Good morning, I welcome you all to this session
of Fluid mechanics. Well, today we will be
discussing about the losses due to geometric
changes, losses due to geometric changes.
Well now before discussing this topic, you
must start with this, at the outset we must
say like this, that we have recognized that
when there is a flow of a real fluid through
a duct, then we know the flow takes place
from a higher energy to a lower energy level.
Energy gradient is the potential gradient
for the fluid flow that means upstream the
energy is more downstream the energy is less
mechanical energy.
So, now in a situation where there is no additional
mechanical energy into the fluid flow and
no taking upper withdrawal of energy. Then
mechanical energy decreases by virtue of certain
energy conversion mechanism, where a part
of the mechanical energy is being converted
to other form of energy, which we call it
as loss, when our attention is mainly focused
on mechanical energy. This thing I am telling
again and again in almost many classes, all
classes and many other classes, because this
is very important thing. Therefore, a fluid
flows in those situations where there is no
mechanical energy is added from outside to
the flow or taken from the flow to the outside.
Then mechanical energy decreases in the direction
of flow, and this decrease in mechanical energy
is known as energy loss.
So, one of the very important factors for
this loss, how the loss takes place? Who is
responsible for this loss? This is because
of the fluid friction, that is, this is because
of the friction between fluid layers. Finally,
between the solid surface and the fluid layer,
friction between the solid surface and fluid
layer, which converts a part of the mechanical
energy into intermolecular energy. So, the
loss takes place because of the fluid friction,
and this is the friction between the fluid
and the solid surface, so this happens for
all real fluid because all real fluid possess
friction between fluid layers, which is the
viscosity of the fluid and between fluid and
the solid surface.
Now, there are certain instances, where even
if we neglect this friction. For example,
if very short duct is there, so that the friction
mechanism is there, but the surface area area
of contact because the frictional force depends
upon the areas area of contact. The area of
contact is such that the total frictional
energy loss, energy loss in friction is very
less or surface is very smooth or fluid having
a very low viscosity whatever may be the thing,
in those cases the frictional loss may be
small.
But we have found out that there are other
type of losses apart from the frictional losses
that take place when the flow of a fluid,
the path of the flow of a fluid is suddenly
changed. This happens when the geometry of
the duct through which the fluid is flowing
is changed abruptly. For example, the fluid
is flowing through a cross-sectional area
of the duct the cross-sectional area is slightly
increased or slightly reduced or there is
a immediate change in the direction of the
duct.
So, if there is a change in the geometry of
the duct a sudden change, there is a sudden
change in the path of the fluid flow. This
causes a loss in energy that means a part
of mechanical energy is lost and those losses
are known as losses due to geometric changes.
In this class we will study two of such important
changes. Two two of such important cases of
losses due to geometric changes, these are
losses due to sudden enlargement; that means
the cross-sectional area of the duct is suddenly
enlarged or abruptly increased in the direction
of the flow.
Another one is losses due to sudden contraction;
that means when the fluid is flowing through
a duct of cross-sectional area, the duct area
is suddenly reduced. So, or contracted so
losses due to sudden or abrupt enlargement
and losses due to sudden or abrupt contraction
of the duct cross-sectional area through,
which the fluid is flowing. These are termed
as losses due to geometric changes and also
these are told as minor losses. The philosophy
of minor losses is that in a long pipe, it
has been found when in real situation the
fluid flows through a long duct long pipe,
sometimes the cross-sectional area of the
pipe is suddenly increased or cross-sectional
area of the pipe is suddenly decreased or
the pipe is bend.
It has been found losses due to these geometric
changes are usually small, compared to the
total loss because of friction. Only in real
situation it happens, so because the friction
loss is much more because of the long length
of the pipe compared to those losses due to
geometric changes. This is the reason for
which these losses are termed as minor losses
whereas, the losses due to fluid friction
between solid surface and the fluid is known
as major losses.
So let us now concentrate on losses due to
geometric changes, losses due to geometric
changes, geometric changes or these as known
as minor losses. First we consider abrupt
enlargement, abrupt enlargement, the losses
due to abrupt enlargement. What is the problem?
Let us physically understand the problem losses
due to abrupt enlargement. Let us consider
a duct like this and it is enlarged abruptly
like this, it is abruptly enlarged like this.
Now, let the fluid flows in this direction,
let the fluid flows with an initially uniform
velocity, all right? Uniform velocity at this
section, let this section is one at this section
at this section the fluid flows with uniform
velocity. Let this this is the section where
fluid flows with uniform velocity V 1, let
us consider V 1 through this duct. This is
the duct now. The duct is suddenly enlarged
to a higher area. So, what will happen the
streamline, which is now fluid as in an uniform
velocity for an ideal fluid or for a viscous
fluid? Now here we consider a viscous fluid
because these losses take place only for viscous
fluid. That I will explain now, but initially
we considered the fluid approaches here with
a uniform velocity.
Now, therefore what happens? The fluid stream
cannot take the abrupt change in the cross
section like this because of their inertia,
so the streamline gradually takes a shape
like that and fills up this tube like this.
So, this is a typical streamline, I am showing.
This may be one streamline, this may be other
streamline. So, these are the typical streamlines
of the flow. Steady flow we consider, so it
goes like this. Let this is, let we consider
a section here at 2 and let us consider a
section here at 1. So, fluid approaches with
an uniform velocity V 1 with a pressure p
1.
After expansion in this enlarged section at
some distance, downstream fluid attains a
uniform velocity going out with V 2 and let
the pressure is p 2. It is a long duct, closed
duct. Now, this is the situation that streamline
takes place it diverges like that. Now, very
simple thing, first is that the cross-sectional
area at one is less than 2. Therefore, according
to continuity V 1 is higher than V 2 obviously
because V 1 a 1 is V 2 a 2 because the same
volume flow rate of fluid should pass, that
what is coming in should go out under steady
state. So, similarly p 2 is greater than p
1.
Since, the velocity here is lower pressure
here is more. Therefore, the fluid is flowing
from a lower pressure to a higher pressure,
but I told you earlier emphatically that fluid
flows from a higher energy to lower energy.
Fluid may flow from a lower pressure to higher
pressure, but the total energy at this section
1 1, is much more than the 2 2 or more than
the 2 2. That difference is lost while during
the flow. Now, this is the situation that
fluid is flowing from a section one to a section
two, where the velocity at section one is
more than that at two and the pressure is
at section one is less than that at two, all
right?
Now, we consider that what happens when the
fluid cross-sectional area, fluid flow cross-sectional
area, that is the cross-sectional area of
the stream tube diverges? So, when it is diverging,
what happens it is found in practice that
this areas in this corner, there is a flow
reversal zone. That means the fluid particle
goes on flowing in the back direction. This
is the flow reversal zone. This zone is the,
this zone is the flow reversal zone and this
zone create a same zone turbulent eddies.
That is why this is known as zone of turbulent
eddies, so turbulent eddies.
So, if you just look from a practical point
of view, as a practical man, you will see
that here the fluid particles are going back.
So, they just make a re circulatory flow type
of thing that a fluid particles come back
in this from downstream to upstream direction.
Again flow like that are circulatory flow
is taking place, which creates at urbulent
eddies turbulent eddies. Eddies are small
re-circulating flows. So, zone of turbulent
eddies are formed. This is found in practice.
So, what is this phenomena? This phenomena
is known as separation loss, which is extremely
important, in your advance fluid mechanics
classes also you will learn, this is known
as separation loss.
Sometimes we call it as boundary layer separation,
but the word boundary layer, I am not bringing
here at this in this context because this
will lead more confusion. Only separation
losses, this is boundary layer separation
losses. Now, what is this in real fluid, what
happens, when a fluid flows against an adverse
pressure gradient. This is an adverse pressure
gradient p 2 to p 1. Now, in p 2to p 1 when
the fluid flows, the fluid is able to flow
because energy gradient is not adverse. Energy
here is higher than energy here, but what
happens in fluid very near to the wall, for
example, here very near to the wall has got
almost 0 velocity.
As you know we have read it at the beginning,
that it is because of the viscosity that fluid
velocity at the wall becoming always becomes
0. So, the fluid particles near the wall looses
their kinetic energy, velocity becomes very
small and ultimately 0. That means we cannot
expect though, there is a uniform velocity
of approach to the fluid, but at any section
the fluid velocity is not uniform. There is
a variation of velocity. So, you can ask me
sir, what is this velocity? I can tell this
is the average velocity, we consider in case
of a turbulent flow, it is more or less uniform.
But, at the wall it suddenly changes to 0;
that means at the wall velocity has to be
0, which is known as no slip condition. That
you have already recognized for a viscous
fluid, the no slip condition is that the relative
velocity between the fluid and the solid is
0.
So, if the solid surface is at absolute rest,
the fluid velocity will be absolutely. So,
fluid will gradually change or suddenly change
from a high velocity to a 0 velocity, which
means the fluid particles very near to the
wall will have very low velocity. That means
they will lose the kinetic energy in surmounting
the adverse pressure yield. So, for those
fluid particles it will be difficult to go
against an adverse pressure gradient. That
means higher pressure to lower pressure, because
they lose their kinetic energy because of
friction fluid friction.
So, therefore particles fluid particles near
the solid surface traces back. That means
they take the course of a back flow in favor
of pressure gradient. That means this pressure
here is higher here is lower. So, they make
a back flow. That is why the flow reversal
takes place. This is known as separation loss
and this back flow in this region near wall
makes a small zone of little circulatory flows,
which is the zone of turbulent eddies. They
create turbulent eddies and because of this
a part of the mechanical energy is being destroyed
in form of intermolecular energy or is being
converted in the form of intermolecular energy,
which we call as the destruction of mechanical
energy.
So, basically this is the phenomena separation
loss of creating zone of turbulent eddies.
While the streamline expands or streamline
takes the flow of fluid takes place from a
lower pressure to a higher pressure always
this separation loss takes places for a real
fluid. This is the reason for which the mechanical
loss, the loss in mechanical energy takes
place. Now, we find out the quantitative value
of that. Now, let us find out let us do this
thing. Let us take a control volume.
Let us take a ,b this is c at this corner
da point here e, f, g, h. Let us take a control
volume a, b, c, d, e, f, g, h. Simply by the
use of momentum theorem to the control volume
and the application of Bernoulli’s equation
with losses, you know that a Bernoulli’s
equation is also applicable for real fluids
with losses, we can find out the magnitude
of this losses. How? Let us apply the control
momentum theorem to the control volume. Now,
the pressure applied here is p 1 pressure
applied in this surface e f is p 2. Now, it
has been found out the pressure here the pressure
here in this zone, where the eddies are formed
is this is the pressure imparted on the fluid
surface. On this surface of the fluid, this
pressure is p dash. Let this pressure is p
dash let this pressure is p dash, this has
been found from experiment.
This is equal to the upstream pressure p dash.
This cannot be proved here, this require a
high level theory of viscous fluids but, it
has been found from experiment this you know
as an information as p 1. Now, if I write
the momentum theorem, what is the momentum
theorem? Net rate of momentum a flux that
means mass, let Q is the volume flow rate
through this duct, into the momentum a flux.
That means net a flux that means rho Q into
a flux minus in flux rho Q V 2 is the momentum
a flux rho, Q V 1 is the momentum in flux.
That means I take this direction as the direction
in which I write the momentum equation. That
must be equal to the net force acting in this
control volume in this direction. What is
that force p 1 into A 1 in this direction?
If I consider the area at section one is a
1 that means are of this duct, plus another
force acting in this direction, which is this
force into the area of this well this stream
tube. That means this is p dash into this
projected area in this plane to find out the
net force in this direction is this p dash
into, if we take this projected area. That
means this, this is this point that means
we take this body, you understand as the control
volume. Therefore, the projected area of this
part is A 2 minus A 1, very simple geometry
and minus P 2 A 2 at this area e f. So, if
you do this fluid mechanics is almost over.
So, now p dash is p 1, with this information,
if you substitute here these information then
p 1 A 1, p 1 A 1 cancels out. So, therefore
only thing is that p 1 A 1 and p 1, so A 2
p 1 minus p 2. So, p 1 minus p 2 into A2 is
equal to, this part I am writing in the left
hand side p 1 A 1 minus. Sorry, this is canceled,
so p 1 A 2 minus p 2a 2, p 1 minus p 2 A 2
is equal to rho q V 2 minus V 1 very simple.
Now, this can be written as Q can be written
as, in terms of V 2 A 2 V 2, so V 2 minus
V 1. So, this a two is cancelled, so we get
p 1 minus p 2 is rho V 2 into V 2 minus V
1. So, this is the main equation which will
be of great help to us to find out the energy
loss. Now, I write the Bernoulli’s equation
between one and two. How to write p 1 by rho
plus V 1 square by 2? Let us consider section
one and section two at the same elevation
level q, do not create any problem or if you
think that you can write g z 1. It is of no
problem, but better if you write g z 1, g
z 1, then again a body force term will come.
Therefore, we better make it simplify because
here the g z 1, g z 2, is not there. We have
not taken, if it is in a inclined position
vertical force. Then a component of weight
will come here in the momentum theorem in
the force net force acting on the control
volume, that you can make it simple. So, here
we are considering a simple case that in a
same elevation head is equal to p 2 by rho
plus V 2 square by 2. Let us consider in terms
of the head per unit weight energy plus h
l. That is the loss of head in due to this
geometric change, that means if we consider
the loss of head due to this geometric change,
we take it at this right hand side.
That means total energy at the inlet is equal
to total mechanical energy plus loss. I have
told that Bernoulli’s equation can be written
for a real fluid where the losses are taking
place with a loss term taking care of at the
downstream section. That means this is the
inlet energy which is higher than the outlet
energy mechanical energy plus loss. That means
h l is nothing but p 1minus p 2 by rho g plus
V 1 square minus V 2 square by 2 g. So, if
you substitute, then p 1 minus p 2 by rho
g as V 2 into V 2 minus V 1 by g minus V 1
square. This is very simple, 2 g it becomes
simply V 1 minus V 2 whole square by 2 g.
So, therefore you see the the loss of head
due to this abrupt enlargement is V 1 minus
V 2 whole square by 2 g, all right? This can
be written, this can be written as h L is
equal to… Now, if we take this V 1 square
as common, V 1 square by 2 g, then what will
that, 1 minus V 2 B by, all right? V 1 whole
square.
What is V 2 by V 1? This can be written as
V 1 square by from continuity, what is V 2
by V 1A 1 by A 2 whole square. Now, consider
a case when A 1 by A 2 tends to 0. Then, what
is the loss V 1 square by 2 g. This is the
case now, that when A 2 is very large compared
to A 1. That means depending upon the value
of A 1 by A 2, the loss will take with if
A 1 by it is one there is no enlargement.
Then what will happen if there is no enlargement.
A 1 is A 2, then 1 minus 1 0 there is no loss,
but if a 1 if a 2 tends to infinity that means
for a given A 1 if A 2… A 2 is the downstream
area is very very large, then the loss term
h l becomes equal to V 1 square by 2 g.
This situation resembles this physical problem
that a pipe discharging into a large reservoir,
that a pipe discharging into a large. A 2
tends to infinity. So, until and unless the
desired reservoir is large A 2 cannot be in
A 1 by A 2 cannot be 0, so a 1 by a 2 zero
means A 2 tends to infinity, that means the
physical situation is that a large reservoir.
In that case what happens the loss of head
due to this flow is V 1 square by 2 g. This
is very simple to understand physically, that
the velocity head with which the fluid is
discharge dis ultimately lost, because in
this at this point in the large reservoir
fluid is at rest. That means the velocity
head is arrested in this large reservoir.
Another very important thing is that, if you
now write the Bernoulli’s equation at the
same point, you will find the interesting
thing. That what, if these two points, you
write the Bernoulli’s equation, you will
find this pressure head. This energy pressure
energy this height, datum energy. If you consider
this as the reference, datum 0 kinetic energy
0 the 2 point same energy, but if you write
the Bernoulli’s equation at this point and
the discharge plane, then you will see at
this plane a extra V 1 square by 2 g is there.
That means here if you write at just discharge
plane, so same pressure let this be the height
h.
So, p 1 by rho g is equal to h that means
h plus V 1 square by 2 g. Let 0 is the datum
head is equal to for anyone of these points
where velocity is 0. So, you write the same
pressure ,but 0 velocity 0 datum, so where
is the equality? Equality is that loss. That
means form this point to this point, there
is loss this loss is V1 square by 2 g. So,
this loss is V 1 square by 2 g. That means,
physically it is velocity head is loss. That
means this kinetic energy of the fluid is
being converted in intermolecular energy.
So, these points have got higher temperature
than this point. So, that if you take care
of the intermolecular energy the energy level
will be same, otherwise the mechanical energy
here is lower than this.
This is the loss and this loss is known as
exit loss. That means, when there is an exit
of a fluid stream with some velocity in a
large reservoir or in atmosphere where the
velocity is arrested. That means this kinetic
energy is completely converted in intermolecular
energy, so that the kinetic energy is lost.
This loss is known as exit loss and it is
magnitude is V square by 2 g where V is the
velocity of discharge and this is a special
case of for the loss due to abrupt enlargement.
Now, I come to losses due to 
sudden contraction losses due to sudden contraction.
Now, we can go quickly. Now, I consider a
contraction. That means the fluid flowing
through like this. That a fluid approaching
in a duct whose area is ultimately suddenly
reduced. Then the fluid let us this here this
is 1, let us this is 2. So what happens, the
fluid now here there is an interesting phenomena,
here there is an interesting phenomena, this
will be the nature of the streamlines. Streamline
will adjust like this.
Now, in that case what happens in a real fluid?
That if there is a contraction of area the
streamline contracts to accommodate that area.
There is a lateral force acting on the streamline
because of this change in momentum by this
duct, which makes this streamline narrowing
down ultimately to an area, which is lower
than the minimum geometrical area. That means
if this is the section two that means this
is the pipe 2 that cross sectional area of
2 pipe 2. That means a 2, so this cross sectional
area that means stream tube narrows down narrows
down to an area this cross-sectional minimum
area, which is lower than the cross-sectional
area of the this duct, that is reduced area
duct.
So, this area is known as vena contracta.
This is a Latin name vena contracta. That
means this is the minimum area where if contracts.
It happens so you can see in hydraulics laboratory.
Also, it happens that, if a fluid is allowed
to flow, that I will also discuss afterwards,
that a fluid is there now. If it is allowed
to pass through an orifice at the side of
this, so fluid fluid particle will just fluid
streamlines, will just you will see change
in their path like that and they will narrow
down to an area, which is lower than these
orifice area. So, these area is known as vena
contracta. It happens just after the orifice
section at the downstream section, very close
to the orifice section, that means the stream
tube is narrowed down to an area minimum area,
which is lower than this geometrical area
.This is known as vena contracta, let this
section is C C.
Now, here during this part of the flow of
the fluid the fluid is flowing in an favorable
pressure gradient. What is meant by favorable
pressure gradient; that if this is p 1, this
is p C, p 1 is always greater than p C, because
v C is greater than v 1, because this cross-sectional
area is less than this cross sectional area.
So, fluid is converging and accelerating because
the cross sectional area is reduced with velocity
is increased because of continuity. So, since
the fluid velocity is increased, so fluid
pressure is reduced. That means p 1 is more
than p C or p C is less than p 1.
So, there is no question of being a back flow
because here the wall nears the wall. Even
if the fluid particles losses their kinetic
energy, but they will be forced to the main
direction or direction of main or bulk flow
because pressure is always acting towards
them. Thereis a favorable pressure gradient,
that pressure here is higher pressure here
is lower. This happens in case of an adverse
pressure gradient when the fluid particle
loses its kinetic energy they will force back
by the pressure forces but here, pressure
forces takes place in the direction of the
flow.
So therefore, in a converging flow or in an
accelerating flow, where the velocity is increased,
that stream tube cross-sectional area decreases.
The question of flow separation does not come.
All fluid particles flow in the direction
of main flow. But what happens the situation
when the fluid attains an area vena contracta,
which is lower than the area of this pipe?
That a 2, then what happens the fluid from
the vena contracta? Again expands and fills
the pipe area 2. Therefore, from C C to 2
it is again an expansionand the fluid particle
here separates flow separation. Therefore,
the energy losses that is taking part in these
zones, not in this zone.
So, it is not because of the contraction,
but because of the expansion, that is following
the contraction and why this expansion takes
place? This is because of the fact that the
stream tube attains an area which is lower
than the area of the geometric area of the
tube. That is vena contracta, then it has
to have an expansion or it has to expand to
fill up the pipe 2. Therefore, it is the expansion
from the vena contracta to the downstream
section 2 2 for which the losses takes place.
Now, therefore it is again the losses due
to expansion, but now here if I write the
h l without anything if I recall the formula.
It will be V c minus V 2 whole square by 2
g because it is same as the same thing that
as if it has an abrupt enlargement from an
area C C to 2 2. So it is the enlargement
loss h l is V C minus V 2.So, this can be
written as v 2 square by 2 g into 1 minus.
What is that V 2 square by 2 g, sorry. This
is V C by V 2 minus 1.
Now, a coefficient of contraction is defined
coefficient of contraction is defined coefficient
of contraction a coefficient of contraction
C C is defined, which is the area ratio A
C, the vena contracta divided by this part.
That means how much it is reduced from the
geometrical area of the pipe C C. So, from
the continuity we can write V C into A C is
equal to V 2 into A 2 because same flow rate
should be accommodated through the tube. Therefore,
V C by V 2 is A 2 by A C that means 1 by C
C. So, one can write therefore, h L as V 2
square by 2 g, where V 2 is this final velocity
in this small tube, into 1 by C C minus 1
whole square.
So, if the coefficient of contraction is given.
Then, I can find out we can find out V 2 square
by 2 gone by C C minus 1 is h l. But one very
interesting thing here which appears in the
mind of a reader that you see in this expression,
nowhere the area A 1 is coming, because this
contraction depends upon area A 1 by A 2.
So, it is indirectly represent in terms of
a contraction coefficient which is A C by
A 2, but an intelligent student will find
that this is an interlinking step. That means
this A C is again related to a 1 and A 2.
So, A 1 though is not coming explicitly in
this equation, but A 1 is implicit in the
C C, which mathematically can be explained
that C C is a function of A 1 by A 2.
So, how to find out the value of C C? That
means A C by A 2 that is a function of A 1
by A 2. So, depending upon this area and this
area, so this area though not coming explicitly
in this equation, it comes implicitly through
the functional dependence of C C A 1 by A
2. So, this is 1 by C C minus 1 whole square.
Now, in a special cases when this A 1 by A
2 is 1, that means there is no contraction
A 1 by A 2 1 the value of C C is 1 that means
h L is 0 . There is no question of loss so
a same diameter.
But you consider a situation, where this A
1 by A 2 the rather A 2 by A 1function of
A 2 by A. It looks nice that A2 by A 1, which
one is A 2? This one is A 2, this one is A
1 is again 0. That means A 2 by A 1 is 0,
when A 2 by A 1 is 0, what is the physical
situation. That means A 1 tends to be infinitely
B rather than A 2. In that case it has been
found that this 1 by C C minus 1 whole square
that is K. Let let 1 by C C minus 1 whole
square K that means h l is K V 2 square by
2 g.
So, when A 2 by A 1 equal 0, then k becomes
0.5. C C become 0.75 or something like that
0.72, so that 1 by C C minus 1 whole square
becomes; oh sorry. Not C C 0.72 or it may
be 0.72, I do not know the value exactly.
That means 1 by C C minus 1 whole square.
So, C C value becomes such, when A 2 by A
10, K becomes exactly 0.5. In that case h
l equal to 0.5 V 2 square 2 g. That means
I am exploring a very special case. When A
2 by A 1 0, which physically resembles that
the fluid coming from a very large reservoir.
A 1 is very large compared to A 2.
That means when A 2 by A 1 is 0 means A 2
A 1 tends to infinity. That means A 1 is very
big compared to A 2. This resembles that the
fluid coming from a big reservoir to a pipe.
When the fluid coming from a big reservoir
enter to a pipe, then it is nothing but a
contraction loss. So, this big reservoir is
A 1 and this pipe is A 2 where A 2 by A 1
is… Sorry, A 2 by A 1 is almost 0. In that
case h L is 0.5 V square by 2 g, if V is the
velocity in the pipe, uniform cross-sectional
area. So, that means this is a special case
for contraction that is 0.5 V square by 2
g, because this factor is this is K.
That means K becomes 0.5. So, this loss is
known as Entry loss. That means if we have
a section here just downstream to this and
if we have a section here, so if we write
the Bernoulli’s equation between this two
sections are very close to here, you understand?
So, if you write the Bernoulli’s equation
between this two sections, here the entry
loss has to be take care of which is 0.5 V
square by 2 g. You have understood? So, this
is a special case of losses due to contraction.
So, these are the losses due to geometric
changes. Now, we will come to another topic,
that is measurement of flow. Measurement of
flow through, we will discuss in this classes
measurement of flow through three devices,
venturi meter measurement of flow by three
devices; venturi meter, Orifice meter and
flow nozzles. When the fluid flows through
a closed duct, how to measure the flow rate,
which is very practical information which
is of very practical interest? There are several
measuring devices, which measures the flow
rate of fluid through a duct. One of such
measurements, I tell you which comes very
apparent that if the flow is a liquid and
it is flowing through a hydraulic circuit.
Where there is an open end that fluid enters
enters through some point and then flowing
through a pipe or a pipe network, there is
an open end where the fluid is coming out.
So, most accurate and easiest method of measuring
the flow rate is that collect the liquid flow,
because liquid can be collected in a measuring
bucket. You weigh the amount that is collected
during interval of time and find out the average
flow rate. This is one of the most direct
and accurate method of flow measurement, but
it is not always possible because the circuit
is not open. So, you cannot collect the fluid
or the fluid flowing is not a liquid that
is a compressible fluid. So, there are several
flow measuring instruments which can measure
the flow rate of a fluid under steady condition
through a fluid circuit. This is not within
the scope of our syllabus to study in details,
all the flow measuring devices, but we will
be studying three flow measuring devices in
this class. One is venturi meter another is
orifice meter, another is flow nozzles.
So, this three flow meters in general work
on the same basic principle. The basic principle
is like that. When the fluid flows through
a duct or through a circuit, this flow measuring
device, when this is inserted in the circuit,
they provide a co axial contraction or co
axial. We can tell co axial contraction. That
means the area is gradually changing that
means they in general gives to a geometric
change in to the flow field, so that the fluid
path of the fluid flow is changed because
of which there occurs a change in pressure
in the fluid. Losses are there obviously and
change in pressure is you say when the fluid
is expanding, fluid is contracting because
of the abrupt enlargement abrupt contraction,
so pressure changes takes place.
So, this change of pressure is measured by
a pressure measuring instrument, usually by
a manometer. Then what is done by the straight
forward application of Bernoulli’s equation.
This pressure drop is equated with the velocity
of flow and finally, with the flow rate. Therefore,
by measured values, by substituting the measured
value of this pressure drop in that equation,
which is developed by the application of Bernoulli’s
equation, we find out the flow rate. So, this
is the basic principle, but in geometrical
shape the venturi meter, orifice meter and
flow nozzles are different though their basic
principles are same. These we will be discussing
in the next class as a flow measuring devices.
Thank you.
