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FARHAN HAIDER
LECTURER
MECHANICAL DEPARTMENT
MEHRANUET JAMSHORO
FLUID MECHANICS 1

 Fluid is a substance that alters its shape in response to
any force however small, that tends to flow or to conform
to the outline of its container, and that includes gases and
liquids and mixtures of solids and liquids capable of flow.
 A fluid is defined as a substance that deforms
continuously when acted on by a shearing stress of any
magnitude.

CHARACTERISTICS OF FLUIDS
 Gas or liquid state
 “Large” molecular spacing relative to a solid
 “Weak” intermolecular cohesive forces
 Can not resist a shear stress in a stationary state
 Will take the shape of its container
 Viscosity distinguishes different types of fluids

DEFINITIO N
 Mechanics: The oldest physical science that deals with both
stationary and moving bodies under the influence of forces.
 Fluid mechanics:
The science that deals with the behavior of fluids at rest (fluid
statics) or in motion (fluid dynamics), and the interactionof
fluids with solids or other fluids at the boundaries.

PROPERTIES OF FLUIDS
 Density = r (RHO)
 The density of a fluid is defined as mass per unit volume (kg/m3 )
• Different fluids can vary greatly in density
• Liquids densities do not vary much with pressure and temperature
• Gas densities can vary quite a bit with pressure and temperature
• Density of water at 4° C (standard): 1000 kg/m3
• Density of Air at 4° C (standard): 1.20 kg/m3
 Alternatively, Specific Volume:
v
m

r
r

1


SPECIFIC WEIGHT
 The specific weight of fluid is its weight per unit volume.
• Specific weight characterizes the weight of the fluid (How heavy is it)
• Specific weight of water at 4° C : 9.80 kN/m3
• Specific weight of air at 4° C : 11.9 N/m3
g
r
 

SPECIFIC GRAVITY
 The specific gravity of fluid is the ratio of the density of the
fluid to the density of water @ 4° C.
• Gases have low specific gravities
• A liquid such as Mercury has a high specific gravity, 13.6
• The ratio is unitless.
• Density of water at 4° C : 1000 kg/m3
O
H
SG
2
r
r


VISCOSITY
 Viscosity is a measureof a fluid's resistanceto flow. It describes the
internal friction of a moving fluid.
 A fluid with large viscosity resists motion becauseits molecularmakeup
gives it a lot of internal friction.
 A fluid with low viscosity flows easily because its molecularmakeup
results in very little friction when it is in motion
 Viscosity Varies from Fluid to Fluid and is dependent on temperature,
 Units of Viscosity are N·s/m2 or lb·s/ft2

 One way is to measure a fluid’s resistance to flow when an external force is
applied. This is Dynamic Viscosity.
 The other way is to measure the resistive flow of a fluid under the weight of
gravity. The result is Kinematic Viscosity. Put another way, kinematic
viscosity is the measure of a fluid’s inherent resistance to flow when no external
force, except gravity, is acting on it.
• Kinematic viscosity is anotherway of representing viscosity
• Used in the flow equations
• The units are of L2/T or m2/s and ft2/s
r

 

Compressibility of Fluids: Bulk Modulus
• Measureof how pressurecompresses the volume/density
• Large values of the bulk modulusindicate incompressibility
• Incompressibilityindicates large pressures are needed to compress the
volumeslightly
• It takes 3120 psi to compress water 1% at atmosphericpressureand 60°
F.
• Most liquids are incompressiblefor most practical engineering
problems.
• Units of the bulk modulusare N/m2 (Pa) and lb/in.2 (psi).
r
r

/
d
dp
E 
P is pressure, and r is the density.

SURFACE TENSION
 At the interface between a liquid and a gas or two immiscible liquids, forces
develop forming an analogous “skin” or “membrane” stretched over the fluid mass
which can support weight.
 This “skin” is due to an imbalance of cohesive forces. The interior of the fluid is
 in balance as molecules of the like fluid are attracting each other while on the
 interface there is a net inward pulling force.
 Surface tension is the intensity of the molecular attraction per unit length along
 any line in the surface.
 Surface tension is a property of the liquid type, the temperature, and the other
fluid at the interface.
 This membrane can be “broken” with a surfactant which reduces the surface
tension.

Surface Tension: Liquid Drop

The pressure inside a drop of fluid can be calculated using a free-body diagram:
Real Fluid Drops Mathematical Model
R is the radius of the droplet, s is the surface tension, Dp is the pressure
difference between the inside and outside pressure.
The force developed around the edge due to surface tension along the line:
s
R
Fsurface 2

This force is balanced by the pressure difference Dp:
2
R
p
Fpressure 
D

Applied to Area
Applied to Circumference

 Now, equating the Surface Tension Force to the Pressure
Force, we can estimate Dp = pi – pe:
 This indicates that the internal pressure in the droplet is
greater that the external pressure since the right hand side
is entirely positive.
R
p
s
2

D

This indicates that the internal pressure in the droplet is greater that the
external pressure since the right hand side is entirely positive.
Surface tension is apparent in many practical problems such as movement of
liquid through soil and other porous media, flow of thin films, formation of
drops and bubbles, and the breakup of liquid jets.
Adhesion > Cohesion Cohesion > Adhesion

 Capillary action is found in thermometer where fluid used
in it automatically rises when comes in contact with higher
temperature or falls down with lower ones.
 Capillary action can be performed to transfer fluid from one
vessel to another on its own.
 Capillary action can be experienced in the half dipped cloth
as well as on lantern.

SUMMARY
 INTRODUCTION
 FLUID CHARACTERISTICS
 FLUID PROPERTIES

EVERY ENDING IS
ACTUALLY A NEW
BEGINNING…

FLUID ENERGY AND
CLASSIFICATION OF
FLUIDS

Fluid energy
 Kinetic energy : Energy due to the motion. Energy in
fluid due to its flow is kinetic energy of the fluid.
 Potential Energy: Energy in the fluid by virtue of its
position . Mostly the position is taken in terms of
altitude.
 Pressure energy: Energy in the fluid by existing
pressure (force/unit area) is called pressure energy.

 The pressure energy in an incompressible fluid volume, like a
pressurized tank with water, can be expressed as
 E = dp / ρ (1)
 where
 E = energy (J/kg)
 dp = pressure difference (Pa, N/m2)
 ρ = density (kg/m3)
 Example - Pressure Energy in a Water Tank
 The pressure on a water tank is 10 bar (106 Pa). With density of
water 1000 kg/m3 the pressure energy can be calculated as
 E = (106 Pa) / (1000 kg/m3)
 = 1000 J/kg


 Pressure energy is the energy stored in a fluid due to the force per unit area
applied onto it whereas kinetic energy is the energy of fluid molecules due to
motion of fluid.
 In the above picture, if the piston is at rest, the system only has pressure energy
whereas if the piston is in motion, the system has both, pressure, as well as,
kinetic energy.
 Note: Whenever a quantity of fluid is transferred , the energy it contains also
transfers.

Two Categories of Fluids
Newtonian (Newtonian Law of viscosity)
 A Newtonian fluid's viscosity remains constant, no
matter the amount of shear applied for a constant
temperature.. These fluids have a linear relationship
between viscosity and shear stress.
 Examples:
 Water
 Mineral oil
 Gasoline
 Alcohol
Shear stress
Measured in (SI unit): pascal
Commonly used symbols: τ

A shear stress, is
applied to the top of
the square while the
bottom is held in
place. This stress
results in a strain,
or deformation,
changing the square
into a
parallelogram.

 Shear forces acting tangentially to a surface of a solid
body cause deformation. In contrast to solids that
can resist deformation, liquids lack this ability, and
flow under the action of the force. When the fluid is
in motion, shear stresses are developed due to the
particles in the fluid moving relative to one another.

In common terms, this means the fluid continues to flow,
regardless of the forces acting on it. For example, water is
Newtonian, because it continues to exemplify fluid
properties no matter how fast it is stirred or mixed.
For a Newtonian fluid, the viscosity, by definition,
depends only on temperature and pressure (and also the
chemical composition of the fluid if the fluid is not a pure
substance), not on the forces acting upon it.

Non-Newtonian
A non-Newtonian fluid is a fluid whose flow properties
are not described by a single constant value of viscosity.
Many polymer solutions and molten polymers are non-
Newtonian fluids, as are many commonly found substances
such as ketchup, starch suspensions, paint, blood and
shampoo.
In a non-Newtonian fluid, the relation between the shear
stress and the strain rate is nonlinear, and can even be
time-dependent. Therefore a constant coefficient of
viscosity cannot be defined.

1. Bingham plastics.
They have linear shear stress and strain relationship which require
a finite yield stress before they begin to flow, i.e., the shear stress-
strain curve doesn’t pass through origin.
Eg- clay suspensions, drilling mud, toothpaste, mayonnaise,
chocolate, and mustard. The classic case is ketchup which will not
come out of the bottle until you stress it by shaking.
Types of non
Newtonian fluids

2
. Pseudoplastic Flow
shear thinning is the non-Newtonian behavior of
fluids whose viscosity decreases under shear strain.
Polymers in solutions such as , sodium alginate,
methylcellulose
Viscosity decreases with an increase in shear.
Caused by the re-alignment of polymer and/or the release of
solvents associated with the polymers.
Examples : paint, and nail polish.

3. Dilatant Flow
Dilatant fluids, also known as shearthickening fluids, are liquids or
solutionswhoseviscosity increases as stress is applied
Shear thickening
Suspensioncontaining high-concentrationof small deflocculated
particles

The relationship of shear
stress-stain for all fluids:

THIXOTROPY
 Is the property of some non-Newtonian pseudoplastic
fluids to show a time-dependent change in viscosity; the
longer the fluid undergoes shear stress, the lower its
viscosity.
 A thixotropic fluid is a fluid which takes a finite time to
attain equilibrium viscosity when introduced to a step
change in shear rate.
 the term is sometimes applied to pseudoplastic fluids
without a viscosity/time component. Many gels and
colloids are thixotropic materials, exhibiting a stable form
at rest but becoming fluid when agitated. (greese)

RHEOPECTIC.
 These show a time-dependent increase in viscosity; the
longer the fluid undergoes shearing force, the higher
itsviscosity. Rheopectic fluids, such as some lubricants,
thicken or solidify when shaken. Examples of rheopectic
fluids include gypsum pastes and printer inks.

Dimensional Analysis
Relationship between physical quantities involved in a fluid
phenomenon by considering their dimensions.
we group these parameters into dimensionless combinations
which enable a better understanding of the flow phenomena.
Dimensional analysis is particularly helpful in experimental
work.
it provides a guide to those things that significantly influence the
fluid flow phenomena.

Dimensional Analysis refers to the physical
nature of the quantity (Dimension) and the
type of unit used to specify it.
Distance has dimension L.
Area has dimension L2.
Volume has dimension L3.
Time has dimension T.
Speed has dimension L/T

Application of Dimensional Analysis
 Development of an equation for fluid
phenomenon
 Conversion of one system of units to another
 Reducing the number of variables required in
an experimental program
 Develop principles of hydraulic similitude for
model study

Dimensional Reasoning & Homogeneity
 Both sides of the equation must be numerically and
dimensionally identical.
 To take a simple example, the expression x + y = z when x
=1, y =2 and z =3 is clearly numerically true but only if the
dimensions of x, y and z are identical. Thus
1 elephant +2 aeroplanes =3 days is clearly nonsense but
1 metre +2 metre =3 metre is wholly accurate.
An equation is only dimensionally homogeneous if all the
terms have the same dimensions.

Fundamental Dimensions
 We may express physical quantities in either mass-length-
time (MLT) system or force-length-time (FLT) system.
This is because these two systems are interrelated through
Newton’s second law, which states that force equals mass
times acceleration,
F = ma 2nd Law of motion
F = ML/T2
F = MLT-2
 Through this relation, we can covert from one system to the
other. Other than convenience, it makes no difference
which system we use, since the results are the same.

Dimensions of Some Common
Physical Quantities
[x], Length – L
[m], Mass – M
[t], Time – T
[v], Velocity – LT-1
[a], Acceleration – LT-2
[F], Force – MLT-2
[Q], Discharge – L3T-1
[r], Mass Density – ML-3
[P], Pressure – ML-1T-2
[E], Energy – ML2T-2

How do dimensions behave in
mathematical formulae?
Rule 1 - All terms that are added or
subtracted must have same dimensions
C
B
A
D 


All have identical dimensions

Rule 2 - Dimensions obey rules of
multiplication and division
L]
[
]
[L
[M]
[L]
]
[T
]
[T
[M]
2
2
2



























C
AB
D

Rule 3 - In scientific equations, the arguments of
“transcendental functions” must be dimensionless.
x
D
x
B
x
C
x
A
3
)
exp(
)
sin(
)
ln(




x must be dimensionless
Exception - In engineering correlations, the argument may have
dimensions

Methods for Dimensional Analysis
 Rayleigh’sMethod
 Buckingham’s ∏-method

Rayleigh’s Method
Functional relationship betweenvariables is expressed
in the form of an exponential relation which must be
dimensionallyhomogeneous.
if “y” is a function of independentvariables
x1,x2,x3,…..xn, then
In exponential form as
)
,.......
,
,
( 3
2
1 n
x
x
x
x
f
y 
]
)
,.......(
)
(
,
)
(
,
)
[( 3
2
1
z
n
c
b
a
x
x
x
x
y 


Procedure
 Write fundamental relationship of the given data
 Write the same equation in exponential form
 Select suitable system of fundamental dimensions
 Substitute dimensions of the physical quantities
 Applydimensional homogeneity
 Equate the powers and compute the values of the
exponents
 Substitute the values of exponents
 Simplify the expression
 Ideal up to three independent variables, can be used
for four.
Rayleigh’s Method

 To illustrate the basic principles of dimensional analysis, let us
explore the equation for the speed V with which a pressure wave
travels through a fluid.
 Certainly the compressibility Ev must be factor; also the density
and the kinematic viscosity of the fluid might be factors. The
dimensions of these quantities, written in square brackets are
V=[LT-1], Ev=[FL-2]=[ML-1T-2], r=[ML-3], ν=[L2T-1]
V=C Ev
a rb νd
Where C is a dimensionless constant, and let solve for the
exponents a, b, and d substituting the dimensions, we get

(LT-1)=(ML-1T-2)a (ML-3)b (L2T-1)d
To satisfy dimensional homogeneity, the exponents of each
dimension must be identical on both sides of this equation.
Thus
For M: 0 = a + b
For L: 1 = -a -3b +2d
For T: -1 = -2a – d
Solving these three equations, we get
a=1/2, b=-1/2, d=0
So that V = C √(Ev/r)
This identifies basic form of the relationship,
Dimensional analysis along such lines was developed by Lord
Rayleigh.

g
m
L
p
Pendulum - What is the period?
c
b
a
L
g
m
k
p 
2
/
1
c
b
0
0
L]
[
2
/
1
0
2
0
1
T]
[
0
0
0
0
M]
[
L]
[
T
L
[M]
T]
[ c
2
a
























c
b
b
a
a
b
g
L
k
p
L
g
km
p 

  2
/
1
2
/
1
0

Buckingham’s ∏ method
 This arranges the variables into a lesser number of
dimensionless groups of variables. Because Buckingham used
∏ (pi) to represent the product of variables in each groups, we
call this method Buckingham pi theorem.
 “If ‘n’ is the total number of variables in a dimensionally
homogenous equation containing ‘m’ fundamental
dimensions, then they may be grouped into (n-m) ∏ terms.
f(X1, X2, ……Xn) = 0
Then the functional relationship will be written as
Ф (∏1 , ∏2 ,………….∏n-m) = 0
The final equation obtained is in the form of:
∏1= f (∏2,∏3,………….∏n-m) = 0
 Suitable where n ≥ 4
 Not applicable if (n-m) = 0

Procedure
 List all physical variables and note ‘n’ and ‘m’.
n = Total no. of variables
m = No. of fundamental dimensions (That is, [M], [L], [T])
 Compute number of ∏-terms by (n-m)
 Write the equation in functional form
 Write equation in general form
 Select repeating variables. Must have all of the ‘m’
fundamental dimensions and should not form a ∏ among
themselves
 Solve each ∏-term for the unknown exponents by
dimensional homogeneity.

 Let us apply Buckingham’s ∏ method to an example problem that of
the drag forces FD exerted on a submerged sphere as it moves through a
viscous fluid.
 Step 1: Visualize the physical problem, consider the factors that are of
influence and listand count the n variables.
f (D, V, ρ, μ) = FD
Here we used D, the sphere diameter, to represent sphere size, and f
stands for “some function”. We see that n = 5.
Note that the procedure cannot work if any relevant variables are
omitted. Experimentation with the procedure and experience will help
determinewhich variables are relevant.

MLT-2 , L , LT-1 , ML-3 , ML-1T-1
We see that M, L and T are involved in this example. So m = 3.
 Step 3: Determine n-m, the number of dimensionless ∏ groups
needed. In our example this is 5 – 3 = 2, so we can write Ф(∏1. ∏2)
= 0
 Step 4: Form the ∏ groups by multiplying the product of the
primary (repeating) variables, with unknown exponents, by each
of the remaining variables, one at a time. We choose ρ, D, and V
as the primary variables. Then the ∏ terms are
∏1 = Da Vb ρc FD
∏2 = Da Vb ρc μ-1

 Step 5: To satisfy dimensional homogeneity, equate the
exponents of each dimension on both sides on each pi equation
and so solve for the exponents
∏1 = Da Vb ρc FD = (L)a (LT-1)b (ML-3)c (MLT-2) = M0L0T0
Equate exponents:
L: a +b -3c +1 = 0
M: c +1 = 0
T: -b -2 = 0
We can solve explicitly for
b = -2, c = -1, a = -2
Therefore
∏1 = D-2 V-2 ρ-1 FD = FD/(ρ V2 D2)

Finally, add viscosity to D, V, and ρ to find ∏2 . Select any power
you like for viscosity. By hindsight and custom, we select the
power -1 to place it in the denominator
∏2 = Da Vb ρc μ-1 = (L)a (LT-1)b (ML-3)c (ML-1T-1 )-1 = M0L0T0
Equate exponents:
L: a +b -3c +1 = 0
M: c -1 = 0
T: -b +1 = 0
We can solve explicitly for
b = 1, c = 1, a = 1
Therefore,
∏2 = D1 V1 ρ1 μ-1 = (D V ρ)/(μ) =R = Reynolds Number
R = Reynolds Number= Ratio of inertia forces to viscous forces
Check that all ∏s are in fact dimensionless

Rearrange the pi groups as desired. The pi theorem states that
the ∏s are related. In this example hence
FD/(ρ V2 D2) = Ф (R)
So that FD = ρ V2 D2 Ф (R)
We must emphasize that dimensional analysis does not provide
a complete solution to fluid problems.
It provides a partial solution only. The success of dimensional
analysis depends entirely on the ability of the individual using
it to define the parameters that are applicable.
If we omit an important variable. The results are incomplete.

No slip condition
 the no-slip condition for viscous fluids assumes
that at a solid boundary, the fluid will have zero
velocity relative to the boundary. The fluid velocity at
all fluid–solid boundaries is equal to that of the solid
boundary.
 If the solid is moving then the velocity of the
adjacent layer of fluid will be the velocity of the fluid.

 What did you see?
 The diet coke floats & the regular sinks.
 Why does the diet float??
 Regular soda contains 39 grams of sugar.
 Diet coke contains 100 mg of Nutra-
sweet.
 More "stuff" (matter) is crammed into the
same amount of space, or VOLUME, and
that increases the MASS.
 The relationship of Mass to Volume is
Density.

ARCHIMEDES PRINCIPLE
 Archimedes' principle states that the upward buoyant
force that is exerted on a body immersed in a fluid, whether
fully or partially submerged, is equal to the weight of the
fluid that the body displaces and acts in the upward
direction at the center of mass of the displaced fluid.
 Any object, wholly or partially immersed in a fluid, is
buoyed up by a force equal to the weight of the fluid
displaced by the object.

 Any floating object displaces its own weight of fluid.
 In other words,for an object floating on a liquid surface(like a boat) or
floating submerged in a fluid (like a submarine in water or dirigible in
air) the weight of the displaced liquid equals the weight of the object.

 Consider a 1-ton block of solid iron. As iron is nearly eight
times as dense as water, it displaces only 1/8 ton of water
when submerged, which is not enough to keep it afloat.
 Suppose the same iron block is reshaped into a bowl. It still
weighs 1 ton, but when it is put in water, it displaces a
greater volume of water than when it was a block.
 The deeper the iron bowl is immersed, the more water it
displaces, and the greater the buoyant force acting on it.
When the buoyant force equals 1 ton, it will sink no farther.

 Metacentre, also spelled metacenter, in fluid mechanics, the
theoretical point at which an imaginary vertical line passing
through the centre of buoyancy and centre of gravity intersects
the imaginary vertical line through a new centre of buoyancy
created when the body is displaced, or tipped, in the water,
however little.

 The water in a tank is pressurized by air, and the pressure is measured by a
Multi-fluid manometer as shown in Figure. The tank is located on a mountain
at an altitudeof 1400mwhere the atmospheric pressure is 85.6 kPa.Determine
the air pressure in the tank if h1 = 0.1 m, h2 = 0.2 m, and h3 =0.35 m. Take
the densities of water, oil, and mercury to be 1000 kg/m3,850 kg/m3, and
13,600kg/m3, respectively.

 A crane is used to lower weights into the sea (density 1025 kg/m3) for
an underwaterconstructionproject figure. Determine the tension in the
rope of the crane due to a rectangular0.4-m 0.4-m 3-m concrete
block(density 2300 kg/m3) when it is
(a) suspendedin the air and
(b) completely immersed in water.

Analytical Method for Metacentric Height:
 Considering a ship floating freely in water. Let the ship
be given a clockwise rotation through a small angle q (in
radians) as shown in Fig. The immersed section has now
changed from acde to acd1e1.

 The original centre of buoyancy B has now changed to a new
position B1. It may be noted that the triangular wedge ocn has
gone under water. Since the volume of water displaced
remains the same, therefore the two triangular wedges must
have equal areas.
 A little consideration will show, that as the triangular wedge
oam has come out of water, thus decreasing the force of
buoyancy on the left, therefore it tends to rotate the vessel in
an anti-clockwise direction.

 Similarly, as the triangular wedge ocn has gone under water,
thus increasing the force of buoyancy on the right, therefore it
again tends to rotate the vessel in an anticlockwise direction.

 It is thus obvious, that these forces of buoyancy will form
a couple, which will tend to rotate vessel in anticlockwise
direction about O. If the angle (q), through which the
body is given rotation, is extremely small, then the ship
may be assumed to rotate about M (i.e., metacentre).
 Let l =length of ship
b=breadth of ship
q=Very small angle (in radian) through which the
ship is rotated
V=Volume of water displaced by the ship

From the geometry of the figure, we find that
am=cn=bq/2
--Volume of wedge of water aom
= ½ (b/2 x am)xl
= ½ (b/2 x bq/2)l (am = bq/2)
= b2ql/8
--Weight of this wedge of water
= gb2ql/8 (g=sp. Wt. of water)
--And arm of the couple = 2/3 b
--Moment of the restoring couple
= (gb2ql/8) x (2/3 b) = gb3ql/12 …(i)

--And moment of the disturbing force
= g . V x BB1 …(ii)
--Equating these two moments (i & ii),
gb3ql/12 = g x V x BB1
--Substituting values of:
lb3/12 = I (i.e. moment of inertia of the plan of the ship) and
BB1 = BM x q in the above equation,
g . I . q = g x V (BM x q)
BM = I/V
BM= Moment of inertia of the plan/ Volume of water
displaced
second area moment, is a geometrical property of an area
which reflects how its points are distributed with regard to an
arbitrary axis

--Now metacentric height,
Note: +ve sign is to be used if G is lower than B and,
–ve sign is to be used if G is higher than B.
BG
BM
GM 
=


A wooden cylinder of length L and diameter L/2 is
floating on water with its axis vertical. Find the
metacentric height if the specific gravity of wood is
0.6
 Solution :Given data: Length of the cylinder = L
 Diameter of the cylinder = L/2
 Specific gravity of wood = 0.6

 For flotation weight of the cylinder should be equal to
weight of the water displaced.
 Now if the depth of immersion is h,
 Weight of water displaced = Weight of the cylinder


h = 0.6L

Answer: - 0.1739 L m. (The negative sign indicates the body is in unstable
equilibrium.)
A

 A flatboat hull that, when fully loaded,weighs 150 kN. Parts (b)–(d)
show the top, front, and side views of the boat, respectively. Note the
locationof the center of gravity, cg. Determine whether the boat is
stable in fresh water. First, find out whether the boat will float.

Because the center of gravity is above the center of buoyancy, we must locate the
metacenter to determine whether the boat is stable.

 Is the boat stable? Yes, it is. Because the metacenter is above the
center of gravity, the boat is stable. That is, ymc > ycg.

Outline
• Hydrostatic Force on a Plane Surface
• Pressure Prism
• Hydrostatic Force on a Curved Surface

Hydrostatic Force on a Plane Surface: General Case
General Shape: Planar
View, in the x-y plane
q is the angle the plane makes
with the free surface.
y is directed along the plane
surface.
The origin O is at the Free
Surface.
A is the area of the surface.
dA is a differential element
of the surface.
dF is the force acting on
the differential element.
C is the centroid.
CP is the center of Pressure
FR is the resultant force
acting through CP

Hydrostatic Force on a Plane Surface: General Case
Then the force acting on the differential element:
Then the resultant force acting on the entire surface:
With g and q taken as constant:
We note, the integral part is the first moment of area about the x-axis
Where yc is the y coordinate to the centroid of the object.
We note h = ysinq
hc

Hydrostatic Force on a Plane Surface: Location
Now, we must find the location of the center ofPressurewherethe ResultantForce Acts:
“TheMomentsof the ResultantForce mustEqualthe Momentof the Distributed Pressure Force”
We note,
Moments aboutthe x-axis:
Then,
Secondmomentof Intertia,Ix
ParallelAxis Thereom:
Ixc is the secondmomentof inertia through the centroid
Substitutingthe parallelAxis thereom,and rearranging:
We, note that for a submergedplane,the resultantforce always acts below the centroidof the
plane.
And, note h = ysinq

Hydrostatic Force on a Plane Surface: Location
Moments aboutthe y-axis: 

A
R
R xdF
x
F
And, note h = ysinq
We note,
Then,
Secondmomentof Intertia,Ixy
ParallelAxis Thereom:
Ixc is the secondmomentof inertia through the centroid
c
c
xyc
xy y
Ax
I
I 

Substitutingthe parallelAxis thereom,and rearranging:

Hydrostatic Force on a Plane Surface: Geometric Properties
Centroid Coordinates
Areas
Moments of Inertia

Hydrostatic Force: Vertical Wall
Find the Pressure on a Vertical Wall using Hydrostatic Force Method
Pressure varies linearlywith depth by the hydrostatic equation:
The magnitudeof pressure atthe bottom is p = gh
The width of the wall is “b” into the board
The depth ofthe fluid is “h” into the board
By inspection,the average pressure
occursat h/2, pav = gh/2
The resultantforce actthrough the center ofpressure,CP:
 
h
h
h
y
h
bh
h
bh
y
R
R
3
2
2
6
2
2
12
3





O
yR = 2/3h
y-coordinate: 3
12
1
bh
Ixc 
2
h
yc 
bh
A 

Pressure Prism: Submerged Vertical Wall
 
1
2 h
h
b
A 

 A
h
F 1
1 g

Trapezoidal
 
 A
h
h
F 1
2
2
2
1

 g
The ResultantForce:break into two “volumes” Locationof ResultantForce:“usesum of moments”
Solve for yA
y1 and y2 is the centroidlocation for the two
volumes where F1 and F2 are the resultantforces of
the volumes.

Hydrostatic Force on a Curved Surface
• General theory of plane surfaces does not apply to curved surfaces
• Many surfaces in dams, pumps, pipes or tanks are curved
•
Isolated Volume
Bounded by AB an AC
and BC
Then we mark a F.B.D. for the volume:
F1 and F2 is the hydrostatic force on
each planarface
FH and FV is the componentof the
resultantforce on the curvedsurface.
W is the weightof the fluid volume.

Hydrostatic Force on a Curved Surface
Now, balancing the forces for the Equilibrium condition:
Horizontal Force:
Vertical Force:
Resultant Force:
The location of the Resultant Force is through O by sum of Moments:
H
H
V
V
c
x
F
x
F
x
F
Wx
x
F



2
2
1
1
Y-axis:
X-axis:

Hydrostatic Forces On Submerged Surfaces
• A plate exposed to a liquid, such as a gate valve in a dam, the
wall of a
liquid storage tank, or the hull of a ship at rest, is subjected to
fluid
pressure distributed over its surface.
• Hydrostatic forces form a system of parallel forces,
• we need to determine the magnitude of the force;
• and its point of application,
Hydrostatic Forces On Submerged ForcesOn Plane Surfaces
center of pressure
39/118

When analyzing hydrostatic forces
on submerged surfaces:
the atmospheric pressure can be
subtracted for simplicity when it
acts on both sides of the structure.

Hydrostatic force on an inclined plane
surface completely submerged in
a liquid

absolute pressure at any point on the plate
• resultant hydrostatic force:
• y-coord of the centroid (or center) of the surface:

where PC is the pressure at the centroid of the surface,
which is equivalent to the average pressure on the surface,

The line of action of the resultant hydrostatic force, in general, does not
pass through the centroid of the surface. it lies underneath where the
pressure is higher.(Center of Pressure)
• Two parallel force systems are equivalent if they have the same
magnitude and the same moment about any point.
• equating the moment of the resultant force
line of action of the resultant force

The Ixx are widely available for common shapes in engineering
handbooks, but they are usually given about the axes passing through the
centroid of the area.
• Fortunately, the second moments of area about two parallel axes are
related to each other by the parallel axis theorem,

For areas that possess symmetry about the y-axis, the center of
pressure lies on the y-axis directly below the centroid.

Special Case: Submerged Rectangular Plate

When the upper edge of the plate is at the free surface and thus s=0

Vertical surface of the liquid block considered is simply the projection
of the curved surface on a vertical plane,
Horizontal surface is the projection of the curved surface on a
horizontal plane

The exact location of the line of action of the resultant force
(e.g., its distance from one of the end points of the curved surface) can be
determined by taking a moment about an appropriate point.

The hydrostatic force acting on a
circular
surface always passes through the
center
of the circle since the pressure forces
are
normal to the surface and they all pass
through the center.

HYDROSTATICS
 Fluid statics is the study of fluid at rest, the concept
includes situations where the fluid s are either actually at
rest or undergo uniform acceleration in a container or to
rotate as a solid mass.
 No shear force is then present as the fluid particles do not
move with respect to one another.

Shear Forces Normal Forces
(pressure)
where F is a force normal to area A

Hydrostatic Law
 Many fluid problems do not involve motion. They concern
the pressure distribution in a static fluid and its effect on
solid surfaces and on floating and submerged bodies.
 Fluids at rest cannot support shear stress.
 Pressure is used to indicate the normal force (compressive)
per unit of area at a given point acting on a given plane
within the fluid mass of interest.

Hydrostatic pressure in gases


HYDROSTATICS LAW : VARIATION OF
PRESSURE IN A STATIC FLUID
 states that “as the distance from a datumsurface level
increases the pressure drops”.
 When an object pushed in a fluid to its bottom, it compresses
the fluid. so there arise an increase in pressure force around it
and is termed as “Hydrostatic force.”
 A projectile soaring higher altitudesfrom earth surface
experiences decrease in pressure and
 similarly underwater missiles experience increase in
hydrostatic force as it penetrates into the depths of sea waters.


Increase in pressure while targeting beneath altitudes

Decrease in pressure while soaring higher
altitudes.

 Similarly, a diver experiences more pressure as he
goes down to the depths of the sea levels, and
 similarly a tumbled mug when pushed to bottom of a
water filled bucket, experience more upward force
and this can be proved by taking a cylindrical fluid
element having an area and length as dA,ds
respectively, at some height from the datum, as
shown in Figure


CASE 1 : If the Cylindrical element is vertical then angle will be 90,

 Case 2: When the cylindrical element is horizontal
then angle will be zero,

Hydraulic Lift
In a closed system, pressure changes from one point are transmitted
throughout the entire system
(Pascal’s Law).

 Find the weight that can be lifted by a hydraulicpress when the force
applied at the plunger is 350 N and has the diameters of 250 mm and
40 mm of ram and plunger respectively, take the specific weight of the
liquid in the press as 1000 N/m3.

 Step 1 : Find area of ram
 Find area of plunger
 Normal pressure due to force applied on plunger (F/A)
 Add pressure caused by liquid level rise of 0.3m
(Sp.weight x height) = total pressure intensity
 Putting total pressure intensity, find weight (P=W/A)
ram.

• PROBLEM 03 – 0096: In the steady flowshown in the sketch below,
find the equation of the streamline passing the point (1, 2, 3)
if u = 3ax, v = 4ay and w = – 7az.
Equation for streamline
Substituting values equation 5

• EXAMPLE 6.6: In a two-dimensional, incompressible flow the fluid velocity
components are given by: u = x – 4y and v = -y - 4x. Show that the flow
satisfies the continuity equation and obtain the expression for the stream
function. If the flow is potential (irrotational) obtain also the expression
for the velocity potential.

1
Internal flows through pipes, elbows, tees, valves,
etc., as in this oil refinery, are found in nearly
every industry.

2
Objectives
• Have a deeper understanding of laminar and
turbulent flow in pipes and the analysis of fully
developed flow
• Calculate the major and minor losses associated
with pipe flow in piping networks and determine
the pumping power requirements
• Understand various velocity and flow rate
measurement techniques and learn their
advantages and disadvantages

3
8–1 ■ INTRODUCTION
• Liquid or gas flow through pipes or ducts is commonly used in heating and
cooling applications and fluid distribution networks.
• The fluid in such applications is usually forced to flow by a fan or pump
through a flow section.
• We pay particular attention to friction, which is directly related to the pressure
drop and head loss during flow through pipes and ducts.
• The pressure drop is then used to determine the pumping power requirement.
Circular pipes can withstand large pressure differences
between the inside and the outside without undergoing any
significant distortion, but noncircular pipes cannot.

4
Theoretical solutions are obtained only for a few simple cases such as fully
developed laminar flow in a circular pipe.
Therefore, we must rely on experimental results and empirical relations for
most fluid flow problems rather than closed-form analytical solutions.
Average velocity Vavg is defined
as the average speed through a
cross section. For fully developed
laminar pipe flow, Vavg is half of
the maximum velocity.
The average
velocity for
incompressible
flow in a circular
pipe of radius R

5
8–2 ■ LAMINAR AND
TURBULENT FLOWS
The behavior of
colored fluid
injected into the
flow in laminar
and turbulent
flows in a pipe.
Laminar: Smooth
streamlines and highly
ordered motion.
Turbulent: Velocity
fluctuations and highly
disordered motion.
Transition: The flow
fluctuates between
laminar and turbulent
flows.
Most flows encountered
in practice are turbulent.
Laminar flow is encountered when
highly viscous fluids such as oils flow
in small pipes or narrow passages.
Laminar and
turbulentflow
regimes of
candle smoke.

6
Reynolds Number
The transition from laminar to turbulent
flow depends on the geometry, surface
roughness, flow velocity, surface
temperature, and type of fluid.
The flow regime depends mainly on the
ratio of inertial forces to viscous
forces (Reynolds number).
The Reynolds number can be
viewed as the ratio of inertial
forces to viscous forces
acting on a fluid element.
Critical Reynolds number, Recr:
The Reynolds number at which the
flow becomes turbulent.
The value of the critical Reynolds
number is different for different
geometries and flow conditions.

7
At large Reynolds numbers, the inertial forces, which are
proportional to the fluid density and the fluid velocity, are
large relative to the viscous forces, and thus the viscous
forces cannot prevent the random and rapid fluctuations of
the fluid (turbulent).
At small or moderate Reynolds numbers, the viscous forces
are large enough to suppress these fluctuations and to keep
the fluid “in line” (laminar).

9
For flow through noncircular
pipes, the Reynolds number
is based on the hydraulic
diameter
For flow in a circular pipe:
In the transitional flow region of 2300 
Re  10,000, the flow switches between
laminar and turbulent seemingly
The hydraulic diameter Dh = 4Ac/p is
defined such that it reduces to
ordinary diameter for circular tubes.

10
8–3 ■ THE ENTRANCE REGION
Velocity boundary layer: The region of the flow in which the effects of the
viscous shearing forces caused by fluid viscosity are felt.
Boundary layer region: The viscous effects and the velocity changes are
significant.
Irrotational (core) flow region: The frictional effects are negligible and the
velocity remains essentially constant in the radial direction.
The development of the velocity boundary layer in a pipe. The developed average
velocity profile is parabolic in laminar flow.

11
Hydrodynamic entry length Lh: The length of this region (from intial
point to the fully developed velocity profile).
Hydrodynamically developing flow: Flow in the entrance region.
This is the region where the velocity profile develops.
Hydrodynamically fully developed region: The region beyond the
entrance region in which the velocity profile is fully developed and
remains unchanged.
Hydrodynamically fully developed
In the fully developed flow
region of a pipe, the
velocity profile does not
change downstream, and
thus the wall shear stress
remains constant as well.

13
The variation of wall shear stress in the flow direction for flow in a pipe
from the entrance region into the fully developed region.
The pressure drop is higher in the entrance regions of a pipe, and
the effect of the entrance region is always to increase the average
friction factor for the entire pipe.

14
Entry Lengths
The hydrodynamic entry length is usually taken to be the distance
from the pipe entrance to where the wall shear stress (and thus
the friction factor) reaches within about 2 percent of the fully
developed value.
hydrodynamic
entry length for
laminar flow
hydrodynamic
entry length for
turbulentflow
hydrodynamic entry
length for turbulent flow,
an approximation
The pipes used in practice
are usually several times
the length of the entrance
region, and thus the flow
through the pipes is often
assumed to be fully
developed for the entire
length of the pipe.
This simplistic approach gives
reasonable results for long
pipes but sometimes poor
results for short ones since it
under-predicts the wall shear
stress and thus the friction
factor.

15
8–4 ■ LAMINAR FLOW IN PIPES
We consider steady, laminar, incompressible flow of a fluid with constant
properties in the fully developed region of a straight circular pipe.
In fully developed laminar flow, each fluid particle moves at a constant axial
velocity along a streamline and the velocity profile u(r) remains unchanged in
the flow direction. There is no motion in the radial direction, and thus the
velocity component in the direction normal to the pipe axis is everywhere zero.
There is no acceleration since the flow is steady and fully developed.
Free-body diagram of a ring-shaped
differential fluid element of radius r,
thickness dr, and length dx oriented
coaxially with a horizontal pipe in fully
developed laminar flow.

17
Free-body diagram of a fluid disk element
of radius R and length dx in fully developed
laminar flow in a horizontal pipe.
Maximim velocity
at centerline
Velocity
profile

18
Pressure Drop and Head Loss
A pressure drop due to viscous effects represents an irreversible
pressure loss, and it is called pressure loss PL.

19
pressure loss for
all types of fully
developed
internal flows
dynamic
pressure
Darcy
friction
factor
Circular
pipe,
laminar
Head
loss
In laminar flow, the friction factor is a function of the
Reynolds number only and is independent of the
roughness of the pipe surface.
The head loss represents the additional height that the
fluid needs to be raised by a pump in order to overcome
the frictional losses in the pipe.

20
The relation for pressure loss (and
head loss) is one of the most
general relations in fluid
mechanics, and it is valid for
laminar or turbulent flows, circular
or noncircular pipes, and pipes with
smooth or rough surfaces.
Horizontal
pipe
Poiseuille’s
law

21
The pumping power requirement for a laminar flow piping system can be
reduced by a factor of 16 by doubling the pipe diameter.
For a specified flow rate, the pressure drop and thus the required
pumping power is proportional to the length of the pipe and the
viscosity of the fluid, but it is inversely proportional to the fourth power
of the diameter of the pipe.

22
The pressure drop P equals the pressure loss PL in the case of a
horizontal pipe, but this is not the case for inclined pipes or pipes with
variable cross-sectional area.
This can be demonstrated by writing the energy equation for steady,
incompressible one-dimensional flow in terms of heads as

23
The friction factor f relations
are given in Table 8–1 for fully
developed laminar flow in
pipes of various cross
sections. The Reynolds
number for flow in these pipes
is based on the hydraulic
diameter Dh = 4Ac /p, where
Ac is the cross-sectional area
of the pipe and p is its wetted
perimeter
Laminar Flow in
Noncircular Pipes

30
8–5 ■ TURBULENT FLOW IN PIPES
The intense mixing in turbulent flow
brings fluid particles at different
momentums into close contact and
thus enhances momentum transfer.
Most flows encountered in engineering practice are turbulent,and thus it is
important to understand how turbulence affects wall shear stress.
Turbulent flow is a complex mechanism dominated by fluctuations, and it is still
not fully understood.
We must rely on experiments and the empirical or semi-empirical correlations
developed for various situations.
Turbulent flow is characterized by
disorderly and rapid fluctuations of swirling
regions of fluid, called eddies, throughout
the flow.
These fluctuations provide an additional
mechanism for momentum and energy
transfer.
In turbulent flow, the swirling eddies
transport mass, momentum, and energy to
other regions of flow much more rapidly
than molecular diffusion, greatly enhancing
mass, momentum, and heat transfer.
As a result, turbulent flow is associated
with much higher values of friction, heat
transfer, and mass transfer coefficients

31
Water exiting a tube: (a) laminar
flow at low flow rate,(c) turbulent
flow

33
At very large Reynolds numbers, the friction factor curves on the Moody chart
are nearly horizontal, and thus the friction factors are independent of the
Reynolds number. See Fig. A–12 for a full-page moody chart.

34
Types of Fluid Flow Problems
1. Determining the pressure drop (or head
loss) when the pipe length and diameter
are given for a specified flow rate (or
velocity)
2. Determining the flow rate when the pipe
length and diameter are given for a
specified pressure drop (or head loss)
3. Determining the pipe diameter when the
pipe length and flow rate are given for a
specified pressure drop (or head loss)
The three types of problems
encountered in pipe flow.

43
8–6 ■ MINOR LOSSES
The fluid in a typical piping system passes through various fittings, valves,
bends, elbows, tees, inlets, exits, enlargements, and contractions in addition
to the pipes.
These components interrupt the smooth flow of the fluid and cause additional
losses because of the flow separation and mixing they induce.
In a typical system with long pipes, these losses are minor compared to the total
head loss in the pipes (the major losses) and are called minor losses.
Minor losses are usually expressed in terms of the loss coefficient KL.
Head loss due
to component

44
When the inlet diameter equals outlet
diameter, the loss coefficient of a
component can also be determined by
measuring the pressure loss across the
component and dividing it by the dynamic
pressure:
KL = PL /(V2/2).
When the loss coefficient for a component
is available, the head loss for that
component is
Minor
loss
Minor losses are also expressed in terms
of the equivalent length Lequiv.
The head loss caused by a
component (such as the angle
valve shown) is equivalent to the
head loss caused by a section of
the pipe whose length is the
equivalent length.

45
Total head loss (general)
Total head loss (D = constant)
The head loss at the inlet of a pipe
is almost negligible for well-
rounded inlets (KL = 0.03 for r/D >
0.2) but increases to about 0.50 for
sharp-edged inlets.

49
The effect of rounding
of a pipe inlet on the
loss coefficient.
Graphical representation of flow
contraction and the associated head
loss at a sharp-edged pipe inlet.

50
All the kinetic energy of the flow is “lost”
(turned into thermal energy) through friction
as the jet decelerates and mixes with ambient
fluid downstream of a submerged outlet.
The losses during changes of direction
can be minimized by making the turn
“easy” on the fluid by using circular
arcs instead of sharp turns.

51
(a) The large head loss in a
partially closed valve is due
to irreversible deceleration,
flow separation, and mixing
of high-velocity fluid coming
from the narrow valve
passage.
(b) The head loss through a
fully-open ball valve, on the
other hand, is quite small.

54
8–7 ■ PIPING NETWORKS AND PUMP SELECTION
A piping network in an industrial facility.

55
For pipes in series, the flow rate is the same in
each pipe, and the total head loss is the sum
of the head losses in individual pipes.

56
For pipes in parallel, the head loss is the same in each
pipe, and the total flow rate is the sum of the flow rates in
individual pipes.

57
The relative flow rates in parallel pipes are established from the
requirement that the head loss in each pipe be the same.
The analysis of piping networks is based on two simple principles:
1. Conservation of mass throughout the system must be satisfied.
This is done by requiring the total flow into a junction to be equal to the
total flow out of the junction for all junctions in the system.
2. Pressure drop (and thus head loss) between two junctions must be
the same for all paths between the two junctions.
The flow rate in one of the parallel branches is proportional
to its diameter to the power 5/2 and is inversely proportional
to the square root of its length and friction factor.

58
the steady-flow
energy equation
When a pump moves a fluid from one reservoir
to another, the useful pump head requirement
is equal to the elevation difference between
the two reservoirs plus the head loss.
The efficiency of the pump–motor
combination is the product of the
pump and the motor efficiencies.
Piping Systems with Pumps and Turbines

66
8–8 ■ FLOW RATE AND VELOCITY MEASUREMENT
A major application area of fluid mechanics is the determination of the
flow rate of fluids, and numerous devices have been developed over the
years for the purpose of flow metering.
Flowmeters range widely in their level of sophistication, size, cost,
accuracy, versatility, capacity, pressure drop, and the operating principle.
We give an overview of the meters commonly used to measure the flow
rate of liquids and gases flowing through pipes or ducts.
We limit our consideration to incompressible flow.
A primitive (but fairly accurate) way of
measuring the flow rate of water through a
garden hose involves collecting water in a
bucket and recording the collection time.
Measuring the flow rate is usually done by
measuring flow velocity, and many flowmeters are
simply velocimeters used for the purpose of
metering flow.

67
Pitot and Pitot-Static Probes
(a) A Pitot probe measures stagnation pressure at the nose of the
probe, while (b) a Pitot-static probe measures both stagnation
pressure and static pressure, from which the flow speed is calculated.
Pitot probes (also called Pitot tubes) and Pitot-static probes are widely used
for flow speed measurement.
A Pitot probe is just a tube with a pressure tap at the stagnation point that
measures stagnation pressure, while a Pitot-static probe has both a
stagnation pressure tap and several circumferential static pressure taps and
it measures both stagnation and static pressures

68
• At a stagnation pointthe fluid velocity is zero and all kinetic energy
has been converted into pressure energy(isentropically). Stagnation
pressure is equal to the sum of the free-stream dynamic pressureand
free-stream static pressure.
•Stagnation pressure is sometimes referred to as pitot pressure
because it is measured using a pitot tube.

69
Measuring flow velocity with a
Pitotstatic probe. (A manometer
may be used in place of the
differential pressure transducer.)
Close-up of a Pitot-static probe, showing
the stagnation pressure hole and two of
the five static circumferential pressure
holes.

70
Obstruction Flowmeters:
Orifice, Venturi, and
Nozzle Meters
Flow through a constriction in a pipe.
Flowmeters based on this principle
are called obstruction flowmeters
and are widely used to measure
flow rates of gases and liquids.

73
The total amount of mass or volume of a fluid that
passes through a cross section of a pipe over a
certain period of time is measured by positive
displacement flowmeters.
There are numerous types of displacement
meters, and they are based on continuous filling
and discharging of the measuring chamber. They
operate by trapping a certain amount of incoming
fluid, displacing it to the discharge side of the
meter, and counting the number of such
discharge–recharge cycles to determine the total
amount of fluid displaced.
A nutating disk flowmeter.
A positive
displacement
flowmeter with
double helical
three-lobe
impeller design.
Positive Displacement Flowmeters

75
•Turbine flowmeters use the mechanical energy of the fluid to rotate
a “pinwheel” (rotor) in the flow stream. Blades on the rotor are
angled to transform energy from the flow stream into rotational
energy.
•Shaft rotation can be sensed mechanically or by detecting the
movement of the blades. Blade movement is often detected
magnetically, with each blade or embedded piece of metal
generating a pulse.
•The transmitter processes the pulse signal to determine the flow of
the fluid. Transmitters and sensing systems are available to sense
flow in both the forward and reverse flow directions.
Turbine Flowmeters

77
Paddlewheel Flowmeters
Paddlewheel flowmeter to
measure liquid flow, with
flow from left to right, and a
schematic diagram of
its operation.
Paddlewheel flowmeters are low-cost
alternatives to turbine flowmeters for
flows where very high accuracy is not
required.
The paddlewheel (the rotor and the
blades) is perpendicular to the flow
rather than parallel as was the case
with turbine flowmeters.

78
Variable-Area Flowmeters (Rotameters)
A simple, reliable, inexpensive, and easy-to-install
flowmeter with reasonably low pressure drop and
no electrical connections that gives a direct reading
of flow rate for a wide range of liquids and gases is
the variable-area flowmeter, also called a
rotameter or floatmeter.
A variable-area flowmeter consists of a vertical
tapered conical transparent tube made of glass or
plastic with a float inside that is free to move.
As fluid flows through the tapered tube, the float
rises within the tube to a location where the float
weight, drag force, and buoyancy force balance
each other and the net force acting on the float is
zero.
The flow rate is determined by simply matching the
position of the float against the graduated flow
scale outside the tapered transparent tube.
The float itself is typically either a sphere or a
loose-fitting piston-like cylinder.
Two types of variable-area
flowmeters: (a) an ordinary
gravity-based meter and (b) a
spring-opposed meter.

79
Ultrasonic Flowmeters
The operation of a transit time ultrasonic
flowmeter equipped with two transducers.
Ultrasonic flowmeters operate using sound waves in the ultrasonic range
( beyond human hearing ability, typically at a frequency of 1 MHz).
Ultrasonic (or acoustic) flowmeters operate by generating sound waves with
a transducer and measuring the propagation of those waves through a
flowing fluid.
There are two basic kinds of ultrasonic flowmeters: transit time and
Doppler-effect (or frequency shift) flowmeters.
L is the distance between the transducers and K is a constant

80
Doppler-Effect
Ultrasonic
Flowmeters
Doppler-effect
ultrasonic flowmeters
measure the average
flow velocity along the
sonic path.
The operation of a Doppler-effect ultrasonic
flowmeter equipped with a transducer pressed
on the outer surface of a pipe.
Ultrasonic clamp-on flowmeters enable one to
measure flow velocity without even contacting
(or disturbing) the fluid by simply pressing a
transducer on the outer surface of the pipe.

81
Electromagnetic Flowmeters
A full-flow electromagnetic flowmeter is a nonintrusive device that consists of a
magnetic coil that encircles the pipe, and two electrodes drilled into the pipe
along a diameter flush with the inner surface of the pipe so that the electrodes
are in contact with the fluid but do not interfere with the flow and thus do not
cause any head loss.
Insertion electromagnetic flowmeters operate similarly, but the magnetic field is
confined within a flow channel at the tip of a rod inserted into the flow.

82
Vortex Flowmeters
The operation of a
vortex flowmeter.
This suggests that the flow rate can be determined by generating vortices in the
flow by placing an obstruction in the flow and measuring the shedding frequency.
The flow measurement devices that work on this principle are called vortex
flowmeters.
The Strouhal number, defined as St = fd/V, where f is the vortex shedding
frequency, d is the characteristic diameter or width of the obstruction, and V is
the velocity of the flow impinging on the obstruction, also remains constant in this
case, provided that the flow velocity is high enough.
The vortex flowmeter has the
advantage that it has no moving
parts and thus is inherently
reliable, versatile, and very
ccurate (usually 1 percent over a
wide range of flow rates), but it
obstructs the flow and thus
causes considerable head loss.

•
In fluid dynamics, the drag coefficient (commonly denoted as: , or
) is a dimensionless quantitythat is used to quantify the drag or
resistance of an object in a fluid environment,such as air or
water. ... The drag coefficient is always associated with a
particularsurface area

•
Determine the maximum discharge of water that can be carried out by a 10 cm
Venturimeterwhich has the coefficient of velocity as 0.95, having inlet pressure 100kpa
And pressure at throat 30kpa.

PRESSURE MEASURING DEVICES
•

INTRODUCTION
Pressure means force per unit area, exerted by a fluid
on the surface of the container.
P=F/A
F - FORCE (in Newton)
A - AREA (in meter²)
Pressureis of two types-
STATIC PRESSURE
DYNAMIC PRESSURE
WHERE,
F

Pressure is of two types-
1- STATIC PRESSURE
2- DYNAMIC PRESSURE
STATIC PRESSURE- when the force in a system under
pressure is constant or static (i.e. unvarying), the
pressure is said to be static pressure.
DYNAMIC PRESSURE- If the force is varying, on the
other hand, the pressure is said to be dynamic pressure.

1 atm = 14.7 Psi at sea level
= 101.3 Kilo Pascal
= 760 mm of Hg
= 10.3 m of water
= 1013 mili bar
1 Pascal = 1N/m2
1 Bar = 100 Pascal

Principle:- The bourdon tube works on a simple
principle that a bent tube will change its shape .
As pressure is applied internally, the tube
straightens and returns to its original form when
the pressure is released .
The tip of the tube moves with the internal
pressure change and is easily converted with a
pointer onto a scale.

Advantages:-
 Inexpensive
 Wide operating range
 Fast response
 Good sensitivity
 Direct pressure measurements
Disadvantages:-
 Sensitive to temperature variations
 Limited life when subjected to shock and vibrations
 Applications:-
 These devices should be used in air if calibrated for air and
in liquid if calibrated for liquids.

Diaphragm Gauge of pressure
measurements
 The diaphragm is a flexible disc, which can be either flat or with concentric
corrugations and is made from sheet metal with high tolerance dimensions.
 The diaphragm can be used as a means of isolating the process fluids, or
for high pressure applications.
 It is also useful in providing pressure measurement with electrical
transducers (LVDT).
 It is used as primary pressure transducers in many dynamic pressure
measuring devices.

Advantages :-
Good for low pressure
Inexpensive
Wide range
Reliable and proven
Used to measure gauge, atmospheric and
differential pressure.

Linear variable differential
transformer(LVDT)
 This type of pressure measurement relies on the movement
of a high permeability core within transformer coils . the
movement is transferred from the process medium to the
core by use of a diaphragm .


• Working principle :-
• The LVDT operates on the inductance ratio
between the coils. The primary coil is located
between two secondary coils and is energized
with an alternating current . Equal voltage are
induce in the secondary coil if the core is in
the center when the core is moved from the
center position the result of the voltages in
the secondary winding will be different.

• This manometer consist of U-shaped
tube in this manometric fluid is filled.
• Water and mercury are used as a
manometric fluid.
• Advantage of using these fluid is that
mass density of these fluid can be
obtained easily and they do not stick to
the tube.

Since, P = ρgh
h = (P₁ - P₂)/ρg
P₁ - P₂ = ρgh
Where, ρ - mass density of fluid
g - gravity
P₁ - unknown pressure
P₂ - atmospheric pressure

The well type manometer is not widely used because
of inconvenience; the reading of only a single leg is
required in it.
It consist of a very large-diameter vessel (well)
connected on one side to a very small-sized tube.
Thus the zero level moves very little when pressure is
applied.

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