to 60
3.
4.
60 to 300
300 to 1000
(S.I)
8.5 to 30
30 to 51
51 to 225
255 to 860
Type of turbine
Pelton wheel with single jet
Pelton wheel two or more
jets
Francis turbine
Kaplan or Propeller turbine
27. What is speciality in Francis Turbine?
The inward flow reaction turbine having radial discharge at outlet is known
as Francis Turbine. In modern Francis turbine, the water enters the number of the
turbine in the radial direction at outlet and leaves in the axial direction at the inlet of
the runner. Thus the modern Francis Turbine ia a mixed flow type turbine.
In case of Francis turbine, as the discharge is radial at outlet, the velocity of whirl at
108
outlet will be zero.
i.
Vw2 = 0
Work done per second = Q [Vw1 V1 ] and
Hydraulic efficiency, = h 
VV
i 1
gH
28. Draw the general layout of a Hydro-electric power plant.
General Layout of a Hydro-Electric Power Plant :
Figure below shows a general layout of a hydro-electric power plant which consists
of,
i.
ii.
iii.
iv.
A dam constructed across a river to store water,
Pipes of large diameters called penstocks, which carry water under
pressure from the storage reservoir to the turbines. These pipes are made
of steel or reinforced concrete.
Turbines having different types of varies fitted to the wheels.
Tail race, which is a channel which carries water away from the turbines
after the water has worked on the turbines. The surface of water in the tail
race is also known as tail race.
PART – B
1. A fluid flow field is given by
V=x2yi+y2zj-(2xyz+yz2)k.
Prove that it is a case of possible steady incompressible fluid flow. Calculate the
velocity and acceleration at the point (2,1,3).
Sol.
109
For the given fluid flow field u=x 2 y
 =y 2 z


u
 2xy
x
u
 2xy
x
w
 2xy  2yz.
z
For a case of possible steady incompressible fluid flow, the continuity equation
should be satisfied.
w=-2xyz-yz 2 
u  w


 0.
x y z
i.e.,
Substituting the values of
u 
w
, and
, we get
x y
z
u  w


 2xy  2yz  2xy  2yz  0
x y z
Hence the velocity field V=x2yi+y2zj-(2xyz+yz2)k is a possible case of fluid flow.
Substituting the value x=2, y=1 and z=3 in velocity field, we get,
V=x2yi+y2zj-(2xyz+yz2)k
=22  1i + 12  3j – (2213+132)k
4i+3j-21k.
Re sul tan t velocity = 4 2  32  ( 21)2  16  9  441  466  21.587 units.
The acceleration components ax,ay and az for steady flow are
ax  u
u
u
u

w
x
y
z
ay  u




w
x
y
z
ay  u
w
w
w

w
x
y
z
110
u  x 2 y,
u
u
u
 2xy,
 x 2and
0
x
y
z
  y 2 z,



 0,
 2yz,
 y2
x
y
z
w  2xyz  yz 2 ,
w
w
w
 2yz,
 2xz  z 2 ,
 2xy  2yz.
x
y
z
Substituting these values in acceleration components, we get acceleration at (2,1,3)
ax=x2y(2xy)+y2z(x2)-(2xyz+yz2)(0)
=2x3y2+x2y2z
=2(2)312+22  123=28+12
=16+12=28 units
ay=x2y(0)+y2z(2yz)-(2xyz+yz2)(y2)
=2y3z2+2xy3z-y3z2
=213+32 -2 2133-1332=18-12-9=-3 units
az=x2y(-2yz)+y2z(-2xz-z2)-(2xyz+yz2)(-2xy-2yz)
=-2x2y2z-2xy2z2-y2z3+[4x2y2z+2xy2z2+4xy2z2+2y2z3]
=-2  22 12 3-221=2321233+[422123+221242+421232+21233]
=-24-36-9+[48+36+72+18]
=-24-36-9+48+36+72+18=105
 Acceleration = axi+ayj+azk=28i-3j+105k.
or Resultant acceleration = 28 2   3   1052  784  9  11025
2
= 11818 =108.71 units.
2. The velocity potential function () is given by an expression
xy 3
x3 y
3

x 
 y2
3
3
(i)
(ii)
Find the velocity components in x and y direction.
Show that  represents a possible case of flow.
Sol Given:
111

xy 3
x3 y
 x3 
 y2
3
3
The partial derivatives of  w.r.t. to x and y are
y3
3x 2 y

   2x 
........(1)
x
3
3
and
3xy 2 x 3


  2y........(2)
y
3
3
(i) The velocity components u and v are given y by equation
 y3
3x 2 y  y 3

u
    2x 

 2x  x 2 y

x
3  3
3
y2
 2x  x 2 y
3

u=

 =-
 3x 2 y x 3
 3x 2 y x 3

x3
  
  2y  
  2y  xy 2   2y
y
3
3
3
3
3


(ii) The given value of , will represent a possible case of flow if it satisfies the
Laplace equation, i.e.,
2 2

0
x2 y 2
From equation (1) and (2), we have
112
Now

  y 3 / 3  2x  x 2 y
x

2
 2  2xy
x 2
and

x3
 xy 2   2y
x
3

2
 2xy  2
y 2

 2  2

  2  2xy    2xy  2   0
x 2 y 2

Laplace eqation is satisfied and hence  represent a possible case of flow.
3. (a) State Buckingham’s -theorem.
(b) The efficiency  of an depends on density , dynamic viscosity  of the fluid,
angular velocity diameter D of the rotor and the discharge Q. Express  in terms of
dimensionless parameters.
(A.M.I.E., Winter, 1976).
Sol. (a) Statement of Bucklingham’s -theorem is given in Article
(b) Given:  is a function of ,,,D and Q

=f  ,,,D,Q  or f1  , ,  , , D,Q   0
Hence total number of variables, =6.
The value of m, i.e., number of fundamental dimensions for the problem is obtained
by writing dimensions of each variables. Dimensions of each variable are

=Dimensionless, =ML-3, =ML-1T-1, =T-1, D=L and Q=L3T-1
m=3
Number of -terms
Equation (i) is written as
= n-m = 6-3=3
f1(1, 2, 3)=0
Each -term contains m+1 variables, where m is equal to three and is also repeating
variable.
Choosing D,  and  as repeating variables, we have
113
1  Da1 .b1 .c1 .
2  Da 2 .b2 .c 2 .
3  Da 3 .b3 .c 3 .Q
1  Da1 .b1 .c1 .
First -term (1).
Substituting dimensions on both sides of 1,
M0L0T0=La1.(T-1)b1.(ML-3)c1. M0L0T0
Equating the powers of M,L,T on both sides
Power of M,
Power of L,
Power of T,
0=c1+0,c1=0
0=a1+0,
a1=0
0=-b1+0,
b1=0
Substituting the values of a1,b1 and c1 in 1, we get
1=D000.=
[If a variable is dimensionless, it itself is a -term. Here the variable  is a
dimensionless and hence  is a -term. As it exists in first -term and hence 1=.
Then there is no need of equating the powers. Directly the value can be obtained.
2nd -term. 2=Da2.b2.c2.
Substituting the dimensions on both sides
M0L0T0=La2.(T-1)b2.(ML-3)c2. ML-1T-1
Equating the powers of M,L,T on both sides
Power of M,
Power of L,
Power of T,
0=c2+1,c2=-1
0=a2-3c2-1, a2=3c2+1=-3+1=-2
0=-b2-1,
b2=-1
114
Substituting the values of a2, b2 and c2 in 2,
2  D2 .b3 .1 . 
3rd -term.

D 
2
3  Da 3 .b3 .c 3 .Q
Equating the powers of M,L and T on both sides
Power of M,
Power of L,
Power of T,
0=c3,
0=a3-3c3+3,
0=-b3-1,
c3=0
a3=3c3+1-3=-3
b3=-1
Substituting the values of a3,b3 and c3 in 3,
Q
D2 
Substituting the values of 1, 2 and 3 in equation (ii)
3  D3 .1 .0 .Q 

 

Q 
Q 
f1  , 2 , 2   0 or =  2 , 2  .
 D  D  
 D  D  
4. Using Buckingham’s - theorem, show that the velocity through a circular
D  
orifice is given by V  2gH  ,
 , where H is the head causing flow, D is
 H VH 
the diameter of the orifice,  is co-efficient of viscosity,  is the mass density and g
is the acceleration due to gravity. (A.M.I.E., Winter, 1977)
Sol. Given:
V is a function of H, D, ,  and g

V=f(H,D, , , g) or f1 (V,H,D, , , g)=0
 Total number of variable, n=6
Writing dimension of each variable, we have
115
V=LT-1, H=L,D=L,=ML-1T-1, =ML-3, g=LT-2.
Thus number of fundamental dimensions, m=3
 Number of -terms = n-m6-3=3.
Equation (i) can be written as f1(1, 2, 3)=0
Each -term contains m+1 variables, where m=3 and is also equal to repeating
variables. Here V is a dependent variable and hence should not be selected as
repeating variable. Choosing H,g, as repeating variable, we get three -terms as
1  Ha1 .g b1 .c1 .V
2  Ha2 .g b2 .c2 .D
3  Ha3 .g b3 .c3 .
First -term.
1  Ha1 .g b1 .c1 .V
Substituting dimensions on both sides
M0L0T0=La1.(LT-2)b1.(MT-3)c1. (LT-1)
Equating the powers of M,L,T on both sides,
c1=0
Power of M,
0=c,
Power of L,
0=a1+b1-3c1+1, a1=-b1+3c1-1=
Power of T,
0=-2b1-1,
b1=-
Substituting the values of a1, b1 and c1 in 1,

1

1
1  H 2 .g 2 .0 .V 
Second -term.
V
.
gH
2  Ha 2 .g b2 .c 2 .D
116
1
2
1
1
1  1
2
2
Substituting the dimensions of both sides,
M0L0T0=La2.(LT-2)b2.(ML-3)c2. L
Equating the powers of M,L,T,
0=c2,
c2=0
0=a2+b2-3c2+1,a2=-b2+3c2-1=-1
0=-2b2, b2=0
Power of M,
Power of L,
Power of T,
Substituting the values of a2,b2,c2 in 2,
2  H1 .g 6 .0 .D 
Third -term.
D
.
H
3  Ha 3 .g b3 .c 3 .
Substituting the dimensions on both sides
M0L0T0=La3.(LT-2)a3.(ML-3)c3. ML-1T-1
Equating the powers of M,L,T on both sides
Power of M,
c3=-1
0=c3+1,
1
1
3
0=a3+b3-3c3-1, a3=-b3+3c3+1=  1  1  3  1  
2
2
2
1
0=-2b3-1,
b3=2
Power of L,
Power of T,
Substituting the values of a3, b3 and c3 in 3,
3  H
=
=
3 2
.g
12

H gH

.1
HV
.1 . 


H
3
2
V
HV gH
g
Multiply and Divide by V 






V

 1 
gH


Substituting the values of 1,2 and 3 in equation (ii),
117
 V D
 
f1 
, , 1
  0 or
 gH H
H

V


D
V
 
   , 1
HV 
gH
H
D  
V= 2gH  ,
.
 H VH 
or
Multiplying by a constant does not change the character of -terms.
5. A pipe of diameter 1.5m is required to transport an oil of sp. Gr. 0.90 and
viscosity 310-2 poise at the rate of 3000 litre/s. Tests were conducted on a 15cm
diameter pipe using water at 20oC. Find the velocity and rate of flow in the model.
Viscosity of water at 20oC=0.01 poise. (Delhi University, 1992)
Sol. Given:
Dis. Of prototype,
Dp=1.5m
Viscosity of fluid,
p=3  10-2 poise
Q for prototype,
Qp=3000 lit/s=3.0 m3/s
Sp.gr. of oil,
SP=0.9
 Density of oil,
P=SP  1000 =0.9  1000 =900 kg/m3
Dia. of the model,
Dm=15cm=0.15 m
o
Viscosity of water at 20 C = 0.1 poise = 1 10-2 poise or m=1 10-2 poise
Density of water or m=1000 kg/m3.
For pipe flow, the dynamic similarity will be obtained if the Reynold’s number in the
model and prototype are equal.
Hence using equation,
m Vm D m P VP DP

m
P

Vm P DP  m

.
.
VP m D m  P
=
But
VP 
For pipe linear dimension is D
900 1.5 1  10 2 900
1


 10   3.0
2
1000 0.15 3  10 1000
3
Rate of flow in prototype
3.0
3.0
3.0  4



 1.697 m/s


2
2
Area of prototype

2.25
D
1.5


 P
4
4
118

Vm  3.0  VP  3.0  1.697  5.091m / s.
Rate of flow through model, Q m  A m  Vm 


2
 Dm   Vm   0.15 2  5.091m 3 / s
4
4
0.0899m 3 / s  0.0899  1000lit / s  89.9lit / s.
6. A ship 300m long moves in sea- water, whose density is 1030 kg/m3, A1:100
model of this ship is to be tested in a wind tunnel around the model is 30m/s and
the resistance of the model is 60N. Determine the velocity of ship in sea- water
and also the resistance of the ship in sea – water. The density of air is given as 1.24
kg/m3. Take the kinematic viscosity of sea – water and air as 0.012 strokes and
0.018 strokes respectively.
Sol. Given:
For prototype,
Length,
Fluid
Density of water
Kinematic viscosity,
Let velocity of ship
Resistance
LP=300m
= Sea – water
=1030kg/m3
vP=0.018 strokes = 0.018  104 m2/s
=VP
=FP
For model
1
 300  3m
100
Velocity,
Vm=30m/s
Resistance,
Fm=60N
Density of air,
m=1.24 kg/m3
Kinematic viscosity of air, vm=0.018 stokes =.018  10-4 m2/s.
Length,
Lm 
For dynamic similarity between the prototype and its model, the Reynolds number
for both of them should be equal.

VP  LP Vm  Lm
v
L

orVP  P  m  Vm
vp
vm
v m LP
.012  10 4
3
1
1
=

 30 

 30  0.2m / s.
4
.018  10
300
1.5 100
119
Resistance
= Mass  Acceleration
V 2 V L
L    L2 V 2
t
1 t
2
2
2 2
L V  P P  LP   VP 

FP



 

Fm  L2 V 2  m m  L m   Vm 
 L3 
Then
But
P 1030

m 1.24

FP 1030  300   0.2 


 
  369.17
Fm 1.24  3   30 

FP  369.17  Fm  369.17  60  22150.2N.
2
2
1
of its prototype is 80 N/cm2.
10
The model is tested in water. Find the corresponding pressure drop in the
prototype. Take density of air = 1.24 kg/m3. The viscosity of water is 0.01 poise
while the viscosity of air is 0.00018 poise.
7. The pressure drop in an airplane model of size
Sol. Given:
Pressure drop in model,
Linear scale ratio,
Fluid in model
Viscosity of water,
Density of air,
Pm=80N/cm2=80104 N.m2
Lr=40
=Water, while in prototype = Air
m=0.01 poise
P=1.24 kg.m3
Let the corresponding pressure drop in prototype = PP.
As the problem involves pressure force and viscous force and hence for
dynamic similarity between the model and prototype, Euler’s number and Reynold’s
number should be considered. Making first of all, Reynold’s number equal, we get
from equation.
120
m Vm D m P VP DP
V

L


or m  P  P  m
m
P
VP m L m  P
But
P 1.24

m 1000
LP

0.01
 L r  40, m 
Lm
 P .00018

Vm 1.24
.01

 40 
 2.755
VP 1000
.00018
Now making Euler’s number equal, we get from equation as
Pm m
Vm
V
V
P
P
 P or m 


VP
P
m
Pm
PP
PP / P
m
p
But



Vm

1.24
 2.755 and P 
VP
m 1000
2.755=
Pm
1.24
P

 m  .0352
PP
1000
PP
Pm 2.755

 78.267
PP .0352
Pm
Pm
80
2
  78.267  or PP 
2 
2
PP
 78.267   78.267 
=0.01306 N/cm 2 .
8. Explain the following non – dimensional numbers.
a) Reynolds number
b) Froude’s number
c) Euler’s number
d) Weber’s Number
e) Mach’s number
121
(a) Reynold’s Number (Re):
It is defined as the ratio of inertia force of a flowing fluid and the viscous force
of the fluid. The expression for Reynold’s number is obtained as
Inertia force (Fi) = Mass X Acceleration of flowing fluid velocity
= p X volume X time
= p
Volume
 Velocity
Time
= p X AV X V {since volume per sec = Area X velocity = A X V}
= pAv2
Viscous force (Fv) = shear stress X Area
{Since  =  du / dy  Force =  X Area} =  X A
V
du V

L
dy L
By definition, Reynolds number
.  A Since
Re 
Fi
pAV2
pVL
V L V L




V
Fv   A

(  / p)
V
L
{ Since /p = V = Kinematic Viscosity} in case of pipe flow, the linear dimension L is
taken as diameter, d Hence Reylod’s number for pipe flow.
V d
pVd
Re 
or
V

(b) Fraoude’s Numbers (Fe) :
The Froude’s number is defined as the square root of the ratio of inertia force of a
flowing fluid to the gravity forces mathematically, it is expressed as
Fe 
Fi
Fg
where Fi
= pAV2
122
And Fg
= Force due to gravity
= Mass X Acceleration due to gravity
= p X Volume X g
= p X L3 X g
{since Volume = L3}
= p X L2 X L X G
=pXAXLXg
{Since L2 =A=Area}
F1 pAV 2
 Fe 

Fg
pALg
(c) Euler’s Number (Eu) : It is defined as the square root of the ratio of the inertia
force of a flowing fluid to the pressure force. Mathematically it is expressed as,
Eu 
F1
Fp
where Fp = Intensity of pressure X area
=pXA
and Fi = p AV2
 Eu 
pAV 2
V2

p A p/ p
V
p/ p
weber’s number (we) it is defined as the square root of the ratio of the inertia force of
flowing fluid to the surface tension force. Mathematically it is expressed as webers
numbers We
F1
Fp
where
fi = inertia force = pAV2
and Fs
= surface tension force
= surface tension force
= surface tension per unit length X length
=XL
123
pAV 2=
p  L2  V 2
 we =

{since A=L2}
 L
 L
pL  V 2
V2
=

{since A=L2 }

 / pL
V

 / pL
Mach’s Number (M) : Mach’s Number is defined as the square root of the ratio of the
inertia force of a flowing fluid to the elastic force. Mathematically, it is defined as
M
inertia force
F
 i
Elastic force Fe
where Fi = pAv2
and Fe
= Elastic force = Elastic stress X Area
= K X A = K X L2 (Since K = Elastic Stress)
pAV 2 p  L2  V 2

K  L2
K  L2
V2
V2


K/p
K/p
M 
But
K / p  C  velocity of sound in the fluid
 M=
V
C
9. Explain the parts and working principles of Pelton wheel / Turbine.
The pelton wheel of Pelton turbine is a tangential flow impulse turbine. The
water strikes the bucket along the tangent of the runner. The energy available at the
inlet of the turbine is only kinetic energy. The pressure at the inlet and outlet of the
turbine is atmosphere. This turbine is used for high heads and is named alter L.A.
Pelton, an American Engineer.
The following fig. shows the lay-out of a hydro-electric power plant which the
turbine is Pelton wheel. The water from the reservoir flows through the penstocks at
124
the outlet of which a nozzle id fitted. The nozzle increases the kinetic energy of the
water flowing through the penstock. At the outlet of the nozzle, the water comes out
in the form of a jet and strikes the buckets (vanes) of the runner. The main parts of
the Pelton turbine
1.
2.
3.
4.
Nozzle and flow regulating arrangement (spear)
Runner and buckets
Casing and
Breaking jet.
1. Nozzle and Flow Regulating Arrangement. The amount of water striking the
buckets (vanes) of the runner is controlled by providing a spear in the nozzle as
shown in fig. the spear is a conical needle which is operated either by a hand
wheel or automatically in an axial direction depending upon the size of the
unit. When the spear is pushed forward into the nozzle the amount of water
striking the runner is reduced. On the other hand, if the spear is pushed back,
the amount of water striking the runner increase.
2. Runner with Buckets. The following fig. shows the runner of a Pelton wheel.
It consists of a circular disc on the periphery of which a number of buckets
evenly spaced are fixed. The shape of the buckets is of a double hemispherical
cup or bowl. Each bucket is divided into two symmetrical part by a dividing
wall which is known as splitter.
The jet of water strikes on the splitter. The splitter divides the jet into two
equal parts and the jet comes out at the outer edge of the bucket. The buckets are
shaped in such a way that the jet gets deflected through 160 or 170. The buckets are
made of cast iron, cast steel bronze or stainless steel depending upon the head at the
inlet of the turbine.
3. Casing. A Pelton turbne with a casing. The function of the casing is to prevent
the splashing of the water and to discharge water to trail race. It also acts as a
safeguard against accidents. It is made of cast iron or fabricated steel plates.
The casing of the Pelton wheel does not perform any hydraulic function.
4. Breaking Jet. When the nozzle is completely closed by moving the spear in
the forward direction, the amount of water striking the runner reduces to
zero. But the runner due to inertia goes on revolving for a long time. To stop
the runner in a short time, a small nozzle is provided which directs the jet of
125
water on the back of the vanes. This jet of water is called breaking jet.
Velocity Triangles and Work done for Pelton Wheel: The shape of the vanes or
buckets of the Pelton wheel. The jet of water from the nozzle strikes the bucket at the
spilitter, which splits up the jet into two parts. These parts of the jet, glides over the
inner surfaces and comes out at the outer edge, fig. shows the section of the bucket at
z-z. The splitter is the inlet tip and outer edge of the bucket is the outlet tip of the
bucket . The inlet velocity triangle is drawn at the splitter and outlet velocity triangle
is drawn at the outer edge of the bucket.
Let
H = net head acting on the Pelton wheel = Hg-hf
Where Hg = Gross head and hf 
4fLV 2
D * 2g
Where D* = Dia. Of Penstock,
D = Diameter of the wheel,
N = Speed of the wheel in r.p.m
d = Diameter of the jet.
Then V1 = Velocity of jet at inlet = 2gH
u = u1 = u2 =
 DN
60
The velocity triangle at inlet will be a straight line where,
Vr1 = V1 – u1 = V1 –u
Vw1 = V1
 = 0 and  = 0
From the velocity triangle at outlet, we have Vr2 = Vr1 and Vw2 = Vr2 cos  - u2.
The force exerted by the jet of water in the direction of motion is given by
equation, as Fx = aV1[Vw1+Vw2]
As the angle  is an acute angle, + ve sign should be taken. Also this is the case of
series of Vanes, the mass of water striking is aV1 and not aVr1. in equation, ‘a’ is
the area of the jet which is given as

a = Area of jet  d 2
4
Now work done by the jet on the runner per second = Fx x u = aV1 [Vw1 + Vw2] x
Nm/s
126
Power given to the runner by the jet =
 aV1[Vw 1  Vw 2 ]  u
kW
1000
Work done /s per unit weight of water striking/s

 aV1[Vw 1  Vw 2 ]  u
kW
weight of water striking/s
=
 aV1[Vw 1  Vw 2 ]  u 1
 [Vw 1  Vw 2 ]  u
 aV1  g
g
The energy supplied to the jet at inlet is in the form of kinetic energy and is
equal to
1
mV 2
2
 K.E. of jet per second
1
= (  aV1 )  V12
2
 Hydraulic efficiency, h 
=
work done per second
K .E. of jet per second
 aV1[Vw 1  Vw 2 ]  u
1
(  aV1 )  V12
2

2[Vw 1  Vw 2 ]  u
V12
Now Vw1=V1, Vr1 =V1-u1 = (V1-u)
 Vr2 = (V1-u)
and Vw2 = Vr2 cos-u2 = Vr2 cos -u = (V1-u)cos-u
Substituting the values of Vw2 and vw2 in equation
h 

2[V1  (V1  u )cos   u ]  u
V12
2[V1  u  (V1  u )cos  ]  u 2(V1  u )[1  cos  ]u

V12
V12
The efficiency will be maximum for a given value of V1 when
d
(h )  0
du
or
or
(1+cos ) d
(2uV1  2u 2 )  0
du
V12
2V1  4u  0
or
or
d  2u(V 1  u )(1  cos  ) 

0
du 
V12

d
[2uV1  2u 2 ]  0
du
or u=
V1
2
127
 1+cos

 0

2
V1


The above equation states that hydraulic efficiency of a Pelton wheel will be
maximum when the velocity of the wheel is half the velocity of the jet of water at
inlet. The expansion for maximum efficiency will be obtained by substituting the
value of u 

V1
2
in equation.
V 
V

2 V1  1  (1  cos  )  1
2
2
Max.h  
2
=
V1
V
(1  cos  ) 1
2
2  (1  cos  )
2
V12
Points to be remembered for Pelton Wheel:
i. The velocity of the jet at inlet is given by V1=Cv 2gH where Cv = Co-efficiency of
velocity = 0.98 or 0.99 H = Net head on turbine.
ii. The velocity of wheel (u) is given by u=  2gH where speed ratio. The value of
speed ratio varies from 0.43 to 0.48.
iii. The angle of deflection of the jet through buckets is taken at 165 if no angle of
deflection is given
iv. The mean diameter or the pitch diameter d of the Pelton wheel is given by
u
 DN
60u
or D=
60
N
v. Jet ratio. It is defined as the ratio of the pitch diameter (D) of the pelton wheel to
the diameter of the jet(). It is denoted by ‘m’ and is given as
m
D
(= for most caes)
d
vi. Number of buckets on a runner is given by
Z  15 
D
=15+0.5 m
2d
vii. Number of Jets. It is obtained by dividing the total rate if flow through the
turbine by the rate of flow of water through a single jet.
128
10. A pelton wheel has a mean bucket speed of 10 metres per second with a jet of
water flowing at the rate of 700 litres/s under a head of 30 metres. The buckets
deflect the je through an angle of 160. Calculate the power given by water to the
runner and the hydraulic efficiency of the turbione. Assume co-efficient of
velocity as 0.98.
Solution, Given:
Speed of bucket,
Discharge
Angle of deflection
 Angle,
Co-efficient of velocity
u = u1 = u2 = 10 m/s
Q = 700 litres/s = 0.7 m3/S, Head of water, H = 30 m
= 160
 = 180 – 160 = 20
Cv = 0.98
The velocity of jet,

V1 = Cv 2gH  0.98 2  9.81 30  23.77 m/s
V1 = V1 – u1 = 23.77 – 10 = 13.77 m/s
Vw1 = V1 = 23.77 m/s
From outlet velocity triangle,
Vr2 = Vr1 = 13.77 m/s
Vw2 = Vr2 cos  - u2
= 13.77 cos 20 – 10.0 = 2.94 m/s
Work done by the jet per second on the runner is given by equation, as
= aV1 [Vw1 + Vw2] x u
= 1000 x 0.7 x [23.77 + 2.94] x 10
( aV1 = Q = 0.7 m3/s)
 Power given to turbine =
186970
 186.97 kW. Ans.
1000
The hydraulic efficiency of the turbine is given by equation, as
h 
2[Vw 1  Vw 2 ]  u 2[23.77  2.94]  10

 0.9454 or 94.54%. Ans.
23.77  23.77
V12
11. The penstock supplies water from a reservoir to the Pelton wheel with a gross
head 500 m. one-third of the gross head is lost in friction in the penstock. The rate
of flow of water through the nozzle fitted at the end of the penstock is 2.0 m3/s. the
129
angle of deflection of the jet is 165. Determine the power given by the water to
the runner and also hydraulic efficiency of the pelton wheel. Take speed ratio =
0.45 and Cv = 1.0.
Solution, Given:
Gross head,
Hg = 500 m
hf 
Head lost in friction,
H8 500

 166.7 ,
3
3
Net head
Discharge,
H=Hg-hf = 500 – 166.7 = 333.30 m
Q = 2.0 m3/s
Angle of deflection
= 165
 Angle
 = 180 – 165 = 15
Speed ratio
= 0.45
Co-efficient of velocity,
Cv = 1.0
Velocity of jet,
V1 = Cv 2gH = 1.0  2  9.81 333.3  80.86 m/s
Velocity of wheel,
u = Speed ratio x 2gH
u  u1  u 2  0.45  2  9.81 333.3  36.387m / s
Vr 1  V1  u1  80.86  36.387  44.473 m/s
Also
Vw1 = V1 = 80.86 m/s
From outlet velocity triangle, we have
Vr 2  Vr 1  44.473
Vr 2 cos   u2  v w 2
44.73 cos 15 = 36.387 + Vw2
Vw 2  44.473 cos 15-36.387 = 6.57 m/s
Work done by the jet on the runner per second is given by equation, as
= aV1[Vw1+Vw2] x u=Q[Vw1 + Vw2] x u
( aV1 = Q)
= 1000 x 2.0 x [80.86 + 6.57] x 36.387 = 6362630 Nm/s
 Power given by the water to the runner in KW
130

Work done per second 6362630

 6362.63 kW. Ans
1000
1000
Hydraulic efficiency of the turbine is given by equation, as
h 
2[Vw 1  Vw 1 ]  u 2[80.86  6.57]  36.387

 0.9731 or 97.31% Ans
80.86  80.86
V12
12. A Pelton wheel is working under a gross head of 400 m. the water is supplied
through penstock of diameter 1 m and length 4 km from reservoir to the Pelton
wheel. The co-efficient of friction for the penstock is given as .008. The jet of
water of diameter 150 mm strikes the buckets of the wheel and gets deflected
through an angle of 165. The relative velocity of water at outlet is reduced by 15%
due to friction between inside surface of the bucket and water. If the velocity of
the buckets is 0.45 times the jet velocity at inlet and mechanical efficiency as 85%
determine,
i.
ii.
iii.
Power given to the runner
Shaft power
Hydraulic efficiency and overall efficiency
Solution, Given:
Gross head,
Diameter of penstock
Length of penstock
Co-efficient of friction
Diameter of jet
Angle of deflection
 Angle,
Relative velocity at outlet
Velocity of bucket
Mechanical efficiency
Let
Hg = 400 m
D = 1.0 m
L = 4 km = 4 x 1000 = 4000 m
f = .008
d = 150 mm = 0.15 m
= 165
 = 180 – 165 = 15
Vr2 = 0.85 Vr1
u = 0.45 x Jet velocity
m = 85 % = 0.85
V* = Velocity of water in penstock, and
V1 = Velocity of jet of water
Using continuity equation, we have area of penstock x V* = Area of jet x V1
131

4
D2  V * 
V* 

4
d 2  V1
d2
0.152

V

 V1  .0225 V1
1
D2
1.02
 (1)
Applying Bernoulli’s equation to the free surface of water in the reservoir and outlet
of the nozzle, we get,
Hg  Head lost due to friction+
400 
V12
2g
4fLV *2 V12 4  .008  4000  V *2 V12



D  2g 2g
1.0  2  9.81
2g
Substituting the value of V* from equation (1) we get
400 
V2
4  .008  4000
 (0.0225V1 )2  1
2  9.81
2g
 .0033 V12  .051 V12 or 400=.0543 V12
V1 
400
 85.83 m/s
.0543
Now velocity bucket,
u1=0.45 V1=0.45 x 85.83 = 38.62 m/s
From inlet velocity triangle
Vr1 = V1 - u1 = 85.83 – 38.62 = 47.21 m/s
Vw1 = Vr2 cos - u2 = 40.13 cos 15-38.62
= 0.143 m/s
( u=u1  u2  38.62)
Discharge through nozzle is given as
Q = Area of jet x Velocity of jet = a x V1

4
d 2  v1 

4
(.15)2  85.83  1.516 m3 / s
Work done on the wheel per second is given by equation (1) as
  aV1[Vw 1  Vw 2 ]  u  Q[Vw 1  vw 2 ]  a
 1000  1.516[85.83  .143]  38.62  5033540 Nm/s
i. Power given to the runner in kW 
Work done per second 5033540

 5033.54 kW Ans.
1000
1000
ii. Using equation (1) for mechanical efficiency,
132
m 
Power at the shaft
S.P

Power given to the runner 5033.54
S.P  m  5033.54  0.85  5033.54  4278.5 kW. Ans.
iii. Hydraulic efficiency is given by equation is given by equation, as
h 

2[Vw 1  Vw 2 ]  u
V12
2[85.83  1.43]  38.62
 0.9014  90.14%EAns.
85.83  85.83
Overall efficiency is given by equation, as 0  m h  0.85  .9014  0.7662 or 76.62%
13. A Pelton wheel is to be designed for a head of 60 m when running at 200 r.p.m.
the Pelton wheel develops 95.6475 kW shaft power. The velocity of the buckets =
0.45 times the velocity of this overall efficiency = 0.85 and co-efficient of velocity is
equal to 0.98.
Solution, Given :
Head,
Speed
Shaft power,
Velocity of bucket
Overall efficiency,
Co-efficient of velocity
H=60m
N = 200 r.p.m.
S.P = 95.6475 kW
u = 0.45 x Velocity of jet
0=0.85
Cv=0.98
Design of Pelton wheel means to find diameter of jet (d), diameter of wheel (D),
Width and depth buckets and number of buckets on the wheel.
i. Velocity of jet,
 Bucket velocity,
But

V1 = Cv  2gH  0.98  2  9.81 60  33.62 m/s
u = u1 = u2 = 0.45 x V1 = 0.45 x 33.62 = 15.13 m/s
 DN
;
where D=Diameter of wheel
60
  D  200
60  15.13
15.13=
or D=
=1.44 m, Ans
60
  200
u
133
ii. Diameter of the jet (d)
0 = 0.85
overall efficiency
0 
But
=
S.P 95.6475 95.6475  1000


W .P  W .P 
  g Q  H
 1000 


95.6475  1000
1000  9.81 Q  60
Q=
95.6475  1000
0  1000  9.81 60

95.6475  1000
 0.1912 m2 / 2
0.85  1000  9.81 60
But the discharge, Q = area of jet x Velocity of jet
0.1912 
d

4
d 2  V1 

4
d 2  33.2
4  0.1912
 0.085 m=85mm Ans.
  33.62
iii. Size of buckets
Width of buckets
Depth of buckets
=5 x d = 5 x 85 = 425 mm.
=1.2 x d = 1.2 x 85 = 102 mm. Ans.
iv. Number of buckets on the wheel is given by equation, as
Z  15 
D
1.44
 15 
 15  8.5  23.5 say 24. Ans.
2d
2  .085
14. The three- jet Pelton turbine is required to generate 10,000 kW under a net
head of 400 m. The blade angle at outlet is 15 and the reduction in the relative
velocity while passing over the blade is 5%. If the overall efficiency of the wheel
is 80%, Cv = 0.98 and speed ratio = 0.46, then find: (i) the diameter of the jet, (ii)
total flow in m3/s and (iii) the force exerted by a jet on the buckets.
If the jet ratio is not to be less than, 10, find the speed of the wheel for a
frequency of 50 hertz/sec and the corresponding wheel diameter.
Solution:
Given:
134
No. of jets
Total power,
Net head,
Blade angle at outlet,
Relative velocity at outlet
=3
P = 10,000 kW
H = 400 m
 = 15
= 0.95 of relative velocity at inlet
Overall efficiency,
Value of
Speed ratio
Frequency,
Vr2 = 0.95 Vr1
o = 0.80
Cv = 0.98
= 0.46
f = 50 hertz/s
or
o 
Now using equation (18.6A),
P
   g Q H 


1000


Where Q = Total discharge through three nozzles and  = 1000 kg/m3

0.80 

Q
10000
1000

9.81
 Q  400 



1000


10000
 3.18m3 / s.
0.8  9.81  400
Discharge through one nozzle =
(i)
3.18
 1.06m3 / s.
3
Diameter of the jet (d).
Discharged through one nozzle = Area of one jet x Velocity
But velocity of jet,
V1  Cv  2 gH  0.98  2  9.81 400  87 m / s

1.06 

d

4
d 2  87
4  1.06
 0.125m  125mm.
  87
(ii) Total flow in m3/s
= 3.18m3/s.
135
(ii)
Force exerted by a jet on the wheel.
u1
Speed ratio
=
2 gH

u1  Speedratio  2 gH  0.46  2  9.81  400  40.75m / s.
Now
and
Vr1 = V1 – u1 = 87 -40.75 = 46.25 m/s
Vr2 = 0.95 Vr1 = 0.95 x 46.25 = 44.0 m/s
Vw1 = V1 = 87 m/s
Vw2 = Vr2 cos  - u2 = 44 x cos 15 - 40.75 ( u1=u2 = 40.75m/s)
= 1.75 m/s
Force exerted by a single jet on the buckets
=  x discharge through one jet x (Vw1 + Vw2)
= 1000 x 1.06 (87+1.75) = 94075 N = 94.075 kN.
(iv) jet ratio
 10 or
D
 10
d
 Dia. of wheel,
D = 10 x d = 10 x 0.125 = 1.25 m
But,
u1 

 DN
60
60  u1 60  40.75
N

 620r.p.m.
 D
  1.25
Now using the relation,
N
60  f
p
Where f = frequency in hertz per second,
p = pairs of poles, N = speed
60  f 60  50

 4.85
N
620
Take the next whole number i.e. 5. Hence pairs of poles are 5.
Now corresponding to five pairs of poles, the speed of the turbine will become as
given below:

p
136
60  f 60  50

 600r.p.m.
p
5
 DN
u
60
N
But
As the peripheral velocity is constant.
diameter of wheel will change.
Hence with the change of speed,
60  u 60  40.75

 1.3m
 N
  600
D 1.30
 10
 Jet ratio becomes  
d 0.12
Hence the given condition is satisfied.
D

15. A Francis turbine with an overall efficiency of 75% is required to produce
148.25 kW power. It is working under a head of 7.62 m. The peripheral velocity =
0.26
2gH and the radial velocity of flow at inlet is 0.96
2gH . The wheel runs
at 150 r.p.m. and the hydraulic losses in the turbine are 22% of the available
energy. Assuming radial discharge, determine:
(i)
(ii)
(iii)
(iv)
The guide blade angle,
The wheel vane angle at inlet,
Diameter of the wheel at inlet, and
Width of the wheel at inlet.
(AMIE, Fluid Power-Winter, 1975.
Solution:
Given:
Overall efficiency,
Power produced,
Head,
Peripheral velocity,
o = 75% = 0.75
S.P. = 148.25 k.W
H = 7.62 m
u1 = 0.26
2 gH  0.26  2  9.81  7.62  3.179 m / s
Velocity of flow at inlet,
V f 1  0.96 2 gH  0.96  2  9.81  7.62  11.738m / s.
137
Speed,
Hydraulic losses
Discharge at outlet
N = 150 r.p.m.
= 22% of available energy
= Radial
Vw2 = 0 and Vf2 = V2
Hydraulic efficiency is given as
k 
=
But


(i)
Total headat inlet  Hydraulic loss
Head at inlet
H  .22 H 0.78 H

 0.78
H
H
Vw 1u1
gH
Vw 1u1
 0.78
gH
0.78  g  H
Vw 1 
u1
0.78  9.81  7.62

 18.34m / s.
3.179
k 
The guide blade angle, i.e., . From inlet velocity triangle,
tan  
Vf 1
Vw1

11.738
 0.64
18.34

 = tan-1 0.64 = 32.619 or 32 37.
(ii) The wheel vane angle at inlet, i.e., 
tan  
Vf1
Vw 1  u1

11.738
 0.774
18.34  3.179
  tan 1 .774  37.74 or 3744.4.
(iii) Diameter of wheel at inlet (D1).
Using the relation,
u1 
 D1 N
60
138
D1 
60  u1 60  3.179

 0.4047 m.
 N
  50
(iv) Width of the wheel at inlet (Bt)
o 
But
W .P. 

o 
or
Q
S.P. 148.25

W .P. W .P.
WH   g  Q  H 1000  9.81  Q  7.62


1000
1000
1000
148.25
148.25  1000

1000  9.81  Q  7.62 1000  9.81  Q  7.62
1000
148.25  1000
148.25  1000

 2.644m3 / s
1000  9.81  7.62 o 1000  9.81 7.62  0.75
Using equation (18.21),
Q = D1 x B1 x Vf1

2.644 =  x .4047 x B1 x 11.738

B1 
2.644
 0.177 m.
  .4047  11.738
16. Francis turbine working under a head of 30 m has a wheel diameter of 1.2 m at
the entrance and 0.6 m at the exit. The vane angle at the entrance is 90 and guide
blade angle is 15. The water at the exit leaves the vanes without any tangential
velocity and the velocity of flow in the runner is constant. Neglecting the effect of
draft tube and losses in the guide and runner passages, determine the speed of
wheel in r.p.m. and vane angle at the exit. State whether the speed calculated is
synchronous or not. If not, what speed would you recommend to couple the
turbine with an alternator of 50 cycles? (Fluid Power Engg., A.M.I.E., Summer
1986)
Solution:
Given:
Head on turbine,
H = 30 m
139
Inlet dia,
Outlet dia.,
Vane angle at inlet,
Guide blade angle,
D1 = 1.2 m
D2 = 0.6 m
 = 90
 = 15
The water at exit leaves the vanes without any tangential velocity.

Vw2 = 0 and V2 = Vf2
Velocity of flow is constant in runner.

Vf1 = Vf2
(i) Speed of turbine in r.p.m.
Using equation (18.24), we have
V22 1
  Vw1u1  Vw 2u2 
2g g
H
1
 Vw1  u1 
g

Vw1  0 
1
u1  u1
g
Vf2 1
30  2  u12
2g g

Vw1  u1 


or

V2  V f 2  V f 1

....  i 
But from inlet velocity triangle, we have
tan  
Vf 1
u1
or u1 
Vf 1
tan 

Vf 1
tan15
 ii 
 3.732V f 1
Substituting the values of u1 in equation (i), we get
30 
or
30 
Vf2
2
2g


1
 3.732V f 1
g

2
or 30 
14.928V f 2
1
g
140
Vf2
1
2g

13.928V f 2
2
g

Vf2  V f 1


Vf1 
30  9.81
 4.44 m / s.
14.928
Substituting the value of Vf1 in equation (ii), we get
u1 = 3.732 x 4.44 = 16.57 m/s
But
u1 

N
 D1N
60
or 16.57 
  1.2  .N
60
16.57  60
 263.72r.p.m.
  1.2
(ii)Vane angle at exit (i.e., )
u2 
 D2  N
60

  0.6  263.72
60
 8.285m / s
V f 2  V f 1  4.44
Now from velocity triangle at outlet,
tan  

Vf 2
u2

4.44
 0.5359
8.285
 = 28.87
(iii) For a turbine, which is directly coupled to the alternator of 50 cycles the
p.N *
synchronous speed (N*) is given by f 
where f = Frequency of alternator in
60
cycles/s, p = Number of pair of poles for the alternator.
Assuming the number of pair of poles = 12, we get
50 

12  N *
60
N* 
60  50
 250r.p.m.
12
141
But the speed of turbine 263.72. And synchronous speed (N*) is equal to 250.
Hence the speed of turbine is not synchronous. The speed of turbine should be 250
r.p.m.
17. A Kaplan turbine working under a head of 20 m develops 11772 kW shaft
power. The outer diameter of the runner is 3.5 m and hub diameter 1.75 m. The
guide blade angle at the extreme edge of the runner is 35. The hydraulic and
overall efficiencies of the turbines are 88% and 84% respectively. If the velocity of
what is zero at outlet, determine:
(i)
Runner vane angles at inlet and outlet at the extreme edge of the runner,
and
Speed of the turbine.
(ii)
Solution:
Given:
Head,
Shaft power,
Outer dia. of runner
Hub diameter,
Guide blade angle,
Hydraulic efficiency,
Overall efficiency,
H = 20 m
S.P. = 11772 kW
Do = 3.5 m
Db = 1.75 m
 = 35
k = 88%
o = 84%
Velocity of whirl at outlet = 0. Using the relation, o 
where W .P. 
WP   g  Q  H

we get ,
1000
1000
0.84 
11772
  g Q H
1000
142
S.P.
W .P.
11772  1000
  = 1000 
1000  9.81 Q  20
11772  1000
Q
 71.428m3 / s.

0.84  1000  9.81  20

Q   Do2  Db2   V f 1
Using equation (18.25),
4
=
71.428 
or

Vf 1 

 3.5
4
2
 1.752   V f 1 
4
12.25  3.0625  V f 1  7.216V f 1
71.428
 9.9m / s.
7.216
From inlet velocity triangle, tan  


Vw1 
Vf 1
tan 

Vf 2
Vw1
9.9
9.9

 14.14m / s
tan 35 .7
Using the relation for hydraulic efficiency,
k 
Vw1u1
gH
0.88 

u1 

Vw2  0 
14.14  u1
9.81  20
0.88  9.81  20
 12.21m / s
14.14
(i) Runner vane angles at inlet and outlet at the extreme edge of the runner are given
as:
tan

Vf 1
Vw1  u1

9.9
 5.13
14.14  12.21
  tan 1 5.13  78.97 or 7858.
For Kaplan turbine, u1 = u2 = 12.21 m/s and Vf1 = Vf2 = 9.9 m/s
143
 From outlet velocity triangle, tan  
Vf 2
u2

9.9
 0.811
12.21
  tan .811  39.035 or 392.
 Do N
1

u1  u2 
(ii) Speed of turbine is given by
12.21 
60
  3.5  N
60
60  12.21
N
 66.63 r.p.m.
  3.50

18. The hub diameter of a Kaplan turbine, working under a head of 12 m, is 0.35
times the diameter of the runner. The turbine is running at 100 r.p.m. If the vane
angle of the extreme edge of the runner at outlet is 15 and flow ratio 0.6, find:
(i)
(ii)
(iii)
Diameter of the runner,
Diameter of the boss, and
Discharge through the runner.
The velocity of whirl at outlet is given as zero.
Solution:
Given:
Head,
Hub diameter,
Speed,
Vane angle at outlet,
H = 12 m
Db = 0.35 x D0 where D0 = Dia . or runner
N = 100 r.p.m
 = 15
Vf1

 0.6
2 gH
Flow ratio
V f 1  0.6  2 gH  0.6  2  9.81  12

 9.2m / s
From the outlet velocity triangle, Vw2 =0
Vf 2
V
tan  = f2 
u2
u2



Vf2  V f 1  9.2 
9.2
u2
9.2
u2 
 34.33m / s
tan15
tan15 
144
But for Kalpan turbine, u1 = u2 = 34.33
Now using the relation,

u1 
 Do  N
or 34.33=
  Do  100
60
60
60  34.33
Do 
 6.55m
  100
Db = 0.35 x D0 = 0.35 x 6.35 = 2.3m
Discharge through turbine is given by equation as
Q


4
 Do2  Db2   V f 1 

4
6.552  2.32   9.2
3

42.9026  5.29   9.2  271.77 m
4
s
18. A propeller reaction turbine of runner diameter 4.5 m is running at 40 r.p.m.
The guide blade angle of inlet is 145 and runner blade angle at outlet is 25 to the
direction of vane. The axial flow area of water through runner is 25 m 2. If the
runner blade angle at inlet is radial determine:
(i)
(ii)
(iii)
(iv)
Hydraulic efficiency of the turbine,
Discharge through turbine,
Power developed by the runner, and
Specific speed of the turbine
Solution:
Given
Runner diameter,
Speed,
Guide blade angle,
Runner blade angle at outlet,
Flow area,
Runner blade angle at inlet is radial
D0 = 4.5 m
N = 40 r.p.m
 = 145
 = 25
a = 25 m2
 = 90, Vr1 = Vf1 and u1 = Vw1
For Kaplan turbine, the discharge is given by the product of area of flow and
velocity of flow.
As area of flow is constant and hence Vf1=Vf2 ( Q = Area of flow x Vf1 = Area
of flow x Vf2)
145
The tangential speed of turbine at inlet, u1 
 Do N
60

  4.5  40
60
 9.42 m / s
Also u2 = u1 = 9.42 m/s.
From inlet velocity triangle
tan  180    
Vf1
u1
or tan  180-145   tan 35 
Vf1
u1
 Vf1 = u1 tan 35 = 9.42 tan 35 = 6.59
Also Vw1 = u1 = 9.42 m/s.
From outlet velocity triangle,
tan =
Vf2
u 2  Vw 2
 where Vf2  V f 1  6.59and u 2  u1  9.42 
6.59
9.42+Vw2

tan25=

Vw2  9.42 

Vw2  14.13  9.42  4.71m / s

V2  V f22  Vw22  6.59 2  4.712  43.43  22.18  8.1m / s
Using equation,
H
6.59
 14.13
tan 25
V22 1
 Vw1u1  Vw 2u2 
2g g
Here – ve sign is taken as the absolute velocity at inlet and outlet (i.e., V 1 and
V2) are in the same direction and hence change of velocity will be with a – ve sign.

H-
8.12
1
9.42  9.42  4.71  9.42 

2  9.81 9.81
H-3.344=

1
88.736-44.368   4.522 m
9.81
H=4.522+3.344=7.866m
(i) Hydraulic efficiency is given by equation as
146
h 

Vw 1u1  Vw 2u2
gH
 9.42  9.42  4.71  9.42 
9.81  7.866
 0.575  57.5%
(ii) Discharge through turbine is given by,
Q = Area of flow x Velocity of flow
= 25 x Vf1 = 25 x 6.59 = 164.75 m3/s.
(iii) Power developed by turbine

Work done per second
1000

1  Vw 1u1  Vw 2u2 
 Weight of water
g
1000

1  9.42  9.42  4.71  9.42 
    g  Q
9.81 
1000

1  9.42  9.42  4.71  9.42 
  1000  9.81  164.75  6867 kW
9.81 
1000
(iv) Specific speed is given by the relation,
Ns 

N P
H
5
4

N 6867
7.866
5
4

40  6867
7.866
5
4
40  82.867
 251.62r.p.m
13.173
The pressure at the exit of the runner of a reaction turbine is generally less
than atmospheric pressure. Thus the water at the exit of the runner cannot be
directly discharged to the tail race. A pipe of gradually increasing area is used for
discharging water from the exit of the turbine to the tail race. This pipe of gradually
increasing area is called a draft tube. One end of the draft tube is connected to the
outlet of the runner while the other end is sub-merged below the level of water is the
tail race. The draft tube, in addition to serve a passage for water discharge, has the
following two purposes also:
1. The turbine may be place above the tail race and hence turbine may be
inspected properly.
2. The kinetic energy (V22/2g) rejected at the outlet of the turbine is converted
147
into useful pressure energy.
19. A turbine is to operate under a head of 25 m at 200 r.p.m. The discharge is 9
cumec. If the efficiency is 90%, determine:
(i)
(ii)
(iii)
Specific speed of the machine,
Power generated, and
Type of turbine.
(AMIE, Winter, 1979)
Solution:
Given
Head,
Speed,
Discharge,
H = 25 m
N = 200r.p.m
Q = 9 cumec = 9 m3/s
Efficiency, 0 = 90% = 0.90 (Take the efficiency as overall )
Now using relation
o 
Power developed
P

  g Q H
Water power
1000
P  0 

  g Q H
1000
0.90  9.81  1000  9  25
 1986.5 kW
1000
(i) Specific speed of the machine (Ns)
Using equation
Ns 
N P
H
5

200  1986.5
4
25
5
4
 159.46r.p.m
(ii) Power generated
P = 1986.5 kW
(iii) As the specific speed lies between 51 and 255, the turbine is a Francis turbine
20. A Pelton turbine develops 3000 kW under a head of 300 m. The overall
efficiency of turbine is 83%. If speed ratio = 0.46, Cv = 0.98 and specific speed is
16.5, then find:
148
(i)
(ii)
Diameter of the turbine, and
Diameter of the jet
Solution:
Given:
Power,
Net head,
Overall efficiency,
Speed ratio
Value of Cv,
Specific speed*,
Using equation,
P = 3000 kW
H = 300 m
0 = 83 % or 0.83
= 0.46
= 0.98
Ns = 16.5
Ns 
N P
H
5
4
5
NH
orN  s
P
4
16.5  300

3000
5
4
 375r.p.m
The velocity (V) at the outlet of nozzle is given by,
V  Cv 2  g  H  0.98 2  9.81  300  75.1m / s
Now speed ratio

u
or u = speed ratio  2gH
2 gH
 0.46  2  9.81  300  34.95m / s
(i) Diameter of the turbine (D)
Using
u
 DN
60
or D =
60  u 60  34.95

 1.78m
 N
  375
(ii) Diameter of the jet (d)
Let
Q = Discharge through turbine in m3/s
Using the relation,
149
o 
P
   g Q H 


1000


0.83 
Q
where   g = 1000  9.81N/m3 for water
300
1000

9.81
 Q  300 



1000


3000
 1.23m3 / s
9.81  300  0.83
But discharge through a pelton turbine is given by,
Q = Area of jet x Velocity
Or
1.23 
d

4
d 2  75.1
4  1.23
 0.142m  142mm
  75.1
21. Obtain an expression for the workdone per second by water on the runner of a
pelton wheel. Hence derive an expression for maximum efficiency of the pelton
wheel giving the relationship between the jet speed and bucket speed?
Velocity Triangles and Workdone for Pelton Wheel:
Figure shows the shape of the vanes or buckets of the pelton wheel. The jet of
water from the nozzle strikes the bucket at the splitter, which splits up the jet into
two parts. These parts of the jet, glides over the inner surfaces and comes out at the
outer edge. Figure shows the section of the bucket z-z. The splitter is the inlet tip and
outer edge of the bucket is the outlet tip of the bucket. The inlet velocity triangle id
drawn at the splitter and outlet velocity triangle is drawn at the outer edge of the
bucket.
Shape of Bucket
Let H = Net head acting on the Pelton wheel
= Hg – hf
150
Where Hg = Gross head and
hf 
4fLV 2
D * 2g
Where D* = Dia of penstock
N = Speed of the wheel in r.p.m
D = Diameter of the wheel
d = diameter of the jet
Then V1 = Velocity of jet at inlet =
DN
u = u1 = u2 =
60
------ (1)
2gH
The velocity triangle at inlet will be a straight line where
Vr1 = V1 – u1 = V1 – u
Vw1 = V1
 = 0 and  = 0
From the velocity triangle at outlet, we have Vr2 = Vr1 and Vw2 = Vr2 cos  - u2
The force exerted by the jet of water in the direction of motion is given by equation
(17.19) as
Fx = pa V1  Vw1  Vw2 
-------- (2)
As the angle  is an acute angle, +ve sign should be taken. Also this is the case of
series of vanes, the mass of water striking is paV2 and not pa Vr1 . In equation (2), ‘a’ is
the area of the jet which is given as
A = Area of jet =
 2
d
4
Now work done by the jet on the runner per second
= Fx x u
151
= p a V1  Vw1  Vw2 
--------- (3)
H.P given to the runner by the jet

p a V1  Vw1  Vw 2   u
75
--------- (4)
Work done/sec per unit weight of water striking

p a V1  Vw1  Vw 2   u
Weight of water striking
p a V1  Vw1  Vw 2   u
paV1  g
1
=  Vw1  Vw 2   u
g

---------- (5)
The energy supplied to the jet at inlet is in the form of kinetic energy and is
equal to ½ mV2.
 K.E of jet per second = ½ (paV1) x V12
 Hydraulic Efficiency.
h 
Work done per second
K.E. of jet per second

paV1  Vw1  Vw 2   u

2  Vw1  Vw 2   u
1
2
paVt   V12
V12
Now Vw1 = V1, Vr1 = V1 – u1 = (V1 – u)
 Vr2 = (V1 – u)
and Vw2 = Vr2 cos  - u2 = Vr2 cos  - u = (V1 – u) cos  - u
Substituting the values of Vw1 and Vw2 in equation.
152
h 


2[V1   V1  u  cos   u]  u
V12
2  V1  u   V1  u  cos   u 
V12
2  V1  u  (1  cos )u
----------- (7)
V12
The efficiency will be maximum for a given value of V1 when d/du (h) = 0
Or
Or
d  2u  V1  u 1  cos   

0
Du 
V12

1  cos  
2
1
V


d
2uV1  2u2  0
du
d
2uV1  2u2  = 0 (since 1 + cos /V12  0)
Du 
Or 2 V1 – 4u = 0
V
Or u  1
2
Or
Equation states that hydraulic efficiency of a Pelton wheel will be maximum
when the velocity of the wheel is half the velocity of the jet of water at inlet. The
expression for maximum efficiency will be obtained by substituting the value of u =
V1/2 in equation
V 
V

2  V1  1  1  cos    1
2
2
 Max, h  
2
V1


2
V1
V
1  cos   1
2
2
2
V1
1  cos  
2
------------ (9)
153
UNIT – V
PUMPS
Pumps: definition and classifications
Centrifugal pump: classifications, working principles, velocity
triangles, specific speed, efficiency and performance curves
Reciprocating pump: classification, working principles, indicator
diagram, and work saved by air vessels and performance curves
Cavitations in pumps
Priming- slip- rotary pumps
Working principles of gear, jet and vane pumps
154
PART – A
1. Explain the main parts of a single stage centrifugal pump with sketches.
Main Parts of a Centrifugal Pump:
The following are the main parts of a centrifugal pump:
1.
2.
3.
4.
Impeller
Casing
Suction pipe with a foot valve and a strainer.
Delivery pipe.
1. Impeller: The rotating part of a centrifugal pump is called ‘impeller’. It consists
of a series of backward curved vanes. The impeller is mounted on a shaft which
is connected to the shaft of an electric motor.
2. Casing: The casing of a centrifugal pump is similar to the casing of a reaction
turbine. It is a air-tight passage surrounding the impeller and is designed in such
a way that the kinetic energy of the water discharged at the outlet of the impeller
is converted into pressure energy before water leaves the casing and enters the
delivery pipe.
3. Suction Pipe with a Foot-Valve and a Strainer:
A pipe whose one end is connected to the inlet of the pump and other end dips
into water in a sump is known as suction pipe. A foot valve which is a non-return
valve or one-way type of valve is fitted at the lower end of the suction pipe. The foot
155
valve opens in the upward direction. A strainer is also fitted at the lower end of the
suction pipe.
4. Delivery pipe: A pipe whose one end is connected to the outlet of the pump and
other end delivers the water at a required height is known as delivery pipe.
Main Parts of a centrifugal pump:
2. What are the different types of casings commonly used for centrifugal pumps?
The following three types of the castings are commonly adopted:
a.
b.
c.
Volute casing as shown in figure. (Previous Q & A)
Vortex casing as shown in figure
Casing with guide blades as shown in figure.
(a) Volute Casing:
Figure above shows the volute casing which surrounds the impeller. It is of
spiral type in which area of flow increases gradually. The increase in area of flow
decreases the velocity of flow. The decrease in velocity increases the pressure of the
water flowing through the casing. It has been observed that in case of volute casing,
the efficiency of the pump increases slightly as a large amount of energy is lost to the
formation of eddies in this type of casing.
Different types of Casing
(b) Vortex Casing:
If a circular chamber is introduced between the casing and the impeller as
shown in figure. The casing is known as Vortex Casing. By introducing the circular
chamber, the loss of energy due to the formation of eddies is reduced to a
considerable extent. Thus the efficiency of the pump is more than the efficiency
when only volute casing is provided.
(c) Casing with Guide Blades:
This casing is shown in figure in which the impeller is surrounded by a series
of guide blades mounted on a ring which is known as diffuser. The guide varies are
designed in such a way that the water from the impeller enters the guide vanes
156
without stock. Also the area of the guide vanes increases, thus reducing the velocity
of flow through guide vanes and consequently increasing the pressure of water. The
water from the guide vanes then passes through the surrounding casing which is in
most of the cases concentric with the impeller.
3. Define the terms:
Suction head, Delivery head, Static head, Manometric head?
1. Suction Head (hs): It is the vertical height of the centre line of the
centrifugal pump above the water surface in the tank or sump from which
water is to be lifted. This height is also called lift and is denoted by ‘hs’.
2. Delivery Head (hd): The vertical distance between the centre line of the
pump and the water surface in the tank to which water is delivered is
known as delivery head. This is denoted by ‘hd’.
3. Static Head (Hs): The sum of suction head and delivery head is known as
static head. This is represented by ‘Hs’ and is written as
Hs = hs + hd
4. Manometric head (Hm): The manometric head is defined as the head
against which a centrifugal pump has to work. It id denoted by ‘Hm’. It is
given by the following expressions:
(a) Hm = head imparted by the impeller to the water – loss of head in the
pump


Vw1u2
g
Vw1u2
g
- Loss of head in impeller and casing
- if loss of pump is zero
(b) Hm = Total head at outlet of the pump – Total head at the inlet of the
pump
 p0 V02
  Pi Vi2

 
 Z0    
 Zi 

 w 2g
  w 2g
Where
Po
= Pressure head at outlet of the pump = hd
w
Vo2
= Velocity head at outlet of the pump
2g
157
Vd2
= Velocity head is delivery pipe =
2g
Zo = Vertical height of the outlet of the pump from datum line and
Pi Vi2
, ,Zi = Corresponding values of pressure head, velocity head and datum head
w 2g
at the inlet of the pump.
i.e. hs,
Vs2
and Zs respectively.
2g
Vd2
(c) Hm = hs + hd + hfs + hfd +
2g
Where hs = Suction head
hd = Delivery head
hfs = Frictional head loss in suction pipe
hfd = Frictional head loss in delivery pipe and
Vd = Velocity of water in delivery pipe.
4. What are the important functions of multistage pumps?
Multistage Centrifugal Pumps:
If a centrifugal pump consists of two or more impellers, the pump is called a
multistage centrifugal pump. The impellers may be mounted on the same shaft or on
different shafts. A multistage pump is having the following two important functions:
1. To produce a high head, and
2. To discharge a large quantity of liquid.
If a high head is to be developed, the impellers are connected in series (or on
the same shaft) while for discharging large quantity of liquid, the impellers (or
pumps) are connected in parallel.
5. Define specific speed of a centrifugal pump. Derive an expression for the same.
Specific Speed of a Centrifugal Pump (Ns): The specific speed of a centrifugal
pump is defined as the speed of a geometrically similar pump which would deliver
158
one cubic meter of liquid per second against a head of one meter. It is denoted by
‘Ns’.
Expression for specific speed for a pump: The discharge, Q, for a centrifugal pump
is given by the relation
Or
Q = Area x Velocity of flow
=  D x B x Vf
Q  D x B x Vf
------ (i)
Where D = Diameter of the impeller of the pump, and
B = Width of the impeller
We know that B  D
 From equation (i), we have
Q  D2 x Vf
----- (ii)
We also know that tangential velocity is given by
u
DN
DN
60
------- (iii)
Now the tangential velocity (u) and velocity of flow (V f) are related to the
manometer head (Hm) as u  Vf  Hm ------- (iv)
Substituting the value of u in equation (iii), we get
Hm  DN
or
D
Hm
N
Substituting the valve of D in equation (ii)
H
Q  m2  Vf
N
Hm
 Hm (Since From equation (iv) Vf 
N2
H3 / 2
 m2
N
Hm3 / 2
Q=K 2
(v)
N

159
Hm )
Where K is a constant of proportionality
If Hm = 1 m and Q = 1 m3/sec. N becomes = Ns
Substituting these values in equation (v) we get
3
12
K
1 k 2  2
Ns Ns

K = Ns2
Substituting the value of K in equation (v), we get
H3 / 2
Q  N2s m2
N
2
NQ
N2s  3
Hm2
Ns 
N Q
3
Hm4
6. What is priming? Why is it necessary?
Priming of a Centrifugal Pump:
Priming of a centrifugal pump is defined as the operation in which the suction
pipe, casing of the pump and a portion of the delivery pipe upto the delivery valve is
completely filled up from outside source with the liquid to be raised by the pump
before starting the pump. Thus the air from these parts of the pump is removed and
these parts are filled with the liquid to be pumped.
The work done by the impeller per unit weight of liquid per sec is known as
the head generated by the pump. The head generated by the pump Vw2 u2 /g metre.
This equation is independent of the density of the liquid. This means that when
pump is running in air, the head generated is in terms of metre of air. But as the
density of air is very low, the generated head of air in terms of equivalent metre of
water head is negligible and hence the water may not be sucked from the pump. To
avoid this difficulty, priming is necessary.
7. Define cavitation. What are the effects of cavitation? Give the necessary
precautions against cavitation?
Cavitation: Cavitation is defined as the phenomenon of formation of vapour bubbles
of a flowing liquid in a region where the pressure of the liquid falls below its vapour
160
pressure and the sudden collapsing of these vapour bubbles in a region of higher
pressure. When the vapour bubbles collapse, a very high pressure is created. The
metallic surface, above which the liquid is flowing, is subjected to these high
pressures, which cause pitting action on the surface. Thus cavities are formed on the
metallic surface and also considerable noise and vibrations are produced.
Cavitation includes formation of vapour bubbles of the flowing liquid and
collapsing of the vapour bubbles. Formation of vapour bubbles of the flowing liquid
take place only whenever the pressure in any region falls below vapour pressure.
When the pressure of the flowing liquid is less than its vapour pressure, the liquid
starts boiling and vapour bubbles are formed. These vapour bubbles are carried
along with the flowing liquid to higher pressure zones where these vapours
condense and bubbles collapse. Due to sudden collapsing of the bubbles on the
metallic surface, high pressure is produced and metallic surfaces are subjected to
high local stresses. Thus the surfaces are damaged.
Precaution against Cavitation: The following precautions should be taken against
cavitation:
i.
ii.
The pressure of the flowing liquid in any part of the hydraulic system
should not be allowed to fall below its vapour pressure. If the flowing
liquid is water, then the absolute pressure head should not be below 2.5 m
of water.
The special materials of coatings such as aluminium-bronze and stainless
steel, which are cavitation resistant materials, should be used.
Effects of Cavitation: The followings are the effects of cavitations:
i.
ii.
iii.
The metallic surfaces are damaged and cavities are formed on the surfaces.
Due to sudden collapse of vapour bubble, considerable noise and
vibrations are produced.
The efficiency of a turbine decreases due to cavitation. Due to pitting
action the surface of the turbine blades becomes rough and the force
exerted by water on the turbine blades decreases. Hence the work done by
water or output horse power becomes less and thus efficiency decreases.
8. What are the different efficiencies of a centrifugal pump?
Efficiencies of a centrifugal pump: In case of a centrifugal pump, the power
is transmitted from the shaft of the electric motor to the shaft of the pump and then
to the impeller. From the impeller, the power is given to the water. Thus power is
161
decreasing from the shaft of the pump to the impeller and then to the water. The
followings are the important efficiencies of a centrifugal pump:
a. Manometric efficiency, mano
b. Mechanical efficiency, m and
c. Overall efficiency, o.
a. Manometric Efficiency (mano): The ratio of the manometric head to the head
imparted by the impeller to the water is known as manometric efficiency.
Mathematically, it is written as
man 

Manometric head
head imparted by impeller to water
Hm
gHm

           (1)
 Vw 2 u2  Vw 2 u2


 g 
b. Mechanical Efficiency (m): The power at the shaft of the centrifugal pump is
more than the power available at the impeller of the pump. The ratio of the
power available at the impeller to the power at the shaft of the centrifugal
pump is known as mechanical efficiency. It is written as
m 
Power at the impeller
Power at the shaft
The power at the impeller

Work done by impeller per second
kw
1000

W Vw 2 u2

kw
g 1000
W  v w 2u2 
g  1000 
m 
S.P
---- (2)
Where W=weight of water lifted and S.P = shaft Power.
c. Overall Efficiency (o): It is defined as the ratio of power output of the pump
to the power input to the pump. The power output of the pump.
162
Weight of water lifted  Hm
kw
1000
WHm

1000

Power input to the pump
= Power supplied by the electric motor
= S.P of the pump
 WHm 
 1000 

 o  
S.P
-------- (3)
Also o = mano x m ----------- (4)
9. What are the main characteristics and operating characteristic curves of a
centrifugal pump?
Main Characteristic curves of a pump
Main Characteristics Curves: The main characteristic curves of a centrifugal pump
consists of variation of head (manometric head, Hm), power and discharge with
respect to speed. For plotting curves of manometric head versus speed, discharge is
kept constant. For plotting curves of discharges versus speed, manometric head (Hm)
is kept constant. And for plotting curves of power versus speed, the manometric
head and discharge are kept constant Figure shows main characteristics curves of a
pump.
Operating Characteristics Curves of a pump
Operating Characteristic Curves: If the speed is kept constant, the variation of
manometric head, power and efficiency with respect to discharge gives the operating
characteristic curves of a pump.
10. What are the main parts of a reciprocating pump?
The main parts of a reciprocating pump are:
163
1.
2.
3.
4.
5.
A cylinder with a piston, piston rod, connecting rod and a crank.
Suction pipe
Delivery pipe
Suction valve and
Delivery valve
11. Define ‘Slip’ of a reciprocating pump.
Slip of a reciprocating pump is defined as the difference between the
theoretical discharge and actual discharge of the pump.
Slip = Qthe - Qact.
The actual discharge of a pump is less than the theoretical discharge due to
friction and leakage in the pipe.
12. What is the main difference between single acting and double acting
reciprocating pump?
In a single acting reciprocating pump, the liquid acts on one side of the piston
only. In a double acting reciprocating pump, the liquid acts on both sides of the
piston.
13. Define negative slip of the reciprocating pump.
If the actual discharge is greater than the theoretical discharge, then the slip of
the pump is called as negative slip.
Negative slip occurs when the delivery pipe is short and suction pipe is long
and the pump is running at high speed.
14. What are the uses of air vessels fitted in a reciprocating pump?
Air vessels are used in a reciprocating pump to obtain a continuous supply of
water at uniform rate, to save a considerable amount of work and to run the pump at
a high speed without separation.
15. Compare the reciprocating pump with the centrifugal pump.
164
Reciprocating Pump
Centrifugal Pump
1. Suitable for small discharge and high Suitable for large discharge and
heads.
smaller heads.
2. The discharge is fluctuating and The discharge is continuous and
pulsating.
smooth.
It can be used for lifting highly viscous
3. It is used for lifting less viscous liquids.
liquids.
4. The reciprocating pump runs at low The centrifugal pump runs at high
speed.
speed.
5. The efficiency is low.
The efficiency is high.
6. It requires larger floor area for It needs smaller floor area for
installation.
installation.
7. The initial and maintenance costs are
The initial and maintenance are low.
high.
8. Air vessels are required.
Air vessels are not required.
9. No need of priming.
Priming is needed if it is not a selfpriming type.
165
PART – B
1. A centrifugal pump delivers 1.27 m3 of water per minute at 1200 r.p.m. The
impeller diameter is 350 mm and breadth at outlet 12.7 mm. The pressure
difference between inlet and outlet of pump casing is 272 kN/m2. Assuming
manometric efficiency as 63%, calculate exit blade angle.
Solution:
Given: Q =1.27 m3/min; N=1200 rpm; D2 =350 mm =0.35 m,
B2 =12.7 mm = 0.0127 m;mano = 63% = 0.63
Pressure difference between outlet and inlet of pump casin, (Pd = Ps) = 272 kN/m2.
Discharge, Q 
1.27
 0.021 m3 / s
60
Manometric head,
Hm 
Pd  Ps
.g
272  103
1000  9.81
=27.73 m
=
Blade velocity at outlet,
u2 
D2N   0.35  1200

 21.99 m / s
60
60
Discharge,
Q  D2B2 .Vf 2
0.021    0.35  0.0127  Vf 2
mano 
g.Hm
Vw 2 .u2
166
 Vf2 =1.52 m/s
 Vw 2 
g  Hm
9.81 27.73

 19.64 m / s
mano  u2 0.63  21.99
From outlet velocity diagram shown in Fig.
tan=
Vf2
1.52

 06468
u2  Vw 2 21.99  19.64
 Exi t angle of impeller blade,
2. The internal and external diameter of the impeller of a centrifugal pump are 20
cm and 40 cm respectively. The speed of the pump is 1400 rpm. Assuming a
constant velocity of flow of 5 m/s throughout, radial entry to impeller vanes and
the exit vane angle of 300. Find:
i) Inlet vane angle.
ii) Work done by impeller per N Weight of water.
Solution:
Given : D1=20 cm = 0.2 m; D2=40 cm = 0.4 m; N = 1400 rpm, Vf1 = Vf2 = 5 m/s;
Radial entry i.e,  =900, VW1=0,=300.
D1N
60
  0.2  1400

60
=14.66 m/s
D N
Vane velocity at outlet, u2  2
60
  0.4  1400
=
60
=29.32 m/s
Vane velocity at inlet, 1 
i) Inlet vane angle, :
From inlet velocity triangle,
tan =
Vf1
5

 0.341
u1 14.66
 =18.83o
ii) Workdone/N weight of water, W:
167
From outlet velocity diagram,
VW2  u2  (Vf 2 / tan )  29.32 
W=
5
 20.66 m/s
tan30o
Vw2 .u2 20.66  29.32

 61.75Nm / N
g
9.81
3. The external and internal diameter of the impeller of centrifugal pump are 0.6 m
and 0.3 m respectively and the width of impeller at outlet is 60 mm. The speed of
the pump is 1440 rpm and it is required to work against the head of 105 m. The
velocity of flow through the impeller is maintained at 4 m/s. The exit vane angle
is 350. Determine the vane angle at inlet, workdone by impeller on water per
second and the manometric efficiency of the pump.
Solution: Refer the velocity diagram shown in Fig.
Given: D2 = 0.6 m, D1=0.3 m, B2=60 mm =0.06 m:
N=1440 rpm, Hm=105 m:
Vf1=Vf2 = 4 m/s; =350.
i)
Inlet vane angle. :
Vane velocity at inlet,
D1N   0.6  1440

60
60
= 22.62 m/s
V
4
tan = f1 
 0.1768
u1 22.62
u1 
=10.03o
ii) Work done by impeller per second, W:
Discharge, Q  D2B2 .Vf 2
  0.6  0.06  4  0.452m3 / s
Vane velocity at exit, u2 
D2N   0.6  1440

 45.24 m/s
60
60
From outlet velocity diagram,
Vw2  u2 
Vf 2
4
 45.24 
 39.53 m/s
tan 
tan350
workdone by impeller on waters,
168
W  QVw 2  u2  103  0.452  39.53  45.24
= 808.33  103 W = 808.33 kW
iii) Monometric efficiency of the pump, mano:
mano 
gHm
9.81 105

 0.576 or 57.6%
Vw2  u2 39.53  45.24
4. A three stage centrifugal pump has impellers 40 cm diameter and 2 cm wide at
outlet. The varies are curved back at 450 and reduce the circumferential area by
10%. It’s manometric efficiency is 90% and overall efficiency is 80%.
Determine, the head generated by the pump when running at 1000 r.p.m.
delivering 50 litre per second. What should be the shaft power in kW.
Solution:
Given:
Number of stages, n1=3; D2=40 cm=0.4 m, B2 = 2cm = 0.02 m;  = 450,
Reduction in circumferential area = 10% = 0.1, therefore, Kb = (1-0.1)=0.9;
mano = 90% = 0.9, 0=80%=0.8; N=1000 rpm;
Q=50 litre/s=
50
 0.05 m3 / s
1000
(i) Total head generated, H:
Velocity of flow,
Q
0.05
Vf 2 

 2.21m
s
 D2B2 K b   0.4  0.02  0.9
Blade velocity at outlet,
u2 
 D2N
60

  0.4  1000
60
 20.94 m
s
From outlet velocity diagram shown in Figure.
Vw 2  u2 
Vf 2
2.21
 20.94 
tan 
tan 45o
169
= 18.73 m/s
mano 
g Hm
Vw 2 u2
i .e. 0.9=

9.81 Hm
18.73  20.94
Hm = 35.98 m/stage
Since the pump is multistage pump in series,
Therefore,
Total head generated,
H=Number of stages x H/ stage
= 3 x 35.98 = 107.94 m
(ii) Shaft power, Ps :
Overall efficiency, o 
 Ps =
gQH
Ps
 g Q H 1000  9.81 0.05  107.94

0
0.8
= 66.18 x 10-3 W = 66.18 kW
5. A centrifugal pump has to deliver 13.5 m3/min of water against a head of 30 m.
The speed of the pump is 1500 rpm. Manometric efficiency of the pump is 80%.
The breadth of impeller is 0.4 times the impeller diameter at outlet. The friction
head loss in pump is 0.03 times the square of absolute velocity of water at outlet.
Find:
(i) Impeller diameter
(ii) Exit angle of vane.
Solution:Given: Q = 13.5 m3/min =
13.5
= 0.225 m3/s; Hm = 30 m; N = 1500 rpm;
60
170
mano = 80% = 0.8; B2 = 0.4 D2; hf = 0.03 x V22
(i) Impeller diameter, D2 :
Manometric efficiency is given by the equation,
mano 

g Hm
;
Vw 2 u2
or
0.8=
g  30
Vw2 .u2
Vw 2 .u2 30

 37.5 m
g
0.8
(It represents the head required to be developed by the pump).
 Friction losses in the pump,
Vw 2 u2
-Hm = 37.5 – 30 = 7.5 m
g
But, hf = 7.5 m = 0.03 V22
 V2 = 15.81 m/s
Q
Q
Velocity of flow at outlet, Vf 2  
A  D2B2
0.225
Vf 2 
 0.179 / D22m / s
  D2  0.4D2
hf 
Blade velocity at outlet,
u2 
 D2N
60

  D2  1500
60
 78.54D2m / s.
Substituting the value of u2 in Equation (i) above, we get,
Vw 2  78.54D2
 37.5
9.81
4.684
Vw 2 
D2
From the outlet velocity diagram shown in figure.
V22  V22  Vf 22
 4.684   0.179 
15.81  

 
2
2
 D   D2 

2

2
2
249.96=
21.94 0.032

D22
D24
171
 249.96 D24 – 21.94 D22 – 0.032 = 0
On solving,
D2 = 0.2963 m or 29.63 cm
(ii) Exit angle of vane, 
V12=
0.179
0.179

 2.039 m/s
2
D2
(0.2963)2
Vw 2 
4.684 4.684

 15.808 m/s
D2
0.2963
u2  78.54D2  7.54  0.2963  23.271 m/s
tan  
Vf 2
2.039

 0.273.
u2  Vw 2 (23.271  15.808)
  15.28
Ans
6. A centrifugal pump while running 1000 rpm discharges 80 litres/sec. against a
net head of 16m. The manometric efficiency of the pump is 85% . If the vane angle
at the outlet is 35 and the velocity of flow is 1.5m/s, estimate the outer diameter
the impeller and its width at the outlet.
Solution:
Given N=1000rpm, q =80 litres/s =80 x 10-3=0.08m3/s; Hm=16 m;
mano=85=0.85;=35, Vf2=1.5 m/s
(i)
Outer diameter of impeller , D2:
From outlet velocity diagram shown in fig P.5.7
u2  Vw 2 
Vf 2
1.5

 2.142m / s;
tan  tan35
Vw 2  u2  2.142
mano 
g.H m
Vw 2 .u2
0.85 
9.81 16
(u2  2.142)u2
172
 u22  2.142u2  184.66  0
 u2 
2.142  (2.142)2  4  184. 2.142  27.262

2
2
=14.702 m/s or (12.56m/s, which is not possible)
u2 
 D2N
60
;
D2 
 D2  0.28208 m
(ii)
60u2 60  14.702

N
  1000
or 28.08 cm
Ans
width of impeller at outplet , B2
Q   D2B2Vf 2
0.08=  0.808  B2  1.5
B2  0.0605 m or 6.05 cm
7. A centrifugal pump delivers 0.1 m3/s of water through a pipe of 0.2m diameter
of length 300m upto a height o 26m. Darc’s coefficient for pipe, f=0.02, inlet losses
in suction pipe are estimated to e 0.4m. Calculate the power required to drive the
pump I its overall efficiency is 73%
Solution;
Given:
Q  0.1m 3 / s, d d  0.2m, Id  300m; Hs  26m; f  0.4hfs  0.4m;
0  73%  0.73
Q
Q
4  0.1


 3.183m / s
A  .d 2   0.22
4 d
Head equivalent to kinetic energy in piple
V 2d 3.1832

 0.52m
2g 2  9.81
Vs  Vd 
hfd 
fId .Vd2
0.02  300  (3.183)2

 15.49m
dd  2g
0.2  2  9.81
Manometric head,
173
Vd2
Hlm  Hs  hfs  hfd 
 26  0.4  15.49  0.52  42.41m
2g
Power required to drive the pump , Ps :
Ps 
=
p.g.Q.Hm
0
 103 kW
1000  9.81 0.1 42.41
 103  56.992kW
0.73
8. A double acting reciprocating pump with air vessel in suction pipe has 200 mm
piston diameter and 400 mm stroke. The suction pipe diameter is 160mm. Find
the crank angles at which there is no flow of water into or from the vessel. The
pump runs at 120 rpm. Assume motion of piston with S.H.M.
Solution:
Dp  200 mm  0.2 m, L  400 mm  0.4m,
Given:
Ds  160 mm  0.16m;
Area of piston 
 2 
.Dp  x(0.2)2  0.0314 m2
4
4
2N 2x120

4 rad / s
60
60
Instant velocity of water from or into air vessel.
A
V1  p .w r sin   A p ..r.sin 
As
Angular velocity of crank, w =
Discharge from or into cylinder
A 
r
 Discharge Q1  A s . p x  2A p.
As 

Net discharge = 0 i.e. QI- Qm=0
r
A p .r sin   2xA p .
0

  39.540 or 140.460
9. A centrifugal pump of 20 cm diameter running at 1430 rpm delivers 0.1 m3/s of
water against a head of 40m with an efficiency of 90% what is its non-dimensional
specific speed? (AU,Nov 2002)
Given D=20cm, N=1430 rpm, Q=0.1m3/s H=40m, mo=90%
174
Solution:
Specific speed Ns 
N Q
=
3
1 4
1430  0.1
3
40 4
=28.43 rpm.
Find the power required to drive a centrifugal pump which delivers 0.04m3/s of
water to a height of 20m through a 15 cm diameter pipe and 100m long. The
overall efficiency of the pump is 70% and coefficient of friction is 0.15 in the
formula hf 
4flv 2
2gd
(AU,Apr 04)
Given Q=0.04m3/s, h=20m, d=0.15m, l=100m, mo=70%, 4f=0.15
Solution:
Velocity of water in the pipe Vd 
Q
a
0.04

2
 0.15 
4
=2.26m/s
=
Fraction head, hf 
=
4flv 2d
2gd
0.15  100   2.26 
2  9.81 0.15
=26.11m
Manometric head, Hm  h  hfd 
Vd2
2g
=20+26.11+
2.262
249.81
=46.37m
Power required to drive the pump=
WQHm
mO
1000  9.81 0.04  46.37
0.7
=25993.7W
=
=25.994 hW
175
10. Two geometrically similar pumps are running at the same speed of 750 r.p.m.
one pump has an impeller diameter of 0.25 m. and lifts water at the rate of 30
litres/sec against a head of 20m. Determine the head and impeller diameter of the
other pump to deliver half the discharge. (AU-Nov 2003)
Given:
N1=N2=750 rpm.
D1=0.25m
Q1=30lit/sec
H1=20m
Q2=
Q1
2
Solution:
Q  AV 
 2 DN
D 
4
60
 QD3N
For the first pump,
Q1 D31 N1 ------(1)
For the second pump
Q2 D32 N2 ------(2)
1 Q1 D13N1


2 Q2 D32N1
Q1 D13

 N1  N2 
Q2 D32
D32 
=
D2 
Q1
 D13
Q2
Q1
2  Q2
0.25
2
1
3
 0.198m
176
Similarly, V=
T1DN
60
 VDN
Also V= 2gH
V H
 DN H
For the first Pump
D1N1 H1      (3)
D2N2  H2      (4)
H1
3 D1N1


4 D2N2
H2
D1
H1

D2
H2

N1  N2 
2
D 
H2   2   H1
 D1 
2
 0.198 
=
  20
 0.25 
 12.55m
11. The diameter and stroke of a single acting reciprocating pump are 200mm and
400mm respectively. The pump runs at 60 rpm and lifts 12 litres of water per
second through a height of 25m. the delivery pipe is 20m long and 150mm in
diameter. Find (i) theoretical power required to run the pump (ii) percentage of
slip (iii) Acceleration head at the beginning and middle of the delivery
stroke.(AU-Nov 2003).
Given:
D=0.2m
L=0.4m
L 0.4

 0.2m
2
2
N  60rpm.
Q actual =12 litres/s
r
=0.012m3 / s
hd=25m
ld=20m
177
dd=0.15m
Solution:
Qth 
ALN
60
  0.2   0.4  60
= 
 0.0126m3 / s
4
60
2
(i)
Power & required =W x Qthx(hs+hd)
=1000x9.81x0.0126(0+25)
=3090.15W
=3.1 kw.
(ii) Percentage slip =
Qth  Qact
 100
Qth
0.126-0.12
 100
0.126
=4.76%
=
(iii) Acceleration head of the beginning of stroke
had 
ld A 2
w r
g ad


cosOO  1

2
2
0.2
20 4  
 2  60 
=

 0.2
2
9.81 
60 

 015 
4
=28.62 m of water.
(iv) Acceleration head at the middle of stroke,
had 
ld A 2
 w cos90o
g ad
=0
12. The length and diameter of a suction pipe of a single acting reciprocating
pump are 5m and 10cm respectively. The pmp has a plunger diameter of
150mm and stroke length of 300mm. the centre of the pump is 4 m above the
water surface in the sump. The atmospheric pressure head is 10.3m of water
and the pump is running at 40rpm. Determine
(i)
Pressure head due to acceleration at the beginning of suction stroke.
(ii)
Maximum pressure head due to acceleration .
178
(iii)
Pressure head in the cylinder at the beginning and at the end of the
stroke (AU- Nov’2004.
Given:
Ls=5m
Ds=0.1m
D=0.15m
L=0.3m
L 0.3

 0.15m
2
2
hs  4m
r
Hafm  10.3m of water
N=40 r.p.m
Solution:
Acceleration head on the suction stroke,
Has 
ls A 2
w r cos 
g as
At the beginning of the strike, =Oo
has 
ls A 2
 w r
g as


cos 0o  1

2
2
  0.15 
5
 2  40 
=
4

 0.15

2
9.81
60 

0.1
 
4
=3.02m of water.
Maximum pressure head due to acceleration
=hs+has = 4+3.02
=7.02m of water
Absolute pressure head of the beginning of stroke,
=Hatn-(hs+has)
=10.3-7.02
=3.28 m of water.
179
Acceleration pressure head at the end of suction stroke,
has 
ls A 2
w r
g as
(
=180o cos 180o  1)
=-3.02m of water.
Absolute pressure head at the end of stroke.
=Hatn-(hs+has)
=10.3-(4-3.02)
=9.32m of water
13. The indicator diagram of a single acting reciprocating pump gives effective
delivery head of 5m and 23m with the crank at inner and other dead centres
respectively. What is the static delivery head of the reciprocating pump? (AU –
April 2005)
Given:
Effective delivery head at IDC;
Hd-had=5m
Effective delivery head at ODC,
Hd+had=23m
Solution:
Hd-had=5
-----(1)
Hd+had=23
-----(2)
(1)+(2)
2hd=28
hd=14m.
14. A double acting reciprocating pump runs at 40 rpm. It has the cylinder of 200
mm diameter and stroke of 400 mm. It delivers water to a height of 1 m through a
pipe of 150 mm diameter and 40 m long. An air vessel is attached at 3 m height
from the centre of cylinder. The coefficient of friction for the pipe is 0.01. Find
the pressure head in the cylinder at the beginning and at the end of delivery
stroke. Assume motion of piston by SHM.
Solution:
Given:
N = 40 rpm; Dp = 200 mm = 0.2 m;
180
L
 0.2 m
L = 400 mm = 0.4, therefore
2
hd  1m;dd  150 mm  0.15 m;
r
Height or air vessel, ldt  3m, coefficient of friction, f = 0.01
ld  Total length of pipe – height of air vessel = 40 – 3 =37m.
Area of piston A p 
 2 
Dp  x0.22  0.03142 m2
4
4
Area of delivery pipe, A d 
 2 
2
xdd   0.15   0.1767 m2
4
4
Angular speed of crank,  
2N 2x40

 4.19 rad / s
60
60
Accelerating speed of crank

2N 2x40

 4.19 rad / s
60
60
Accelerating pressure head in delivery pipe up to the height of air vessel,
had 1 
ldl A p 2
3
0.01342
. . r cos  
x
 4.192 x0.2cos 
g Ad
9.81 0.1767
= 1.909 cos 
Mean velocity of water in delivery pipe,
v
2LAN 2x0.4x0.01342x40
Q
p


A d 60xA d
60x0.01767
Head loss due to friction.
hfd 
4f.ld.V2 4x0.01x37x0.952

 0.454 m
dd.2g
0.15x2x9.81
Pressure head in delivery pipe,
181
Hd  hd  had1  hfd  11  1.909 cos   0.454
 11.454  1.909 cos 
Head at the beginning of delivery stroke    0  ,
Hd  11.454  1.909 cos 0  13.363 m
Head at the end of delivery of stroke    180 
Hd  11,454  1,909 cos 180  9.265 m
***************
182