UNESCO-NIGERIA TECHNICAL & VOCATIONAL
UNESCO
EDUCATION REVITALISATION PROJECT-PHASE
PROJECT
II
NATIONAL DIPLOMA IN
MECHANICAL ENGINEERI
ENGINEERING
NG TECHNOLOGY
FLUID MECHANICS
COURSE CODE: MEC214
YEAR 22 SE MESTER I
THEORY
Version 1: December 2008
TABLE OF CONTENT
Table of Contents
Week 1
1.0
Fluids
1.1
Definition
1.2
Classification of fluids
1.2.1 Gases
1.2.2 Liquids
1.3
The difference between liquids & gases
1.4
Properties of fluids
Week 2
2.0
Types of fluids
2.1
Ideal and real fluids
2.2
Newtonian fluids
2.3
Non-Newtonian fluids
2.4
Specific gravity of liquids
Week 3
3.0
Pressure equation
3.1
Pressure of liquids vary with depth
3.2
An orifice tank
4.0
Pressure measuring instrument
4.1
The barometer
4.2
Piezometer
4.3
U-tube manometer
4.4
Inverted U-tube manometer
5.0
Questions and solutions on pressure measurement
6.0
Derivation of equations
6.1
Parallel axes theorem
6.2
Total thrust acting on a vertical plate
WEEK 7 : Understanding the Archimedes principle and its applications
7.1 Buoyancy of floating bodies
7.1 Archimedes principles
7.1.1 Centre of buoyancy
7.1.2 Buoyant force
7..2 Hydrometer
7.3 Floating bodies
WEEK 8: Analysis and the practical determination of metacentric height
8.1 Determination of metacentric height of a vessel
8.2 Experimental method
WEEK 9: Understanding the principle of conservation of mass
9.1
The principle of conservation of mass
9.2 Conservation of mass
9.2.1 The discharge
9.3 Application of continuity equation
WEEEK 10: Understanding the conservation of energy
10.1 The conservation of energy
10.2 The Bernoulli’s equation
WEEK 11: To know the momentum equation and its applications
11.1: Introduction and understanding the concept of steady streamline and
stream tube
WEEK 12: To understand the concept of momentum and fluid flow
12.1 Momentum and fluid flow
12.2 Momentum equation for two and three dimensional flow along a
streamline
WEEK 14: FLUID DYNAMICS
14.1 Types of flow
14.2 Determination of laminar and turbulent flow
WEEK 15 : To know the types fluid power machines and their application
15.1 The fluid power machines
15.2 The pumps
15.2.1 Classification of pump
15.2.2 Centrifugal pump
15.3 Hydraulic turbines
15.3.1 Types of hydraulic turbines
15.3.2 Impulse turbine (pelton wheel)
15.4 Reaction turbine
15.4.1 Types of reaction turbine
15.5 Hydraulic press
15.6 Air Compressors
15.6.1 Types of air compressors
15.6.2 Reciprocating compressor
15.6.3 Rotary compressor
FLUID MECHANICS
WEEK 1
1.1.1 Define a fluid.
1.2
The different types of fluids.
1.3
Distillation or classification of fluids.
1.4
Properties of fluids. (Density, specific weight, relative density or specific
gravity, specific volume pressure, viscosity, adhesion, cohesion, surface
tension compressibility, capillarity.
FLUIDS
Liquids have definite volumes: may vary likely with temperature and pressure.
Definition – Fluid is a substance which deforms continuously under the action
of shearing forces, however small they may be i.e it offers negligible
resistance to change of shape and is capable of the following. Thus a fluid can
either be a liquid or gas.
Distinction or classification of fluids.
1.1
1.22a Liquids: - Liquids have definite volumes and the volume may vary likely with
temperature and pressure; and normally have free surface.
1.23.1 Gases: - Gases normally occupy the volume of their containers and they can
readily be compressed. The fundamental difference between liquid and gas is
that liquid is hard to be compressed while gas can easily be compressed.
1.23.2 Liquids:- e.g. water, oil, petrol, kerosene, paint, e.t.c.
1.23.3 Gases:- Oxygen, carbon dioxide, Nitrogen e.t.c
1.3
Liquids
Gases
1. Difficult to compress, may be regarded as 1. A gas is comparatively easy to
incompressible.
compress.
2. A given mass of liquid occupies a fixed 2. a. Changes of volume with
volume irrespective of the size or shape of its pressure are large, cannot normally
container.
be neglected and are related to
change of temperature.
2b. A given mass of gas has no fixed
volume and will expand continuously
unless restrained by a containing
vessel.
3. A free surface is formed of the container is 3. It will completely fill any vessel in
greater than that of the liquid.
which it is placed and does not form
a free surface.
4. The molecular structure of liquid are less
densely packed, making the structure looser
than that of solids. Thee individual molecules
have greater freedom of movement causing
change of structure..
4. Gasses have no formal structure,
the spaces between molecules are
large and the molecules can move
freely.
1.4.1 Properties of fluids.
Density:- The density of a substance is ρ that quantity of matter contained in
unit volume of the substance.
Density, ρ =
mass of subs tan ce
volume of subs tan ce
Density of a substance can be expressed in three different ways namely;
(i)
(ii)
(iii)
Mass density
Specific weight
Relative density.
(i)
Mass density:Mass density is defined as the ρ mass of the substance per unit volume.
ρ=
(
m
kg .m − 3
v
Typical values of density at
)
P = 1.013 x 105 Nm−2 , T = 288.15k
Water, 1000 Kg m−3 , air, 1.23 Kg m −3
1.42
(ii)
Specific weight:- Specific weight, ω , is defined as the weight per unit volume.
ω=
weight Mg m
=
=
x g = ρg
volume
v
v
Typical values: water, 9.81 x 103 Nm−3 ;
Air, 12.07 Nm−3
1.43
(iii)
Relative density or Specific gravity:Relative density, is defined as the ratio of the mass density of a substance to some
standard mass density.
For solids and liquids the standard mass density chosen is the maximum density of
water(which occurs at 4c at temperature and pressure)
σ =
ρ subs tan ce
ρH 2 0 at 4°C
For gases, the standard density may be that of Hydrogen at a specified temperature and
pressure, but the term is not used frequently.
Relative density is dimensionless, since it is pure number.
Typical value: Water 1.0; oil 0.9.
1.44
Specific Volume:- This is defined as the reciprocal of either mass density or reciprocal of
either mass density or specific weight; i.e volume per unit mass or volume per unit weight.
Unit = m
3
N
1.45
Pressure:- Pressure is the intensity of a force measured per unit area of the surface on which
the force acts. The pressure exerted by a fluid has the following important properties:i.
ii.
The pressure exerted by a fluid “at a point “ that is, on a very small area, is the same in
all directions.
The direction of the resultant pressure exerted by a fluid at rest on a solid surface is
always perpendicular (normal) to the surface.
h
AB is the surface area.
P is the force exerted.
X
P=P
A
1.46 VISCOUSITY
Viscosity is the internal resistance to movement of one layer of the fluid over an adjacent one.
It is the property of a fluid rather than that of a static fluid. Relative movement between two
layers requires shear forces parallel to the surface over which they act.
1.47 Vapour Pressure:All liquids have the tendency to evaporate when they are bounded by interface. Evaporation
means that there is a continuous movement of molecules in the form of vapour pressure
above the liquid surface. The magnitude corresponds to the rate of molecules escaping from
the surface. The number of liquid molecules going out of the liquid continues to increase until
it reaches the point where the vapour pressure is sufficient to maintain a balance i.e the
number of molecules leaving the liquid is equal to the number entering it. At this stage, when
there is a balance, the liquid saturated with vapour and the vapour pressure at this point is
called saturation pressure.
If ABCD (FIG 1) represents an element in a fluid with thickness perpendicular to the diagram,
then the force P will act over an area A equal to BC x s. The force per unit area p/a is the shear
stress φ and the deformation, measured by the angle ф (the shear strain), will be proportional
to the shear stress.
In a solid,ф will be a fixed quantity for a given value of φ, since a solid can resist shear stress
permanently. In a fluid, the shear strain ф will continue to increase with time and the fluid will
flow. It is found experimentally that in a true fluid, the rate of shear strain (or shear strain per
unit time) is directly proportional to the shear stress.
(1) Cohesion
(2) Adhesion
(3) Viscosity
1.48 Adhesion:- Adhesion is the physical attraction of the physical attraction of the of one material
for the surface of another. (see Portland cement)
The most widely accepted and investigated theory of adhesion is the wet ability – a adsorption
theory. This states that for maximum adhesion the adhesive must come into complete imitate
contact with the surfaces of the adherent. That is the adhesive must completely wet the
adherent.
Cohesion:- A phenomenon in which the same material in contact makes them cling
together(with two different materials the similar).
According to kinetic theory, cohesion is caused by attraction between particles at the atomic or
molecular level.
Surface tension, which causes liquids to form spherical droplets is caused by cohesion.
1.49 Surface Tension:- The property that causes the surface of a liquid to behave as if it were
covered with a weak elastic skin; this is why a needle can float on the smallest possible area
because of cohesive forces between molecules at the surface.
1.410 Capillarity:- This is the spontaneous movement of liquids up and down narrow tubes or
capillarity. The movement is due to unbalanced molecular attraction at the unbalanced
molecular attraction at the boundary between the liquid and the tube.
If liquid molecules near the boundary are more strongly attracted to molecules in the
materials of the tube than to other nearby liquid molecules, the liquid will rise in the tube. If
liquid molecules are less attracted to the material of the tube than to other liquid molecules,
the liquid will fall.
WEEK 2
2.1 Types of fluids.
2.2 Newtonian fluids.
2.3 Non-Newtonian fluids.
2.4 State Newton’s law of viscousity.
2.5 Explain the effects of viscousity in fluids.
2.1
TYPES OF FLUIDS
There are two types of fluids namely, ideal and real or non-ideal fluids. The ideal fluid does not
exhibit viscous properties and cannot sustain frictional and shear stresses when in motion. The
forces sustaining its motion are purely pressure forces. Since it cannot sustain frictional
stresses, it cannot dissipate mechanical energy into heat. On the other hand, the real fluid
possesses viscous properties sustains frictional and shear stresses and dissipates mechanical
energy into heat. In practice, the ideal fluid does not exist, but the flow of many real fluids can
be analysed by assuming that they are ideal especially if their viscousities are low.
h
X
Fig. 2.1:
2.2
Newtonian fluids:- The fluids that obey Newton’s law of viscousity and for which N has a
constant value are known as Newtonian fluids. These fluids are the ones whose viscousities
are not dependent on the rate of shear or the duration and or the duration and conform to
Newton’s basic law of viscous resistance.
τ =N
dv
dy
Most common fluids fall into this category, for which shear stress is e.g. Gases and thin liquids
Shear
Stress
Plastic
Bingham Plastic
Pseudo-Plastic
Newtonian fluid
Dilatant
Ideal fluid T = 0
Rate of Shear,
Fig 1: Variations of shear stress with velocity gradient.
du
dy
2.3
Non- Newtonian fluids:Fluids which do not obey Newton’s law of viscousity are known as non-Newtonian and fall into
one of the following groups.
(1)
Plastics, for which the shear stress must reach a certain minimum value before flow
commences. Thereafter, shear stress increases with the rate of shear according to
the relationship.
 du 
τ = A + B  
 dy 
n
Where A, B and n are constants.
If n=1, the material is known as a Bingham plastic (e.g. sewage sludge)
(2)
Pseudo-plastic, for which dynamic viscousity decreases as the rate of shear increases
(e.g. colloidal solutions, clay, milk, blood, solution of cement)
(3)
Dilatant substances in which dynamic viscosity increases as the rate of shear
increases(e.g. Quicksand)
(4)
Thixotropic:- substances for which the dynamic viscousity decreases with the time
for which shearing forces are applied (e.g. thixotropic jelly paints, e.g. printer’s ink)
(5)
Rheopectic materials for which the dynamic viscosity increases with the time for
which shearing forces are applied.
(6)
Visco-elastic materials, which behave in a manner similar to Newtonian fluids under
time in variant conditions but if the shear stress changes suddenly, behave as if
plastic.
Examples of Non-Newtonian fluids are long chained hydrocarbons.
2.4
Liquid
Specific gravity
1. Alcohol, ethyl.
0.79
2. Benzene
0.88
3. Carbon tetrachloride 1.59
4. Kerosene
0.81
5. Mercury
13.6
6. Crude oil
0.85 – 0.93
7. Lubricating
0.85 – 0.88
8. Water
1.00
Types of Fluids
There are two types of fluids, namely, ideal and real or non-ideal fluids. The ideal fluid does not
exhibit viscous properties and cannot sustain frictional and shear stresses when in motion. The
forces sustaining its motion are purely pressure forces.
Since it cannot sustain frictional stresses, it cannot dissipate mechanical energy into heat. On the
other hand, the real fluid possesses viscous properties sustains frictional and shear stresses and
dissipates mechanical energy into heat.
In practice, the ideal fluid does not exist, but the flow of many real fluids can be analysed by
assuming that they are ideal especially if their viscosities are low.
WEEK 3
2.1 Explain how a fluid can exert pressure due to its own weight.
2.2 Derive an expression for the pressure at a point in a fluid.
2.3 Explain why the pressure in a fluid varies with depth.
h3
A
h2
h1
B
C
Fig 3.1
Consider a cylindrical container as shown in fig 3.1. A fluid of known density is poured into it to a
level l1. The area of the cross-section is A.
Pr essure =
But density ρ =
Force Weight W
=
=
Area
Area
A
Mass(M )
Volume(V )
Mass (M) = Mass(M) x acceleration due to gravity (g)
M= ρ V
But Weight(W) = Mass(M) x acceleration due to gravity (g)
W = Mg
Pr essure (ρ ) =
Mg ρvg
=
A
A
But volume (V) = Area (A) x height (h)
W = Mg
Pr essure (P ) =
Mg
ρvg
=
A
A
But Volume(V ) = Area (A ) x height (h )
V = Ah
Pr essure =
ρAhg
A
= ρgh
3.3
Pressure in a fluid varies with depth
h3
A
h2
h1
B
C
Fig. 2.3: An Orifice Tank
WEEK 4
Describe the working principles of the following fluid measuring instruments.
Explain their uses
3
4
12
The barometer.
13
Piezometer.
14
U- tube manometer.
15
The inverted U- tube.
Describe the working principles of the following fluid measuring instruments.
Explain their uses
4.1
(1) The Barometer:A practical method of measuring pressure of the atmosphere of the atmosphere is by means
of the mercury barometer (fig 4.1)
A
a
Fig 2
h
The space A at the top of a closed vertical tube dipped in an open bath of mercury is emptied
of air as far as possible. The pressure in space A is then equal only to the vapour pressure of
the mercury vapor. However, this vapour pressure is low and may often be neglected. The
space A is therefore almost at zero pressure.
The pressure of the atmosphere on the mercury in the open bath forces the liquid up the
tube until the sufficient to balance the pressure of the atmosphere.
Let h be the height of the mercury column above the level mercury in the bath; that is h is the
“height” of the mercury barometer. Let Ws be the specific weight of mercury and a specific
weight of mercury and a the cross-section area of the tube;
The volume of mercury column = ax h
Weight of mercury column = Ws x a x h
Pressure exerted by the mercury at the level of the bath
=
weight of column
area of tube
=
ws x a h
a
= ws h
This is the pressure exerted by the atmosphere on the mercury surface in the bath. The height
h of the Barometer is therefore directly proportional to the pressure of the atmosphere.
The density of mercury is 13.6 x 103 Kg
m3
, therefore specific weight of mercury
ws = 13.6 x 103 x 9.81
Hence, if h is in metres, atmospheric pressure = ws h
Thus, 1 meter of mercury, h = 1 corresponds to a pressure of 133.3 KN/m².
For large pressures the unit used for the atmospheric pressure is the atmosphere.
1 atmosphere (atm) = 101.3kN/m².
= 1.013bar
= 760mm of mercury
= 10.4m of water
2.41 (ii) Piezometer:The relationship between pressure and head is utilized for pressure measurement in the
manometer or liquid gauge. The simplest form is the pressure tube or Piezometer shown in
fig. 4.2
h1
h2
•
A
•
B
Fig. 3: Pressure Tube or Piezometer
It consists of a single vertical tube open at the inserted into a pipe or vessel containing liquid
under pressure which rises in the tube to a light depending on the pressure. If the
atmosphere, the pressure measured is ‘gauge’ pressure.
Pressure of A = Pressure due to column of height h
ρ A = ρgh1
Similarly, pressure at B = ρb = ρgh2
2.42 (iii) U-tube manometer:D
Fluid P,
Liquid Q1
Mass Density ρ
•
A
h2
h1
B
C
Fig 4 U-tube manometer
The U-tube gauge, fig. 4.3, can be used to measure the pressure of either liquids or gases. The
bottom of the U-tube is filled with a manometric liquid Q which is of greater density ρman and is
immiscible with the fluid P, liquid or gas of density P, whose pressure is to be measured if B is the
level of the interface in the left-hand limb and C is a point at the limb and C is point at the same level
in the right-hand limb,
Pressure ρ B at B = Pressure ρC at C
For the left-hand limb
Pressure Pb = Pressure Pa at A + Pressure due to depth h1 of fluid P
= ρ A + ρgh1
For the right-hand limb.
ρC = Pr essure ρ D at D + pressure due to depth h2 of liquid Q
ρ D = Atmospheric pressure = zero gauge pressure
and so ρC = 0 + ρ man gh2
Since ρ B = ρC
ρ A + ρgh1 = ρman gh2
ρ A = ρ man gh2 − ρgh1
(iv)
The inverted U- tube:-
Fluid, density
ρman
X
X
h
b
•B
a
•
A
Liquid, density ρ
Fig 4.4 inverted U-tube manometer.
The inverted U-tube shown in fig 4.4 is used for measuring pressure difference in liquids. The
top of the U-tube is filled with a fluid, frequently air, which is less dense than that connected
to the instrument. Since the fluid in the top is at rest, pressures at level xx will be the same in
both limbs.
For the left- hand limb
ρ xx = ρ A − ρga − ρ man gh
For the right-hand limb
ρ xx = ρ B − ρg (b + h )
ρ B − ρ A = ρg (b − a ) +gh(ρ − ρ man )
Thus
or if A and B are at the same level ,
ρ B − ρ A = (ρ − ρ man )gh
If the top of the tube is filled with air ρ man is negligible compared with ρ and
ρ B − ρ A = ρgh .
If the tip of the tube is filled with air ρ man is negligible compared with ρ and
ρ B − ρ A = ρgh .
On
the other hand, if the liquid in the top of the tube is chosen so that ρ man is very nearly
equal to ρ , and provided that the liquids do not mix, the result will be a very sensitive
manometer giving a large value of h for a small pressure difference.
WEEK 5
Compressibility and the Bulk Modulus.
All materials whether solid, liquid, or gases are compressible, that is the volume V of a given mass
will be reduced to V-δv when a force is exerted uniformly all over its surface. If the force per unit
area of surface increases to P+δp, relationship between change of pressure and change of volume
depends on the bulk modulus of the material.
Bulk modulus =
change in pressure
volumetric strain
Volumetric strain is the change in volume divided by the original volume. Therefore, volumetric
strain =
dv
V
Bulk modulus =
K = -v
Vdρ
dv
dρ
(The minus sign indicates that the volume decreases as pressure
dv
increases.
Therefore,
Change in volume
Change in pressure
=
Original volume
Bulk mod ulus
− δv
δρ
=
V
K
In the limit, as sp => 0
K = -v
dρ
dv
……………………….1.1
considering unit mass of a substance,
V =
1
ρ
…………………. 1.2
Differentiating,
Vdρ + ρdv = 0
v
dv = -   dρ
ρ
Substituting for V from equation 1.2
 1 
dv = -  2 dρ
ρ 
……………………1.3
p
Putting the values of v and dv obtained from equations (1.2) and (1.3) in equation (1.1)
K = ρ
dρ
dρ
……………………..1.4
The value of K is shown by equation (1.4) to be dependent on the relationship between
pressure and density and, since density is also affected by temperature, it will depend on how
the temperature changes during compression. If the temperature is constant, conditions are
said to be isothermal, while if no heat is allowed to enter, or leave during compression,
conditions are adiabatic. The ratio of the adiabatic bulk modulus to the isothermal bulk
modulus is equal to r, the ratio of the specific heat of a fluid at constant pressure to that at
constant volume.
For liquids, γ is approximately unity and the two conditions need not be distinguished; for
gases, the difference is substantial (for air γ =1.4)
Perfect gas: A perfect gas is defined as a substance that satisfies the perfect gas law.
PVs = RT…………. (1) and that has constant specific heats, specific volume; R is the gas
constant; and T is the absolute pressure; R is the gas constant; and T is the absolute
temperature. The perfect gas must be distinguished from the ideal fluid.
An ideal fluid is frictionless and incompressible. The perfect gas has viscosity and can therefore
develop shear stresses, and it is compressible according to equation (i) above.
It may be written as P = PRT
Where P = m/v
PV = RT
Replacing V by
1
ρ
;- ρ x
1
ρ
= RT
ρ = PRT
VISCOUSITY
The criteria which determines whether flow is viscous or turbulent is the quantity
ρvl
, known as
µ
the Reynolds number.
Re =
ρvl
µ
Where ρ = density of the fluid.
V = the critical velocity.
L = the characteristic length of the system
U = coefficient of dynamic viscosity
Experiments carried out with a number of different fluids in straight pipes of different diameters
have established that if the Reynolds number is calculated by making l equal to the pipe diameter
and using the mean velocity v, then, below a critical value of
ρvd
= 2000
µ
Flow will normally be laminar (viscous).
This value of the Reynolds number applies only to flow in pipes. Reynolds number is a ratio of forces
and hence, a pure number and may also be written as

µ
ρ
1
vl
=
, where V is the kinematic Viscosity  v =  i.e
ρ
µ
v
v

A
4cm
Total mass = 3kg
3cm
G
5cm
B
Example: Find the moment of inertia of fig below
a
X
ρ
Q
X
δ2
d
K.E = ½w²∑mr²
M = 2+1+3+4 = 10kg
R = 3+1+2+2 = 8m
I = ∑mr²
= M1r1² + M2R2² + M3R3² +M4R4²
= 2 x 3² + 1 x 1² + 3 x 2² + 4 x 4
= 2 x9 + 1 + 3 x 4 + 4 x 4
= 18 + 1 + 12 + 16
= 47kgm²
Radius of gyration
M = M1 + M 2 + M3 + M4
½w² x Mk² = ½w² x ∑mr²
Mk² x ∑mr² = I
I = Mk²
K = √(I/M)
Moments of inertia (page 710)
K.E = ½mv²
V = wr
k.e = ½m(wr)²
= ½mw²r²
K.E (1) = ½w² x m(1)r(1)²
K.E (2) = ½w² x m(2)r(2)²
K.E (3) = ½w² x m(3)r(3)²
K.E (4) = ½w² x m(4)r(4)²
K.E = K.E1 + K.E2 + K.E3 + K.E4
K.E = ∑½w² x mr²
Example : To find the moment of inertia (1) and the radius of gyration (k) of a uniform thin rod about
an axis through one end perpendicular to the length of the rod.
S
S
d1
Z
dz
Q
P
a
Let P = mass per unit length of the rod
Mass of element of PQ = ρδx
Second moment of mass PQ about xx = mass x (distance)²
= ρ·δx·x² = ρ·x²·δx
d2
Total second moment for all such elements can be written I = ∑ρx²δx
As x --- 0 ρ²x² = ρ
I = Mk²
M = aρ
I = k² = a²/3
= k² = ⅓ x a²
= √⅓ x a = a/√3
Hence k² = a²/3
K = a/√3
Example 2:
Find I for a rectangular plate about an axis through its e.g. parallel to one side as shown in fig
2x
Let ρ = mass per unit area of plate
Mass of strip PQ = b·δx·x²ρ i.e. mass x (distance)²
Total second moment for all strips covering the figure
I = ∑ b·ρ·x²·δx
I = b·ρ·d³
12
I = Mk² = b·ρ·d³
12
K = d²
12
I = Md²
12
but M = ρbd
Example 3: Find I for a rectangular plate 20cm x 10cm of mass 2kg about an axis 5cm from are 20cm
side as shown.
Mass of strip = b·δx·ρ
Ρ = 0.01kg/cm²
Area of strip = 20·δx
Mass of strip = 20·δx·ρ
2nd moment of mass of strip about xx
≈ 20·5x·ρx²
I = ∑x²dx·20ρ
Total thrust on a vertical plate immersed in liquid
Consider a thin strip at a depth Z below the surface of the liquid.
-
Pressure at P = w z (where w = weight of unit volume of the liquid)
∴ Thrust on strip PQ ≈ wz - 9.52
Then the total thrust on the whole plate
z =d 2
≈ ∑ aw zd 2
z =d c
If δ 2 → 0, total thrust =
d2
∫ awzd
d1
=
 z2  d2
aw 
 2  d1
=
d 2 d 2 
aw 2 − 1 
2
2
=
aw 2
d 2 − d12
2
[
]
2
=
aw
(d 2 − d1 )(d 2 + d1 )
2
=
 d + d1 
aw(d 2 − d1 ) 2

 2 
 d 2 + d1 

 is the depth half-way down the plate i.e. it indicates the depth of the centre of gravity of
 2 
−
the plate denoted by z
−
Total thrust
=
aw(d 2 − d1 ) z
=
a(d 2 − d1 )w z
−
Also, a (d 2 − d1 ) is the total area of the plate
∴
Total thrust = area of plate x pressure at the c3 of the plate
Depth of Centre of Pressure
The centre of pressure is the point of application of the resultant force due to fluid pressure on one
face of an immersed surface.
O
b
x
−
x
−
dp
y
dx
C
G
C
P
Fig. 6
−
To determine the depth
y of the centre of pressure we use the principle of moments. The sum of
the moments about the water surface 0-0 fig “K” of the forces on all the thin strips such as D − must
z
−
equal the moment p x
y of the resultant thrust p about 0-0.
Force on strip DE = pressure x area of strip
dp = p x bdx
= wx . bdx, sin ce p = wx
Moment of this force about 0-0 is dp . x
=
wxbdx . x
=
wx 2bdx
−
2
∫ wx bdx and this is equal to the moment P x y of the resultant force
Total moment about 0-0 =
P about 0-0
Hence
-
P . y = w ∫ x 2bdx, since w is a constant.
But
∫ x bdx is the total second moment of area I of the area A about 0-0
2
AB
Thus
−
P . y = w x I AB
−
−
Now P = wA x , where x is the distance of the centroid G below 0-0
Thus
−
y=
wI AB wI AB I AB
=
= −
P
wA−
Ax
x
Second moment of area about 0 - 0
First moment of area about 0 - 0
=
Let I G = Second moment of area A about axis through G parallel to water surface 0-0
AK2
=
Where K is the corresponding radius of gyration. Then the parallel theorem for 2nd moments of area
stated:-2
I AB = I G + A x
-2
−
y =
IG + A x
-
Ax
-2
AK 2 + A x
=
−
=
Ax
−
−
K2
−
−
+x
x
 K2
−
−
−x
+
x
 −

 x

But GC = y − x = 
K2
GC =
−
x
WEEK 6
6.1 State the parallel axes theorem.
6.2 Derive an expression for the total thrust acting on a vertical plate submerged in a liquid.
6.3 Identity the point where the resultant thrust acts.( depth of centre of pressure).
Example 1 To find the moment of inertia(I) and the radius of gyration (k) of a uniform
thin rod about an axis through one end perpendicular to the length of the rod.
Let ρ ═ mass per unit length of rod.
Mass of element PQ ═ ρ σ x
Second moment of mass PQ about XX ═ mass ×(distance)2
═ ρ . σ x.x2 ═ ρ x2 σ x 2
Total
second
moment
 x3 

3
a
2
∫ ρ x dx = ρ 
→ O,
as X
o
I
═MK
2
for
═
ρ
a 3
3
but
For the left- hand limb Pxx = Pa – Pga – Pmangh
For the right-hand limb
Pxx = Pb - Pg (b+h)
all
such
ρ
x
3
M ═ a
ρ
═
═
elements
ρ
a
3
can
be
written
I ═
∑ ρx σ
2
x
Thus Pb – Pa = Pg (b-a) + gh (P – Pman)
Or if A and B are at the same level, Pb – Pa = (P- Pman) gh
If the top of the tube is filled with air Pman is negligible compared with P and Pb –Pa = Pgh.
On the other hand, if the liquid in the top of he fluid is chosen so that Pmax is very near equal to P,
and provided that the liquid do not mix, the result will be a very sensitive manometer giving a large
value of h for a small pressure difference.
Compressibility and the bulk modulus.
All materials whether solid, liquid, or gases are compressible, that is the volume V of a given mass
will be reduced to V-δv when a force is exerted uniformly all over its surface. If the force per unit
area of surface increases to P+δp, relationship between change of pressure and change of volume
depends on the bulk modulus of the material.
Bulk mod ulus =
change in pressure
Volumetric strain
Volumetric strain is the change in volume divided by the original volume. Therefore, volumetric
strain = dv
Bulk modulus =
K =
- Vdp
dv
Vdp
dv
(The minus sign indicates that the volume decreases as pressure
increases.
Therefore, change in volume = change in pressure
Original volume
− δv SP
=
v
K
In the limit, as sp => 0
bulk modulus
K=
Vdp
……………………….1.1
dv
considering unit mass of a substance,
v=
1
…………………. 1.2
P
Differentiating,
Vdp + PdV = 0
Dv = - (v) dp
P
Substituting for V from equation 1.2
dv = - (1) dp
……………………1.3
p
Putting the values of v and dv obtained from equations (1.2) and (1.3) in equation (1.1)
K=P
dp
……………………..1.4
dp
The value of K is shown by equation (1.4) to be dependent on the relationship between pressure and
density and, since density is also affected by temperature, it will depend on how the temperature
changes during compression. If the temperature is constant, conditions are said to be isothermal,
while if no heat is allowed to enter, or leave during compression, conditions are adiabatic. The ratio
of the adiabatic bulk modulus to the isothermal bulk modulus is equal to r, the ratio of the specific
heat of a fluid at constant pressure to that at constant volume.
For liquids, γ is approximately unity and the two conditions need not be distinguished; for gases, the
difference is substantial (for air γ =1.4)
Perfect gas: A perfect gas is defined as a substance that satisfies the perfect gas law.
PVs = RT…………. (1) and that has constant specific heats, specific volume; R is the gas
constant; and T is the absolute pressure; R is the gas constant; and T is the absolute temperature.
The perfect gas must be distinguished from the ideal fluid.
An ideal fluid is frictionless and incompressible. The perfect gas has viscosity and can therefore
develop shear stresses, and it is compressible according to equation (i) above.
It may be written as P = PRT
Where P = m/v
PV =RT
Replacing V by 1/p; P x 1/P = RT
P = PRT
A
4cm
Total mass = 3kg
3cm
G
B
5cm
X
P
2cm
example : Find the moment of inertia of fig below
X
d
ρ
b
G
Q
−
d
2
δ2
d
2
K.E = ½w²∑mr²
M = 2+1+3+4 = 10kg
R = 3+1+2+2 = 8m
I = ∑mr²
= M1r1² + M2R2² + M3R3² +M4R4²
= 2 x 3² + 1 x 1² + 3 x 2² + 4 x 4
= 2 x9 + 1 + 3 x 4 + 4 x 4
= 18 + 1 + 12 + 16
= 47kgm²
Radius of gyration
M = M1 +M2 +M3 +M4
½w² x Mk² = ½w² x ∑mr²
Mk² x ∑mr² = I
I = Mk²
K = √(I/M)
Moments of inertia (page 710)
K.E = ½mv²
V = wr
k.e = ½m(wr)²
= ½mw²r²
K.E (1) = ½w² x m(1)r(1)²
K.E (2) = ½w² x m(2)r(2)²
K.E (3) = ½w² x m(3)r(3)²
K.E (4) = ½w² x m(4)r(4)²
K.E = K.E1 + K.E2 + K.E3 + K.E4
K.E = ∑½w² x mr²
Example 3: Find I for a rectangular plate 20cm x 10cm of mass 2kg about an axis 5cm from are 20cm
side as shown.
Mass of strip = b·δx·ρ
Ρ = 0.01kg/cm²
Area of strip = 20·δx
Mass of strip = 20·δx·ρ
2nd moment of mass of strip about xx
≈ 20·5x·ρx²
I = ∑x²dx·20ρ
= 20ρ
QUESTIONS PART A
1.
2)
A particular fluid stored in a container has a volume of 3m3 and mass of 40kg. Calculate
i)
Specific weight
ii)
Specific volume
iii)
Relative density
A container fig. Q2 shown below has the following fluids stored in it. What would be the
pressure exerted by air (Pair), if pressure (PA) at the bottom of the container is 200 KN/m2
and the density of oil is 600 Kg/m3, that of water is 1000 Kg/m3. Take specific gravity of
mercury to be 13.6.
Air
3m
Oil
2m
Water
500mm
Mercury
A
3)
4)
What would be the pressure in KN/m if the equivalent head is measured as 600mm of
a)
Mercury of specific gravity 13.6
b)
A liquid of density 520 Kg/m3
What would be
a)
the gauge pressure
b)
the absolute pressure of water at a depth of 12m below the free surface.
Assume that the density of water to be 1000 Kg/m3 and the atmospheric pressure is 101
KN/m2.
5)
101 KN/m2
P1 = 0.87 x 103 Kg/m3
4m
Pm = 13.6 x 103 Kg/m3
3m
A
Fig. Q5
Fig. Q5 is a container that has two immiscible fluids in it. Calculate the pressure at point A
(at the bottom of the container).
6)
7)
8)
The diameters of the pistons of a hydraulic jack are
a)
100mm and 10mm respectively. Find the effort required to lift a load 20KN. The
pistons are on the same level.
b)
The diameters of the pistons of a hydraulic jack are in the ratio 1:5. The pistons are
on the same level and a force 200N is applied to the smaller piston. Determine the
load that could be raised by the larger piston.
What would be the pressure in KN/m2, if the equivalent head is measured as 4000mm of
a)
Mercury of specific gravity 13.6
b)
Water of density 103 Kg/m3
c)
Oil of specific weight 7.9 KN/m3.
The pressure at a point in the sea is 8 atmospheres. What is the depth of the point, if the
specific gravity of sea water is 1.17?
P
A
10m
B
Fig. Q9
Fig. Q9 is special Jar that contains water. If the pressure required to remove the cork at B is
150 Kpa, calculate the value of the pressure P that must be applied at A.
10)
A U-tube manometer is arranged as shown in fig. Q10 to measure the pressure difference
between two points A and B in a pipeline conveying water of density ρ = 103 Kg - m-3.
The density of the manometer liquid Q is 13.6 x 103 Kg. m-3 and point B is 0.3m higher than
point A. Calculate the pressure difference when h = 0.5m.
Fluid P of density ρ
B
A
b
a
h
C
++
++
++++
++
++++
++
++++++
++
++ ++
++ ++++ ++++ ++ ++ ++++ ++
++ ++++
++ ++
+++ ++ ++++++
++ ++
++ ++
++ ++++
++ ++
++ ++ ++
Manometric liquid Q of
density ρ man
D
Fig. Q10
PART B
1)
If the critical velocity of air in a pipe of 125mm diameter is 0.234 m/s, what would be the
Reynolds’ number.
Calculate also the critical velocity of water in the same pipe. Take the kinematics viscosity of
air as 1.46 x 10-5 m2 and that of water as 1.10 x 10-6 m2
s
2)
s
Oil of specific gravity 0.9 and viscosity 0.17 NS/m2 is pumped through a 75mm diameter
pipeline at the rate of 2.7 kg/s.
Show that the critical velocity is not exceeded.
3)
Water flows through a pipe 25mm in diameter at a velocity of 6 m/s. Determine whether
the flow would be laminar or turbulent assuming that the dynamic viscosity of water is
1.30 x 10-5 kg
ms
and its density 10- 5 kg
viscosity 9.6 x 10- 2
kg
ms
m3
. If oil of specific gravity 0.9 and dynamic
is pumped through the same pipe, what type of flow will occur.?
WEEK 7
UNDERSTANDING THE ARCHIMEDES PRINCIPLE AND ITS
APPLICATION
7.0 BUOYANCY OF FLOATING BODIES
When ever a body is immersed wholly or partially in a fluid it is
subjected to an upward force which tends to lift (buoy) it up. This
tendency for an immersed body to be lifted up in the fluid, due to
an upward force opposite to action of gravity is known as
buoyancy. The force tending to lift up the body under such
condition is known as buoyant force or force of buoyancy or up
thrust. The magnitude of buoyant forces can be determined by
Archimedes principle.
7.1 Archimedes principles:
This states that when a body is wholly or partially immersed in
fluid, it experiences an up thrust which is equal to the weight of
fluid displaced. This is mainly about ships and boats: how they
managed to float upright, and why they sometimes do not.
Consider the equilibrium of a stationary floating object such as a
ship. Two forces act upon it its own weight W which is acting,
vertically upwards excerted by the surrounding fluid. The upthrust
is governed by the archemedes principle.
7.2 (I) CENTRE OF BUOYANCY
The up thrust on a body always acts through the centre of gravity
of the fluid displaced by the body. This point is called the centre
of buoyancy.
Consider a cylinder in which the upthrust act along the vertical
centre line, because of the symmetry of the cylinder about the
centre line, it would be unreasonable to expect it to act anywhere
else.
Consider another example say ship. It has a
A much complex shape. The side view has no symmetry, so we
cannot easily tell where the line of action of the upthrust will be. If
we take moment about any point such as G, the centre of gravity
of the ship, we can see there is anticlock wise moment acting. The
ship can be in equilibrium only if the weight and the upthrust are
equal and opposite and also acts along the same restical line.
7.2
(ii) BOUYANT FORCE
The vertical upward force which the fluid or water exerted on the
body that were immersed in the fluid is known as the buoyant
force.
7.3 HYDROMETER
A given floating body will float at different height in liquids of
different densities for example, because the temperature and
salinity of the sea varies from place to place, its density also
values, and so a ship will float at different depths in different part
of the world. The depth at which a given body floats can be used
to indicate the density of the liquid. An instrument does it is called
a hydrometer.
The hydrometer is usually shaped as shown in the diagrams
below, weighted at the bottom so that it floats upright. The stem
is graduated.
(i)
To indicate the density of the liquid
In the left hand diagram the hydrometer is shown floating in a
liquid of density ρ0, a volume V0 of the hydrometer is submerged
so that V0 of the liquid is displaced. When the hydrometer is
floated in a liquid whose density is grater the diameter floats a
distance H higher, so that submerged volume is now (V0 – Ah),
where A is the area of X-section of the stem.
The upthrust on the hydrometer must be equal and opposite to its
weight W.
Therefore, W = e0gV0 = e.g. (V0-Ah)
ρ0/ρ = 1- Ah
V0
h = V0 (ρ – ρ0)
A
ρ
By making the area A rather small, that is, by making the steam
narrow, we obtain a sensitive hydrometer, one which gives a large
change in height in for only a small difference (ρ-ρ0) in densities.
7.4 FLOATING BODIES
Consider a wooden board that float in the position shown below.
It is noted that the centre of buoyancy is below the centre of
gravity. To observe how the body can float stable even with B
below G, consider the situation where the plank has been turned
through an angleθ.
From fig (i) the weight W=mg acts through the center of gravity G
and the upthrust R acts through the center of buoyancy B of the
displaced fluid in the same straight line as W. for body tilted at
angle θ the volume of liquid remains unchanged but the shape of
volume and its center of gravity changes while the center of
buoyancy move from B to B, and a turning moment, WX is
produced.
The metacentric m is the point at which the line of action of the
upthrust R for the as placed position cuts the original vertical
through the center of gravity of the body G.
The metecentric height is the distance GM for angle of tilt θ.
Righting moment =WX
Where x= GMθ.
, provide that the angles of tilt is small so that sin θ=θ in rad.
Righting moment =w. GM. Sinθ.
=w. GMθ.
Note that comparing fig (ii) and (iv), it can be seen that,
(I)
if m lies above G, a righting moment W.GMθ is produced,
equilibrium is stable and GM is regarded as -ve.
(II)
If M lies above G, an overturning moment WGMθ is produced
equilibrium is unstable and GM is regarded as –ve.
(III)
If M coincides with G, the body is in neutral equilibrium.
Therefore the stability of a floating body depend on the
metacentric height i.e. the distance G M.
WEEK 8
DETERMINATION OF METACENTRIC HEIGHT OF A VISSEL.
The metacentric height about the longitudinal axes can be
determined if a load p is moved transversely a distance x across its
deck causing the vessel to move through a small angleθ. The
overturning movement due to movement of load
the Righting movement =W.Gm sinθ.
Where Gm= metacentric height and
p=px
W=p+w, (w, weight of vessel and p=load).
Note that: The effect of moving the load p and a distance x is
produce an overturning movement on the vessel and a righting
movement is produced for equilibrium of the vessel in tilled
position.
For small angle sin θ = θ
:.
Righting movement=W.GMθ
Where W=total weight of vessel (w+p)
For equilibrium in tilled position, Righting moment = overturning
moment.
WGMθ=PX.
GM= p x
W sinθ
Where G.M= metacentric height
θ=radians.
EXAMPLE 1
A cylindrical drum with flat ends, whose diameter is 600mm, floats
in fresh water with its vertical, if the total mass m of the drum
(including content) is 225kg, to what depth h will it be
submerged?
Since the cylinder is floating in equilibrium, its weight must be
exactly balanced by the upthrust, and this is equal to the weight of
the water displaced.
Weight of cylinder=weight of water displaced
Mg=water x (volume displaced) g
225xg=1000 x (π x 0.32 xh) g.
h=0.796mm or 796mm
EXAMPLE 2
In an experiment to determine the metacentric weight for a
passenger lines of displacement 28.000 tone (i.e. its mass is
28,000 tones). An object of mass & tone is moved a distance 10m
across the width of the ship. The ship is observed to turn through
an angle of o.380. Find the metacentric height.
For the conversion of angle,
θ = 0.380
= 0.38 x π/180 rod.
The metacentric height, GM= PX.
Wθ
= 8 x 10
28.000x66x10-3
=0.43m
EXAMPLE 3
The crew of a small motor cruiser all moves to one side of the
vessel to wave to a passing boat. The displacement of the cruiser
is 40 tomes. The combined mass of the crew is 400kg and thing
are initiated distributed evenly over the width of the boat which is
4m. If the metacentric height of
The boat is 0.25m, through what angle does it turn when the crew
all move to one side
The angle, θ= PX
WGm
= 400X2
40,400X0.25
= 0.08 rad
:. Q=460.
WEEK 9:
UNDERSTANDING THE PRINCIPLE OF
CONSERVATION OF
MASS
9.1 PRINCIPLE OF CONSERVATION OF MASS
The theory of fluid flow tests on two principles namely the
principle of conservation of mass and principle of conservation of
energy. The principle of conservation of mass stated that “matter
can neither be created nor destroyed but can be transformed from
one form to another”
This principle can be applied to the flow of fluid. Using the
principle of conservation of mass, it can be said that in a given
period of time, the same mass of fluid that enter a system must be
equal to the amount of fluid flowing out.
9.2
CONSERVATION OF MASS
Consider a flow through a pipe of cross-area A1, density ρ1, and
valuation V1 at section (1) i.e. point of entry and ρ2, A2, V2 at
section (2) i.e. at outlet mass flow rate at section (1) = ρ1A1V1
Mass flow rate at section (2) = ρ2A2V2
By conservation of mass, mathematically,
ρ1A1V1 = ρ2A2V2 = m- -
-
(1)
Where m = mass flow rate
For incompressible flow (in general gasses and vapour are
compressible, liquids are incompressible) the density is practical
constant. The equation (1) becomes.
A1V1=A2V2 = Q
-
-
-
(2)
Where Q = discharge or volume flow rate m3/s.
The equation (2) is called the equation of continuity of flow.
9.3 DISCHARGE
The volume of liquid passing through a pipeline at any given cross
section in unit time is referred to as discharge or volumetric flow
rate, Q. Assume that the cross section area of the pipe is A and
the fluid flow at uniform velocity V, then the discharge or
volumetric flow rate Q is given by.
Q = AV (m2 x m2/s)
The unit is cubic meters per second m3/s
Give the density of the fluid passing through the pipe of cross
section in unit time, the mass flow rate can found,
Mass flow rate, m = ρAV = ρQ (kg/m3/s)
The unit is kilogram per second (kg/s).
Air is discharged from a compressor outlet at a velocity of 3mk
with a mass flow rate of 7.68 x 10-3 kg/s. if air has density of
1.293kg/m3, determine the volumetric flow rate and the diameter
of the outlet.
SOLUTION: Velocity of the air = 3m/s mass flow rate m = 7.68 x 10-3 kg/s the
density of air = 1.293kg/m3 Q =?
From the date.
Q = AV.
But m = ρAV and A = m
ρV
∴ Q=3
= 7.88 x 10-3
1.293 x 3
9.4 APPLICATION OF CONTINIUTY EQUATION.
The continuity equation can be applied to determine the relation
between the flows into and out of a junction consider the figure
below: -
For steady condition
Total inflow to junction = Total out let from junction
ρ1Q1 = ρ2 Q2 + ρ3 Q3
For an incompressible fluid, ρ1 = ρ2 = ρ3 so that Q1 = Q2 + Q3 or
A1V1 = A2V2 + A3V3.
I general if we considered flow toward he junction as positive and
flow from the junction as negative them for steady flow at any
junction the algebraic sum of all mass all mass flows must be zero.
WEEK 10
UNDERSTANDING THE CONSERVATION OF ENERGY
10.1 CONSERVATION OF ENERGY
Consider an element of fluid shown below will posses potential
energy due to its height Z above datum and Kinetic energy due to
its velocity V and the same way as any other object.
For an element of weight, mg.
Potential energy = mg Z
Potential energy per unit weight = mgz = Z - - - - - - - - - - (i)
mg
Kinetic energy = ½ mg V2 = ½ mV2
g
Kinetic energy per unit weight = ½ mV2 - - - - - - - - - - --(ii)
Force exerted on AB = PA
After a weight of fluid has flowed along the stream tube, section
AB will have moved to AB
Volume passing AB = mg /ρg = m/ρ
∴ Distance AA1 = m/ρA
Work done per unit weight = P/ρg - - - - - - - - - (c)
The term p/ρg is called the flow work or pressure energy.
By equation a, b, and c, it can be seen that the 3 terms of
Bernoulli equation are the pressure energy per unit weight, Kinetic
energy per unit weight and the Potential energy per unit weight.
Thus, Bernoulli’s equation state that for steady flow of a
frictionless fluid along a straight a streamline, the total energy per
unit weight remain constant from point to point although its
division between the three forms of energy may vary:
Pressure energy
K.E per
P.E per
Total energy
Per unit weight
unit weight
unit weight
per unit weight = constant
P1
+ V2 +
ρg
2g
Z1 = H - - - - - - - - (IV)
Each of these has the dimension of length or head and they are
often referred to as his pressure head p/ρg, the velocity head
V2/2g the potential head Z and total head it. Applying this equation
between two points 1 and 2 on stream line, equation d becomes.
P1 + V12 + Z1 = P2 + V22 + Z2 - - - - - - (v)
ρg
2
ρg
2
Or total energy per unit weight at 1 = total energy per unit
Weight at 2.
QUIZ 1.
A jet of 50mm diameter impinges on a curved vane and is deflected
through an angle of 175. The vane moves in the same direction as
that of jet with a velocity of 35 m/s. If the rate of flow is 170 litres
per second, determine the component of force on the vane in the
direction of motion. How much would be the power developed by
the vane and what would be the water efficiency. Neglect friction.
QUIZ 2.
A jet of water , 50 mm in diameter , issues a velocity of 10 m/s
and impinges on a stationary flat plate which destroys its forward
motion. Find the force exerted by the jet on the plate and the work
done.
QUIZ 3.
A jet of water of diameter 75 mm moving with a velocity of 20 m/s
strikes a fixed plate in such a way that the angle between the jet and
the plate is 60. Find the force exerted by the jet on the plate.
i. In the direction normal to the plate , and
ii in the direction of the jet
WEEK 11: TO KNOW THE MOMENTUM EQUATION AND ITS PRACTICAL
APPLICATIONS
11.1 INTRODUCTION AND UNDERSTANDING OF CONCEPTS OF STEADY
FLOW, STREAMLINE AND STREAMTUBE
We now come to the subject of fluid dynamics: the study of fluids in motion.
We only need to think of the swirling flow that can be seen in the spray behind
a road vehicle travelling in wet weather to realize that the motion of fluids can
be very complex. Flow of this kind is virtually impossible to analyze in detail.
Fortunately, there are many common situations in which analysis is possible
because the flow is steady; that is, the parameters of the flow at any point (the
speed and direction of motion, the pressure, the density etc.) do not vary with
time. Flows of liquid along a pipe, of air over the wing of aircraft or of water
through a pump are just a few of the many cases which can be treated by
steady-flow analysis. For each of the following cases, do you think the flow is
likely to be steady or unsteady?
(a) Flow of water trough the nozzle of a fire-hose
(b) Flow of air over a waving flag
(c) Flow of natural gas along a pipeline.
(a) Steady; (b) Unsteady; (c) Steady
In most instances the rate of flow of water through a fire-hose or of gas along a
pipeline is probably roughly constant, so the flow will be steady. In case of the
flag, even though the speed of the wind may be constant, the flow around the
waving flag is continuously varying as the shape of the flag changes, so the
flow is unsteady. In this book we shall be confining our attention to instances
of steady flow.
Whether the flow is steady or unsteady, it is often useful to draw a diagram of
the pattern of flow of a fluid. A common way to do this is to draw lines which
at every point have the direction of the fluid velocity. These lines are called
streamlines. A typical diagram is shown in the next frame.
The diagram above shows an aerofoil (a shape like the wing of an aircraft or
the blade of a turbine), with the flow around it illustrated by streamlines. A
picture like this is very helpful when we wish to visualize the flow.
When fluid is flowing past a solid boundary, such as the surface of an aerofoil
or the inner surface of a pipe, obviously the fluid cannot flow into or out of the
wall, so its velocity close to the boundary wall must be parallel to the wall.
So, close to the solid boundary, is it possible to have streamlines that are not
parallel to the boundary?
No, the streamlines must be parallel to the boundary wall
At very point the streamlines have the direction of the fluid velocity: this is
how they are defined. Close to the boundary the velocity is parallel to the wall,
so the streamlines must be parallel to the wall also. (There is only one
exception to this: close to a point where the velocity of the fluid is zero, the
streamlines close to a wall are always parallel to it.)
Because the fluid is everywhere moving in the same direction as the
streamlines, fluid can never cross a streamline
Do you think it is possible for streamlines to intersect one another? (What is
the direction of motion of the fluid at the point where the lines cross?
No streamlines cannot intersect one another
If two streamlines were to cross, then at the point of intersection a particle of
fluid would have to be moving in two directions at once, which is of course
impossible.
It is often useful to consider a part of the total flow in isolation from the rest. A
common way of doing this is to imagine a tubular surface, formed by
streamlines, along which the fluid flows. This tubular surface is called a stream
tube.
The diagram above shows a stream tube. Remembering that the ‘wall’ of the
tube is formed of streamlines, can fluid flow through the wall of the tube?
Fluid cannot flow through the ‘wall’ of a stream tube
We have already seen that flow is always along and never across a streamline,
so the fluid cannot escape through a stream tube’s wall, which is entirely
composed of streamlines. In this respect a stream tube is just like a solidwalled pipe.
It differs from a pipe in that if the flow is not steady the streamlines will not
always be in same place; but when the flow conditions at each point do not
vary with time the position of the streamlines do not very, so the walls of the
stream tube are effectively fixed. When the flow does not vary with time we
call it…
WEEK 12:
UNDERSTANDING THE CONCEPT OF MOMENTUM AND FLUID
FLOW
12.1 MOMENTUM AND FLUID FLOW
In mechanics, the momentum of a particle or objectify is defined as the
product of its mass M and its velocity V :
Momentum = MV .
The particles of a fluid stream will possess momentum, and, whenever the
velocity of the stream is changed in magnitude or direction, there will be a
corresponding change in the momentum of the fluid particles. In accordance
with Newton’s second law, a force is required to produce this change, which
will be proportional to the rate at which the change of momentum occurs. The
force may be provided by contact between the fluid and a solid boundary (e.g.
the blade of a propeller or the wall of a bend in a pipe) or by one part of the
fluid stream acting on another. By Newton’s third law, the fluid will exert an
equal and opposite force on the solid boundary or body of fluid producing the
change of velocity. Such forces are known as dynamic forces, since they arise
from the motion of the fluid and are additional to the static forces due to
pressure in a fluid; they occur even when the fluid is at rest.
To determine the rate of change of momentum in a fluid stream consider a
control volume ABCD (as shown below). As the fluid flow is assumed to be
steady and non-uniform in nature the continuity of mass flow across the
control volume may be expressed as.
l 2 A2V2 = l 1 A1V1 = m
Fig 12.1 control volume ABCD
i.e there is no storage within the control volume and m is the fluid mass flow.
The rate at which momentum exits the control volume across boundary
CD may be defined as
l 2 A2 V2V2
Similarly the rate at which momentum enters the control volume across AB
may be expressed as
l 1 A1V1V1
Thus the rate of change of momentum across the control volume may be seen
to be
l 2 A2V2V2 = l1 A1V1V1
Or, from the continuity of mass flow equation,
Note that this is the increase of momentum per unit time in the direction of
motion, and according to Newton’s second law will be caused by a force F ,
such that
This is the resultant force acting on the fluid element ABCD in the direction of
motion.
By Newton’s third law, the fluid will exert an equal and opposite reaction on its
surroundings.
12.2
MOMENTUM EQUATION FOR TWO-AND THREE- DIMENSIONAL
FLOW ALONG A STREAMLINE
The momentum equation F = m(V2 − V1 ) was derived for one-dimensional flow
in a straight line, assuming that the incoming and outgoing velocities V 1 and V2
were in the same direction. The figure below shows a two-dimensional
problem in which v1 makes an angle θ with the x -axis, while V2 makes a
corresponding angle
φ.
Since both momentum and force are vector
quantities, they can be resolved into components in the x and y directions
and equation F = m(V2 − V1 ) applied. Thus, if FX and FY are the components of
the resultant force on the element of fluid ABCD,
Fig 12.2 Element of fluid in pipe ABCD
FX = Rate of change of momentum of fluid in x direction
= Mass per unit x change of fluid in x direction
= m(V2 cos φ − V1 cos θ ) = m(VY 2 − VX 1 )
Similarly,
FY = m[V2 sin φ − V1 sin θ ] = m(VY 2 − VY 1 )
These components can be combined to give the resultant force,
F =  F

2
x
2
+F
Y


Again, the force exerted by the fluid on the surroundings with be equal and
opposite. For three dimensional flows, the same method can be used, but the
fluid will also have component velocities Vz1 and Vz2 in the z direction and the
corresponding rate of change of momentum in this direction will require a
force
(
Fz = m V z − VZ 1
)
QUIZ
Derive the momentum equation and the resultant force for a fluid flow in pipe
for both two and three dimensional flow.
WEEK 13: THE APPLICATION OF MOMENTUM EQUATION
13.0 APPLICATION OF MOMENTUM EQUATION
13.1 IMPACT OF FREE JETS
INTRODUCTION
A fluid jet is a stream of fluid issuing from a nozzle with a high velocity and
hence a high kinetic energy. When a jet impinges on a plate or vane, it exerts a
force on it [due to change in momentum].this force can be evaluated by using
impulse momentum principle. The following cases of impact of jet will be
considered.
13.1.1 FORCE EXERTED BY THE JET ON THE STATIONARY PLATE
1. When flat plate is held normal to the jet;
2. When flat plate is held inclined to the jet;
3. When plate is curved.
13.1.2 FORCE EXERTED BY THE JET ON THE MOVING PLATE.
1. When plate is held normal to the jet;
2. When plate is held inclined to the jet;
3. When plate is curved.
13.1.3 FORCE EXERTED BY THE JET ON A STATIONARY PLATE
Force exerted on a stationary flat plate held Normal to the jet
The Fig. below shows a fluid jet striking a stationary flat plate held
perpendicular to the flow direction. Let a and v be cross sectional area and
velocity of the jet respectively. The jet after striking this plate [vertical],will get
its direction changed through 90 ; but, it will move on and off the plate with
velocity v, if we neglect the friction between the jet and the plate as is possible
when the plate is smooth. If the friction is considered, the velocity of liquid
coming off the plate will be slightly less than v.
The force exerted by the jet on the plate [assuming is smooth] in the
direction of jet. [X direction],
Fx =Rate of change momentum [in the direction of force]
= [Initial momentum-final momentum]
= Impulse momentum principle.
= [mass/sec] x [velocity of jet before striking the plate- velocity of jet
after
striking the plate].
= ρaV [V-0]
Or
Fx = ρaV2
Where ρ = mass density of liquid; a = area of jet and d = diameter of the jet.
It may be noted that a jet leaves in the direction normal to X-axis, the final
velocity in the X-direction is zero
.
13.1.4 FORCE EXERTED ON A STATIONARY FLAT PLATE
HELD INCLINED TO
THE JET
The fig. below shows the stationary flat plate inclined at θ0 to the direction of
horizontal jet. If a and v are the cross-sectional area and velocity of the jet
respectively, then the mass of liquid per second striking the plate
= ρ x aV
After striking the plate (assuming it smooth), the jet leaves the plate with
velocity equal to initial velocity (V).
Let us apply the impulse-momentum equation in the direction normal to the
plate. Force in normal direction, Fn = ρaV (Vsin θ - 0)
= ρaV2 sin θ
This normal force can be resolved into two components; component Fx parallel
to the direction of jet and component Fy, normal to the direction of the jet
Fx = Fn sin θ = ρaV2 sin θ x sin θ = ρaV2 sin2 θ
Fy = Fn cos θ = ρaV2 sin θ x cos θ
13.1.5 FORCE EXERTED ON STATIONARY CURVED PLATE
i.
Jet strikes the curved plate at the centre:
Consider a fluid jet striking a stationary curved plate (smooth) at the centre
shown below. The jet after striking the plate comes out with the same velocity,
in the tangential direction of the curved plate.
The velocity at the outlet of the plate can be resolved into the following two
components:
i.
Component of velocity in the direction of jet = - V cos θ (-ve sign
indicates that the velocity at the outlet is in a direction opposite to that
of the fluid jet)
ii.
Component of velocity perpendicular to the jet = v sin θ
Applying impulse-momentum equation, we have:
Force exerted by the jet (in the direction of jet).
Fx = ρaV (V1x – V2x)
Where
V1x = initial velocity in the direction of jet = V
V2x = final velocity in the direction of jet = - V cos θ
∴
Fx = ρaV (V – (-V cos θ) = ρaV ( V + V cos θ)
Or
Fx = ρaV2 (1 + cos θ)
Similarly,
Fy = ρaV (V1y – V2y)
Where
V1y = initial velocity in the direction of y = 0
V2y = final velocity in the direction of y = V sin θ
∴
Fy = ρaV (0 – V sin θ) = - ρaV2 sin θ
NOTE
The resultant of the force F = Fx2 + Fy2
Example 1: A jet of water, 75mm in diameter, issues with a velocity of 30m/s
and impinges on a stationary flat plate which destroys its forward motion. Find
the force exerted by the jet on the plate.
Solution . Diameter of jet, d = 75mm = 0.075m
Velocity of jet,
V = 30m/s
The force exerted by the jet on a stationary vertical plate is given by
F = ρaV2
Where,
ρ = Mass density of water = 1000 kg/m3
a = Area of jet = π/4 x d2 = π/4 x 0.0752 = 0.004418 m2
∴
F = 1000 x 0.004418 x 302
= 3976.2 N (Ans)
Example 2: A jet of water strikes with a velocity of 35m/s a flat plate inclined at
300 with the axis of the jet. If the cross-sectional area of the jet is 25cm2,
determine:
i.
The force exerted by the jet on the plate.
ii.
The components of the force in the direction normal to the jet.
Solution . Velocity of jet, V = 35m/s
Inclination of the plate with the jet axis, θ = 300
Area of the jet, a = 25cm2 = 25 x 10-4m2
i.
The force exerted by the jet F:
F = ρaV2 sin θ
= 1000 x (25 x 10-4) x 352 x sin 300
= 1531.25N (Ans.)
ii.
The component of force, F:
Fx = F sin θ = 1531.25 x sin 300
= 765.625N (Ans.)
Fy = F cos θ = 1531.25 x cos 300
= 1326.1N (Ans.)
QUIZ 1.
A jet of 50mm diameter impinges on a curved vane and is deflected
through an angle of 175. The vane moves in the same direction as
that of jet with a velocity of 35 m/s. If the rate of flow is 170 litres
per second, determine the component of force on the vane in the
direction of motion. How much would be the power developed by
the vane and what would be the water efficiency. Neglect friction.
QUIZ 2.
A jet of water , 50 mm in diameter , issues a velocity of 10 m/s
and impinges on a stationary flat plate which destroys its forward
motion. Find the force exerted by the jet on the plate and the work
done.
QUIZ 3.
A jet of water of diameter 75 mm moving with a velocity of 20 m/s
strikes a fixed plate in such a way that the angle between the jet and
the plate is 60. Find the force exerted by the jet on the plate.
i. In the direction normal to the plate , and
ii in the direction of the jet
Week 14: FLUID DYNAMICS
Fluid dynamic is concerned with the study of fluids in motion.
14.1 TYPES OF FLOW
1.
Laminar flow: in this case, fluid particles move along in layers or
laminar with one layer sliding over and adjacent layer. The flow is
governed by Newton’s law of viscosity (for one-dimensional flow) it
is also called streamline viscous flow.
2.
Turbulent flow: - This occurs when the fluid particle move in very
irregular paths causing an exchange of momentum from one
portion of a fluid to another. This is also called a non viscous flow.
Laminar flow tends occur when the fluid velocity s small or the
fluid velocity is large or both the turbulent flow sets up greater
shear stresses and causes more mechanical energy to be
converted to thermal energy.
3.
Steady Flow: This is characterized by a steady mass flow rate and
by that across any section at right angles to the flow all properties
are constant with respect to time. True steady flow is found only in
laminar flow. Steady, turbulent flow is said to exist when the mean
velocity of flow at a section remains constant with time.
4.
Unsteady Flow: This occurs when conditions at any point change
with time. An example is the flow of a liquid being pumped
through a fixed system at an increasing rate.
5.
Uniform Flow: This occurs when, at every point the velocity vector
is identical in magnitude and direction at any given instant. That is
du
/ds = o in which time is held constant and S is the displacement
in any direction. This equation states that there is no change in
the velocity vector in any direction through out the fluid at any
instant.
6.
Non-Uniform Flow: This occurs when the velocity vector vari8es
from place to place at any instant i.e. du/ds ≠ o AN example is the
flow of a liquid through a tapered or curved pipe.
To determine whether a particular flow is laminar or turbulent
depends on three particular i.e. velocity, viscosity and density of
the and the cross-sectional area through which the fluid flows.
14.2 Determination of laminar and turbulent flow
Fig 14.1 long tube carrying water
The sketch shows a long tube along which water flows from left to
right. A thin stream of coloured dye is introduced into the water on
the centre-line of the tube. Would you expect the stream of dye to
be straight, as shown below, or confused, as shown below?
Either is possible
When the water is flowing slowly, the motion is orderly, and all the
particles move in straight lines parallel to the axis of the tube. The
stream of dye will then be straight. This kind of flow is called
laminar flow.
By contrast, in a fast-moving flow of water the particles
move in a random way, in addition to the general left to right flow.
The dye rapidly gets thoroughly mixed, and after only a short
distance the water becomes a uniform pale colour. This is called
turbulent flow. But what we do mean here by ‘fast’ and ‘slow’? In
particular, at what does the flow change from orderly to random –
from laminar to turbulent?
This question was investigated by the British engineer
Osborne Reynolds at Manchester during the early 1880s in one of
the classic experiments of fluid mechanics.
He
used
a
thin
filament of coloured dye in water flowing along a glass tube, so
that the pattern traced by the dye could be observed. The water
was stored in a tank in which was allowed to settle for sometime
before each experiment, so that there will be no eddies remaining
in the tank that might affect the behaviour of the water in the
pipe.
Fig 14.2 Reynolds apparatus
Reynolds observed what we have already discoursed: that there
are two kinds of flow. Furthermore, he showed, when the value of
the dimensionless ratio
ρud
µ
Is less than about 2000, the flow is always laminar. (Here u
represent the u represent the mean velocity of the fluid, ρ and µ
are the density and viscosity, and d is the inside diameter of the
tube.) When the value of this ratio is greater than 2000, the flow
is usually turbulent, unless special precautions are taken.
We have already met the ratio
ρud
µ
It is now universally known as Reynolds number. It is denoted by
Re.
Example 1:
Water at 500C flows at mean speed of 0.5m/s along a tube of
5mm diameter. The density of the water is 988 kg/m3 and the
viscosity is 0.548 x 10-3N s/m2. Determine whether the flow is
turbulent or laminar?
Solution
To determine the character of the flow, need to calculate the value
of Reynolds number:
ρud
µ
= 988 x 0.5 x (5 x 10-3)
0.548 x 10-3
= 3830
and this is greater then 2000, so the flow is likely to be turbulent.
Example 2:
A glycerine at temperature 250C is flowing down a pipe of inner
diameter 20mm at a mean speed of 0.5m/s. the density of the
glycerine is 1250kg/m3, and the viscosity µ is 0.942N s/m2.
Determine whether the flow is laminar or turbulent.
Solution:
We need to calculate the Reynolds number, and this time we find
ρud
µ
= 13.3
and this is less than 2000, so the flow is laminar.
Quiz 1:
A pipeline 700mm in diameter carries a flow of methane gas at
150C and gauge pressure 3 bars. The density of the gas is
2.70kg/m3 and its viscosity is 1.15 x 10-5 N s/m2. The mean
velocity of flow of the gas is 0.95 m/s.
Quiz 2
Find the friction factor of the conditions given in questions 1, and
estimate the pressure drop in a 100m length of the pipeline. The
roughness value of the inner surface of the pipeline is k =
0.28mm.
The characteristic no relating this parameters is refers to as the
Reynolds number and is given by,
Reynolds no.
=
ρVd
γ
Where ρ = density of fluid
V = velocity of fluid
d = cross section diameter of pipe
γ = dynamic viscosity of fluid
The following condition holds for the Reynolds no for turbulent and
laminar flow.
Reynolds no < 2100 means a laminar flow
Re
> 2300 means a turbulent flow
Re = 2100 to 2300 means transition from lamina to turbulent
WEEK 15 FLUID POWER MACHINE
Fluid power machine could be described as how energy is added to a
system in order to move the system or convert from one form of energy
to another. This could be classified as either positive displacement or
dynamics. In positive displacement, energy is periodically added to
increase the fluid velocity while in dynamics energy is continuously
added to increase the fluid velocity.
15.1 PUMP
A pump is a machine which provides energy to a fluid in a fluid system. It
assists to increase the pressure energy or kinetic energy or both of the
fluid by converting the mechanical energy.
15.1.1CLASSIFICATION OF PUMP
On the basic of transfer of mechanical energy the pumps can be
classified as follows:1.
Rotor dynamics pumps
i.
Radial flow pump
ii.
Axial flow pump
iii.
Mixed flow pump and
iv.
Centrifugal pump
2.
Positive displacement pump
i.
Reciprocating pump
ii.
Plunger pump
iii.
Piston pump
iv.
Diaghran pump
15.1.2CENTRIFUGAL PUMP
A centrifugal pump consists of the following main component that is
impeller, casing, suction pipe and delivery pipe. The impeller is a wheel
(or rotor) with a series of backward curved vanes (or blades). It is
mounted on a shaft which is usually coupled to an electric motor. The
casing is an air tight chamber surrounding the pump impeller. It contains
suction and discharge arrangements, supporting for bearings and
facilitate to house the rotor assembly. It has a provision to fix stuffing
box and house packing material which prevent external leakage. The
pipe which connect the center/eye of the impeller to sump from which
liquid is to be lifted is known as suction pipe. In order to check the
formation of air pocket the pipe is laid air tight.
The centrifugal pump works on the principles that when a certain mass
of fluid is rotated by an external source, it is thrown away from the
central axis of rotation and a centrifugal head is produce which enable it
to rise to higher level. As the impeller rotate at an angular velocity with
the aid of a shaft and the electric motor, it creates a vacuum that force
the liquid into the housing which subsequently discharge out through
the discharge pipes.
FIG 15.1 CENTRIFUGAL PUMP
15.2 HYDRAULIC TURBINES
Hydraulic turbine is a prime mover (a machine which uses the raw
energy of a substance and converts in to mechanical energy) i.e uses
energy of flowing water and converts it into the mechanical energy. This
mechanical energy is used in running an electric generator which is
directly coupled to the shaft of the hydraulic turbine. From this electric
generator, we get electric power which can be transmitted over along
distances by means of transmission lines.
15.2.1
TYPES OF HYDRAULIC TURBINES
15.2.2
Impulse turbine (pelton wheel)
In impulse turbine the pressure energy of water is converted into kinetic
energy when passed through the nozzle and forms the high velocity jet
of water. The formed water jet is used for driving the wheel.
Pelton wheel among various impulse turbines is a tangential flow
impulse turbine. It consists of a rotor, at the periphery of which
amounted equally spaced double hemispherical buckets. Water is
transferred from a high head source through penstock which is fitted
with a nozzle through which water flows out at a high speed jet. A
needle spear moving inside the nozzle controls the water flow through
the nozzle and the same time provides a smooth flow with negligible
energy loss. All the available potential energy is thus converted into
kinetic energy before the jet strike the buckets of the runner.
A pelton turbine is provided with a casing, the function of which is to
prevent the splashing of water and to discharge water to the tail race.
15.3 REACTION TURBINE
In reaction turbine water enters the wheel under pressure and flow over
the curve blades. When the water flows over the curved blades, the
pressure head is transformed into velocity head. Water leaving the blade
has a large relative velocity but small absolute velocity. Thus practically
whole of the initial energy of water of water is given to the runner.
In these turbines water leaves the runner at atmospheric pressure. The
difference of pressure between the entrance and exit points of the
runner is known as reaction pressure. This reaction pressure is
responsible for the motion of the runner and hence, these turbines are
known as Reaction or pressure turbines.
FIG 15.2 REACTION TURBINE
15.3.1
i.
TYPES OF REACTION TURBINES
Francis turbine. In these type of turbines water under pressure enters
the runner through the guide blade radially in inward direction and
leaves axially i.e. parallel to the axis of shaft. This turbine as already
explained is a medium head [30 to 200 m] turbine and requires medium
quantity of water.
The water under pressure, from the penstock enters the casing which
surrounds the fixed blades and rotating runner. These turbines generally
use scroll casing, which helps in transforming a part of pressure energy
into kinetic energy. Thus this difference of pressure is responsible for the
rotation of runner.
FIG 15.3 Pelton wheel turbine
ii.
Kaplan turbine. Here the runner also has two vanes attached to a hub or
a boss and are mounted in such a way that their angles can be adjusted
while the turbine is in motion. These blades are usually adjusted
automatically by means of a servo motor governing. With the change of
angle of the runner blade at varying load condition, quantity of water
passing through the blades also changes to suit the varying load
conditions. This type of turbine thus has a speciality that it gives good
efficiency even at different load conditions and water flows over the
blades without shock.
Fig 15.4 Kaplan turbine
15.4 Hydraulic Press
The hydraulic press is a device used for lifting heavy loads by the
application of much smaller force. It is based on pascal’s law, which
states that intensity of pressure is transmitted equally in all directions
through a mass of fluid at rest.
Working principle: The working principle of a hydraulic press may be
explained with the help of the fig. below. Consider a ram and plunger,
operating in two cylinders of different diameters, which are interconnected at the bottom, through a chamber, which is filled with some
liquid.
Fig 15.5 Hydraulic pressure
Let W = Weight to be lifted
F = Force applied on the plunger,
A = Area of ram, and
a = Area of plunger
Pressure intensity produced by the force F, P =
F
Area of plunger
= F
a
As per Pascal’s law, the above intensity P will be equally transmitted in all
directions.
∴ The pressure intensity on ram = p = F = W
a
or
A
W=F x A
a
15.5 AIR COMPRESSOR
Air [or gasses] are required at high pressure for many application e.g. in
chemical, industrial and construction field. It is used for operating small
pneumatic hand tools in workshops, for fabrication works, and also for
paint spraying, mining operations etc.
i.
ii.
iii.
iv.
v.
Compressed air has been accepted as more useful power medium and
replacing steam, because of the following advantages.
Compressed air can transmitted from one point to another with ease.
Use of compressed air is safe.
The tools using compressed air do not overheat in use and are cheaper
and easy to maintain.
Pneumatic tools are lighter than similar ones using steam.
15.5.1TYPES OF AIR COMPRESSORS
Air compressor could either be positive displacement or non positive
displacement. Positive displacement compressors are mainly
reciprocating air compressor, rotary air compressor and screw air
compressors. Where non positive displacement type of compressors are
centrifugal compressors and axial-flow compressors.
15.5.2RECIPROCATING COMPRESSOR
A reciprocating compressor is a machine which increases pressure of air
by compression process and delivers it at a higher pressure.
Earlier compressors followed steam engine of large bore and stroke, few
cylinder, slow speed, double acting, horizontal construction. With the
development of internal combustion engine, air compressors nowadays
are of small bore, short stroke, multi cylinder, high speed, vertical
construction. The reciprocating air compressors resemble an internal
combustion engine in construction. In this a piston moves in a cylinder
through the motion of a crank shaft. Inlet and outlet vales are provided,
as in I.C. engines, internal lubrication will also done like I.C. engines, with
the result of small amount of oil is always present in the air deliver by
the compressor.
Prime-movers employed on air compressors are generally electric motor
and diesel engine. Electric motors are used were electric supply is
available and compressor is stationary. Where as diesel engine are used
to drive portable compressor and for stationary unit when electric is
available.
15.5.3ROTARY AIR COMPRESSORS
Rotary machines which are used for supplying air or any other fluid are
called fans or compressors. A fan is a machine which moves gaseous
fluids with a pressure ratio of up to 1:15. A compressor is a machine
which moves gaseous fluids with a pressure of ratio of more than 1:15.
Rotary compressors have adiabatic compression. Their speed is very high
and are small in size, as compared to reciprocating compressors, due to
high speed and can be directly coupled to turbine, electric motors etc.
Rotary compressors deliver more clean air and uniform delivery of air is
obtained.