Chapter 4

• Introduction to Fluid Dynamics
• Control Surface/Area/Volume
• Flow Type
• Streakline, Streamline & Pathline
Introduction to fluid dynamics
• Fluid dynamics is a study of fluid in motions.
• It has several subdisciplines including aerodynamics (the
study of air and other gases in motion) and
hydrodynamics (the study of liquids in motion).
• Fluid dynamics has a wide range of applications e.g :
forces and moments calculation on aircraft, determining
the mass flow rate of petroleum through pipelines,
predicting weather patterns etc.
• The solution to a fluid dynamics problem typically
involves calculating various properties of the fluid, such
as velocity, pressure and flow rate.
Mean or Average Velocity
• The velocity in the pipe normally is not constant
across the cross section.
• Crossing the centreline of the pipe, the velocity is zero
at the walls increasing to a maximum at the centre
then decreasing symmetrically to the other wall.
• This variation across the section is known as the
velocity profile or distribution. A typical one is shown
in the figure below.
Steady flow - A flow in which the fluid
properties such as velocity, temperature
and pressure at a point in the system do
not change over time.
Unsteady flow - A flow in which at least
one variable at a fixed point in the flow
changes with time .
Uniform flow - The velocity of the fluid has
the same magnitude and direction at any
position in the fluid flow.
Non-uniform flow – The velocity
magnitude and direction are not equal at
any position in the fluid flow.
Compressible flow – ?
Incompressible flow - ?
Streakline, Streamline & Pathline
Streamlines — A streamline is
formed by tangents of the
velocity field of the flow.
Pathlines — A pathline can be
formed from fluid particles of
different colour originated
from the same points, such as
a line formed after the
introduction of ink into a
shallow water flow.
Streaklines — A streakline
represents a locus made by a
miniature particles or tracers
that passes at a same point.
In a steady flow, fluid particle normally moves along a
streamline…so streaklines, streamlines and pathlines are the
same (coincide).
Control Surface/Area/Volume
 In fluid mechanics, it is more convenient to work with control volume.
 Control volume = a specific region, space or area chosen for study.
 1D analysis – although simple (almost not describe the actual situation) but
it gives a useful engineering estimate.
• Mass flow rate, m
• Volume flow rate, Q
• Continuity Equation
• Derivation of the Continuity Equation
Mass flow rate, m
• The amount of mass flowing through a cross section
per unit time is called the mass flow rate and is
 . The dot over a symbol is used to indicate
denoted by m
time rate of change.
• If we want to measure the rate at which water is
flowing along a pipe, a very simple way of doing this is
to collect all the water coming out of the pipe in a
bucket over a fixed time period.
• Measuring the weight of the water in the bucket and
dividing this by the time taken to collect this water
gives a rate of accumulation of mass or the mass flow
• For example an empty bucket weighs 2.0 kg. After 7
seconds of collecting water the bucket weighs 8.0 kg, so:
•Performing a similar calculation, if we know the
mass flow is 1.7 kg/s, how long will it take to fill
a container with 8 kg of fluid?
Volume flow rate, Q
• More commonly we need to know is the discharge or the
volume flow rate (or simply called flow rate).
• The volume flow rate, Q is defined as the volume of the fluid
flowing through a cross-section per unit time or.
Volume, 
or Q  velocity area  VA,
Time, t
• The SI unit for Q is m3/s.
• However, in practice the unit commonly used is l/s (litre per
• Multiplying this by the density of the fluid gives us the mass flow rate,
m  Q  kg / s
• Consequently, if the density of the fluid in the above example is 850
kg/m3 then:
Example 1
A garden hose attached with a nozzle is
used to fill a 10-gal bucket. The inner
diameter of the hose is 2 cm, and it
reduces to 0.8 cm at the nozzle exit. If it
takes 50 s to fill the bucket with water,
(a) the volume and mass flow rates of
water through the hose,
(b) the average velocity of water at the
nozzle exit.
•Matter cannot be created or destroyed but it
is simply changed into a different form of
•This principle is know as the conservation of
mass and it is used in the analysis of flowing
fluids to derive the continuity equation.
• Generally;
Mass entering per unit time = Mass leaving per unit time + Increase of mass
in the control volume per unit time
• For steady flow there is no increase in the mass within the
control volume ,
Mass entering per unit time = Mass leaving per unit time
Derivation of the Continuity Equation
• Consider a streamtube such as shown
at the figure in the left side.
• A liquid is flowing from left to right and
the pipe is narrowing in the same
direction. No fluid flows across the
boundary so mass only enters and
leaves through the two ends of this
streamtube section.
• By the continuity principle, the mass
flow rate must be the same at each
section - the mass going into the pipe is
equal to the mass going out of the pipe.
• So we can write:
1A1V1 = 2A2V2
 Or for steady flow,
1A1V1 = 2A2V2 = constant
 This equation is called as continuity equation.
 When the fluid is considered as incompressible (i.e. the density does not
change or ρ1 = ρ2 = ρ), therefore,
A1V1= A2V2= Q
 This is the form of the continuity equation most often used.
• Introduction to Bernoulli Equation
• Limitations on the Use of the Bernoulli Eqn.
• Derivation of the Bernoulli Equation
• Other forms of Bernoulli Equation
• Stagnation pressure
• Bernoulli Equation’s Applications
- orifice in reservoir
- orifice in pipes
- pitot tube
- venturi meter
- weirs
Introduction to Bernoulli Equation
• Bernoulli Equation is the most popular equation in fluid
• The Bernoulli equation is an approximate relation between
pressure, velocity, and elevation, and is valid in regions of
steady, incompressible flow where net frictional forces are
• It is derived from Euler equation (Leonhard Euler – the
greatest Swiss mathematician).
• The Bernoulli equation is one of the most frequently used
and misused equations in fluid mechanics. Its versatility,
simplicity, and ease of use make it a very valuable tool for
use in analysis, but the same attributes also make it very
tempting to misuse. Therefore, it is important to
understand the restrictions on its applicability and observe
the limitations on its use.
Limitations on the Use of the Bernoulli Eqn.
1.Steady flow - applicable to steady flow.
2. Frictionless flow - Every flow involves some friction, no matter how
small, and frictional effects may or may not be negligible. The situation
is complicated. But in general, frictional effects are negligible for short flow
sections with large cross sections, especially at low flow velocities.
(viscousity assumed as constant or no fiction losses occur).
3. Incompressible flow - One of the assumptions used in the derivation
of the Bernoulli equation is that ρ = constant and thus the flow is
incompressible. This condition is satisfied by liquids and also by gases
at Mach numbers less than about 0.3 since compressibility effects and
thus density variations of gases are negligible at such relatively low
4. Flow along a streamline -The Bernoulli equation is applicable along a streamline,
and the value of the constant C, in general, is different for different streamlines.
Others : no shaft work like pump/turbine, etc. (Refer cengel)
Derivation of the Bernoulli Equation
• Consider the motion of a fluid particle in a flow field in steady flow.
Other forms of Bernoulli Equation
2. Total head - by dividing each term of
the Bernoulli equation (1) by g.
pressure head + velocity head + elevation head = C
3. Total pressure - by multiplying each
term of the Bernoulli equation (1) by
density ρ.
1. Total energy
static pressure + dynamic pressure + hydrostatic pres. = C
commonly used
The Bernoulli equation between two points can be written as :
Or in the form of total head eqn. :
Stagnation pressure
• The stagnation pressure represents the pressure at a
stagnation point, where the fluid is brought to a
complete stop isentropically ( or fluid particle velocity
equal to zero).
• Eg. of stagnation point :
The stagnation pressure is the sum of the static and
dynamic pressures and it is expressed as
When static and stagnation pressures are measured at
a specified location, the fluid velocity at that location
can be calculated from :
Bernoulli Equation’s Applications
• The Bernoulli equation can be applied to a great many
• In the following sections we will see some examples of
its application in flow measurement from tanks, within
pipes as well as in open channels.
Bernoulli Equation’s Applications
Its include
•the pitot tube
•and also the devices to measure flowrate
in pipes and conduits such as orifice and
venturi meters.
The operation of each is based on the
same physical principles - an increase in
velocity causes a decrease in pressure.
The difference between them normally is a
matter of cost and accuracy.
Bernoulli Equation’s Applications
Flow Through Orifice at Reservoir
• Orifice = small opening.
• E.g of application : drain outlet at
• Purpose : to calculate the rate/discharge at
which a fluid will flow from a tank so that
the time it takes to drain can be calculated.
•Consider a flow of liquid from a tank through
a hole in the wall side close to the tank base.
The general arrangement and a close up of
the hole and streamlines are shown in the
figure below.
•At the streamlines, we can see the flow contract
after the orifice to a minimum value.
•This convergence is called the vena contracta (the
point in a fluid stream where the diameter of the
stream is the least ).
•At this point, the velocity and pressure are
uniform across the jet and it is necessary to know
the amount of contraction to allow us to calculate
the actual flow.
•This can be done by using a coefficient of
contraction, Cc to calculate the actual area and
this will be shown later.
• We can predict the velocity at the orifice using the
Bernoulli equation.
• Apply it along the streamline joining point 1 on the surface
to point 2 at the centre of the vena contracta.
• We take point 1 to be at the free surface of water so that
P1=Patm (at the atmosphere), also P2=Patm (water discharges
into the atmosphere). V1≈0 (the tank is large relative to the
outlet), and z1=h and z2=0 (we take the reference level at
the center of the outlet).
• Then the Bernoulli equation simplifies to
• Leaving
• This is the theoretical value of velocity. It will be over
estimate of the real velocity because friction losses
have not been taken into account.
• To incorporate friction, we use the coefficient of
velocity, Cv to correct the theoretical velocity.
Vactual = CvVtheoretical
• Each orifice has its own coefficient of velocity, they
usually lie in the range of 0.97 - 0.99.
• To calculate the discharge through the orifice
we multiply the area of the jet by the velocity.
The actual area of the jet is the area of the
vena contracta not the area of the orifice.
• We obtain this area by using a coefficient of
contraction for the orifice
Aactual= Cc Aorifice
 So the discharge through the orifice is given by
Q  AV
 AactualVactual
 Cc AorificeCvVtheoretical
 Cc Cv AorificeVtheoretical
 Cd AorificeVtheoretical
 Cd Aorifice 2 gh
 Where coefficient of discharge, Cd is given by Cc multiply by Cv.
 The discharge coefficient Cd depends on the configuration of the
General steps to solve problem
involving Bernoulli Eqn.
Normally the Bernoulli equation is used to determine
the flowrate, pressure or velocity at a given point. The
following steps might be useful to solve the problems
involving Bernoulli equation.
i. Consider the figure and identify point 1 and 2.
ii. Write the Bernoulli eqn between the two points.
iii. List out and substitute all known information in
Bernoulli eqn.
iv. Use continuity eqn if needed.
v. Solve the equation to determine the required value.
Example 1
A large tank open to the
atmosphere is filled with water
to the height of 5 m from the
outlet tap. A tap near the
bottom of the tank is now
opened, and water flows out
from the smooth and rounded
outlet. Determine the water
velocity at the outlet.
Example 2
An orifice 50 mm in diameter is discharging water
under a head of 10 m. If Cd = 0.6 and Cv = 0.97, find
actual velocity and actual discharge of the jet at
vena contracta.
Example/Exercise 3
Orifice Meter
Orifice plate
• A plate which has an opening in
it, smaller than the internal
diameter of the pipeline.
• Installed in a pipe which
equipped with pressure
measuring devices.
Advantages :
• Simple design
• Low cost (compared to venturi
Disadvantage :
• Low Cd/larger energy losses
• A typical value for a sharp edged
orifice being 0.65.
• The orifice plate produces a constriction of the flow
as shown, the cross sectional area A2 of the flow
immediately downstream of the plate being
approximately the same as that of the orifice.
• The theoretical discharge can be calculated from
Bernoulli and Continuity equation but the actual
discharge may be as little as two-third of this value.
A coefficient of discharge also must be introduced.
Fig 6.2: Orifice meter eqquiped with U-tube manometer
Example 1
For figure below, find the flowrate in pipe.
Example 2
An orifice meter having an inside diameter of 2.5 cm is located in a 8
cm pipe. Water is flowing through the pipe and the mercury
manometer measures the differential pressure over the pipe. When
the manometer reading is 35 cm, what is the flow rate of water per
Exercise 1 (from tutorial)
An orifice plate is used to measure the rate of air flow
through a 2 m diameter duct. The mean velocity in the
duct will not exceed 15 m/s and the water tube
manometer, having a maximum difference between
water levels of 150 mm, is to be used. Assuming the
coefficient of discharge to be 0.64, determine a
suitable orifice diameter to make full use of the
manometer range. Take the density of air as 1.2
Exercise 2
• What diameter orifice hole, d, is needed if under ideal conditions the
flowrate through the orifice meter is to be 113L/min of seawater
with p1 – p2 = 16.34 kPa. The contraction coefficient is assumed to be
Pitot tube
 The pitot tube is a simple and
inexpensive way used to measure
the velocity of a fluid stream.
 Consists of a simple L- shaped tube
with opened end,facing into the
oncoming flow.
 Use the principle of stagnation
point - a particle is brought to rest
at stagnation point result a greater
pressure at this point.
pitot tube
- Attached with piezometer – to
measure static pressure while
pitot tube is to measure dynamic
pitot tube static
• In the Pitot-static tube, the inner tube is
used to measure the impact/dynamic
pressure when the outer sheath has holes
in its surface to measure the static
Example 1
 A piezometer and a pitot tube are
tapped into a horizontal water pipe
to measure static and stagnation
pressures. For the water column
heights, determine the velocity at
the center of the pipe.
Ans : 1.53 m/s
Example 2
The specific gravity for manometric fluid as shown in figure
below is 1.07. Calculate the flow rate, Q if the fluid flowing
in the pipe are;
i. water
ii. air
Exercise 1
An airplane flies 44.7 m/s at an elevation of 3048 m in a standard atmosphere.
Determine the pressure at point (1) far ahead of the airplane, the pressure at the
stagnation point on the nose of the airplane (point 2), and the pressure
difference indicated by a pitot-static probe attached the the fuselage.
Air flows through a pipe at a rate of 200 L/s. The pipe consists
of two sections of diameters 20 cm and 10 cm with a smooth
reducing section that connects them. The pressure difference
between the two pipe sections is measured by a water
manometer. Neglecting frictional effects, determine the
differential height of water between the two pipe sections. Take
the air density to be 1.20 kg/m3. Ans 3.7cm
Air at 110 kPa and 50°C flows upward through a 6-cmdiameter inclined duct at a rate of 45 L/s. The duct diameter is
then reduced to 4 cm through a reducer. The pressure change
across the reducer is measured by a water manometer. The
elevation difference between the two points on the pipe where
the two arms of the manometer are attached is 0.20 m.
Determine the differential height between the fluid levels of the
two arms of the manometer.
Venturi Meter
Advantages :
• Higher Cd/small energy
losses Disadvantages :
• Difficult to manufacture
• expensive
• Venturi meter consists of a short
converging conical tube leading to
a cylindrical portion, called the
throat, of smaller diameter than
that of the pipeline, which is
followed by a diverging section in
which the diameter increases
again to that of the main pipeline.
• Normally is made from bronze –
or cast iron (for huge size) with
bronze lining in inner section for a
smooth surface.
• Equipped with pressure
measurement devices such as
piezometer, manometer or
pressure gauge.
Orientation of venturi in pipeline system
Horizontal with pressure gauge
Horizontal with manometer
Inclined with manometer and
• The pressure difference from which the volume rate of flow can be
determined is measured between the entry section 1 and the throat
section 2, often by means of a U-tube manometer (as shown).
• The axis of the meter may be inclined at any angle.
Fig 6.1: Inclined venturi meter and U tube
• Assuming that there is no loss of energy, and applying Bernoulli’s equation to
section 1 and 2,
P1 V12
P2 V22
+ z1 =
+ z2
ρg 2g
ρg 2g
V22 - V12 = 2g [(P1 - P2 )/ ρg + (z1 - z2 )]
• For continuous flow,
A1V1 = A2V2
V2 = (A1 / A2 )V1
• Substituting in equation 6.1
V12 [(A1 /A2 ) - 1] = 2g [(P1 - P2 )/ ρg + (z1 - z2 )]
V1 =
2 2
2g[(P1 - P2 )/ρg + (z1 - z 2 )]
• Volume rate of flow
Q  A1V1  A1 A2 /(A1 - A2 )
 (2gH )
where H  (P1 - P2 )/g + (z1 - z 2 )or m  Area ratio  A1 /A2
Q  A1 /(m - 1)
 (2gH )
(eq 6.2)
• In practice some loss of energy will occur between section
1 and 2. The value of Q given by the equation 6.2 is a
theoretical value which will be slightly greater than the
actual value.
• A coefficient of discharge, Cd is therefore introduced.
Actual discharge, Qactual = Cd x Qtheoretical
• The value of H in the equation can be found from the
reading of the U-tube gauge (Fig6.1).
• Assuming that the connections to the gauge are filled with
the fluid flowing in the pipeline, which has density ρ, and
that the density of the manometric liquid in the bottom of
the U-tube is ρman.
• Then, since pressure at the same level XX must be the same in both
Px  P1  g ( z1  z)  P2  g ( z2  z  h)  manhg
• Expanding and rearranging,
H  ( P1  P2 ) / g  ( z1  z2 )  h( man /  1)
• Equation 6.2 can be written
Q  A1 /(m - 1)
 ρman
1 
 ρ
eq 6.3
 Note that eq 6.3 is independent of z1 and z2,
so that the manometer reading h for a given
rate of flow Q is not affected by the
inclination of the meter,
 If however, the actual pressure difference
(p1-p2) is measured and equation 6.1 and 6.2
used, the values of z1 and z2 and, therefore,
the slope of the meter must be taken into
A pipe inclined at 45° to the
horizontal converges over a
length l of 2 m from a diameter d1
of 200mm to a diameter d2 of
100mm at the upper end. Oil of
relative density 0.9 flows through
the pipe at a mean velocity v1 at
the lower end of 2 m/s. find the
pressure difference across the 2
m length, ignoring any loss of
energy, and the difference in level
that would be shown on a
mercury manometer connected
across this length. The relative
density of mercury is 13.6 and
leads to the manometer are filled
with the oil.
Air is flowing through a venturi meter whose diameter is 6.6 cm
at the entrance part (location 1) and 4.6 cm at the throat
(location 2). The gage pressure is measured to be 84 kPa at the
entrance and 81 kPa at the throat. Neglecting frictional effects,
show that the volume flow rate can be expressed as
and determine the flow rate of air. Take the air density to be 1.2
Example 1
A venturi meter having a throat diameter d2 of 100 mm is
fitted into a pipeline which has an diameter d1 of 250
mm through which oil of specific gravity 0.9 is flowing.
The pressure difference between the entry and the
throat tappings is measured by a U-tube manometer,
containing mercury. If the difference of level indicated
by the mercury in the U-tube is 0.63 m, calculate the
theoretical volume rate of flow through the meter.
Example 2
Water flows through a pipe
reducer as is shown in figure. The
static pressure at point (1) and
(2) are measured by the
differential manometer containing
oil of specific gravity, SG which
less than one. For this manometer,
show that the manometer reading,
h is given by
Exercise 1
• A venturi meter with an entrance diameter of 0.3 m and a throat diameter of 0.2
m is used to measure the volume of gas flowing through a pipe. Assuming the
specific weight of the gas to be constant at 19.62 N/m3, calculate the volume
flowing when the pressure difference between the entrance and the throat is
measured as 0.06 m on U-tube manometer. Given the discharge coefficient Cd is
Exercise 2
JP-4 fuel with SG=0.77 flows through the venturi meter with
velocity of 4.57 m/s in the 15 cm pipe. If viscous rffects are
negligible, determine the elevation h, of the fuel in the open
tube connected to the throat of the venturi meter.
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