Good afternoon, I welcome you to this session
of fluid mechanics. Well last class we were
discussing about the discharge through an
orifice, what is an orifice we just discussed
and what is the discharge through an orifice
under a constant head.
So, let us recall little bit of the earlier
discussion. Well if this is a tank where there
is an orifice at the side, sorry this is wrong.
So, this is the orifice, small hole there
is no projection, small hole and if there
is a constant head maintained, that means
orifice, through orifice the water is discharged
to maintain a constant head that means water
inflow has to be there. So, under a constant
head, head means that if I consider an axis
line, gives the axis of the sharp edged orifice.
Then this height is h under a constant height
from the axis of the orifice and if this orifice
height, that means this height is much less
compared to this head above the orifice; that
means the height of the water surface from
the axis of the orifice. Then we have found
out by applying the Bernoulli’s equation
first of all we see that how the streamline
contracts, that is the surface the we have
recognized that streamlines are like this,
they are contracting like this and ultimately
they come to a section very close to the orifice,
downstream to the orifice just after the orifice
section is C C, where the cross-sectional
area of the stream tube is less than the cross-sectional
area of these orifice, if the cross-sectional
area of the orifice is a, then at C C where
this stream tube converges and then becomes
again parallel, this is the minimum area section,
let this section is C C after that it goes
on falling because of the gravity.
Then this area is a C then a C is less than
a, this is the contraction now if we apply
the Bernoulli’s equation at any point on
the free surface 1 and at any point on this
section C C or at any point on a streamline,
let the point may be 1 dash let the point
may be here 1 double dash, one point will
be here we will always get after using the
Bernoulli’s equation, the velocity of discharge
at C C of the z that V C is root over 2 g
h. Very simple, that means we can tell this
potential energy compared to point 1 is totally
converted into kinetic energy, that velocity
for an ideal fluid without any loss. If we
write this for this and this point, so this
point is the pressure energy because this
point in the same axis, so this point and
any point at the C C plane in the orifice
is at the same elevation level. So, therefore
this is in the form of pressure energy, here
the energy is in the form of pressure energy
because of this height over the point up to
the free surface.
So this is converted to the velocity so V
C is equal to root over 2 g h, then we found
out that this is the theoretical velocity.
So V C actual, which is due to the friction
of the fluid at the orifice h is less than
this theoretical value and this is multiplied
the taken care of by a factor C v that is
known as coefficient of velocity whose value
is less than 1 and value depends upon the
height from the centre of the orifice, that
is the head and the size of the size and shape
of the orifice. So, therefore Q is equal to,
Q actual will be equal to a c that means the
volumetric flow rate into V actual that means
C v into root over 2 g h.
Now, with a definition of coefficient of contraction
as the area of the vena contracta to the area
of the orifice, we can write a c in terms
of, let area of the orifice as a o, o suffix
for orifice c suffix for vena contracta section.
Then we can write C c into C v into a0 root
over 2 g h. This we designate as a theoretical
volume flow rate Q theoretical, that means
if there was no friction in the fluid so that
the total potential head or pressure head
at this point is converted into kinetic head
and if there was no contraction of the streamlines
then this could have been the theoretical
discharge. This is multiplied by C c into
C v, that is the coefficient of contraction
and coefficient of velocity and this is combinedly
known as coefficient of discharge. That means
the theoretical discharge as defined by the
orifice area multiplied by root over 2 g h,
should be multiplied by a factor which is
less than 1 to count for the actual discharge,
this we discussed in the last class.
Now, here one thing, we assume that the orifice
size is so small compared to the height above
the orifice that means, if we consider a central
line to the orifice so any point on this line
or any other point along the height of the
orifice, will have the same liquid head a
that means with respect to this h there is
no variation of the point, that is no variation
in the head of the liquid at different points
of the orifice. That means any point on its
axis is a representative point of the entire
cross-section of the orifice, so that we can
tell the orifice the entire orifice is under
the head h, so that the velocity at each and
every point is uniform.
But it is not so in case of a large orifice,
if we consider a orifice like this, consider
a large orifice that means a orifice like
this a orifice like this, it is a very large
with respect to the head over, that means
let us consider a constant head is maintained
by an inflow arrangement Q, an orifice is
such that we define the H 1 is the height
maintained that height of the free water surface
from this top part of the orifice and from
the bottom part it is H 2.
So H 1 and H 2 are different, earlier if we
define some H 1, H 2 from the top and the
bottom so these height is so small compared
to h. So, H 1 and H 2 are very small, sorry
very, very much equal. That means this height
of the orifice is less, very small compared
to height of the water surface from any point
or any level of the orifice, but here the
orifice is so large, I mean this is the height
of the orifice, which is H 2 minus H 1. So,
H 2 is much bigger than H 1, so H1 or H 2
are not very equal, that means the size of
the orifice is comparable to the height of
the liquid surface, free surface of the liquid
above the orifice. They are comparable to
H 1 to H 2, this is the height of the liquid
within that case, what happens is very simple
a simple integration.
Let us consider the orifice to be if, you
see from here a view a rectangular in shape
that means whose width is B, then what happen
difference is like this, if we consider an
axis you can consider but difference is that
each and every point the head of the liquid
is varying. So, therefore, the velocity at
each and every point will be changing, that
means if we apply the Bernoulli’s equation
between a point here and a point here, we
will get a velocity root over 2 g h, which
is where h is this height, which is varying
from point to point, which starts from H 1
from the upper point of the orifice and goes
to H 2. So therefore, velocity at this point,
at liquid particle at here will be much higher
than the liquid particle here, so there will
be a non-uniform velocity distribution.
So therefore, what we will have to do, we
will have to integrate. The simple procedure
is like that, we consider an instantaneous
height, that means at any level that at any
point. So, let this be the height h at this
point and it has got, perpendicular to this
direction if you take a view so a small elemental
strip you consider. So, this is the h, alright
this is the h and this you consider as d h,
that means now you consider the discharge
through a small elemental strip of the orifice
which is at a height h from the free surface,
this height is h, height h from the free surface,
this is the height h and whose thickness is
d h, that means whose area is B d h, a small
strip rectangular strip. let dQ is the flow
rate through it, what is the formula c d,
now we know c d into a area of the orifice,
what is the area of the orifice B d h into
root over 2 g h. We apply the formula that
coefficient of discharge into area of the
orifice into root over 2 g h, where h is the
head of the liquid that means the height of
the free water surface from the orifice.
So, what happens the for this, we have to
find for the entire orifice we have to integrate
dQ, that means what, we will have to do, we
will have to integrate this. C d B d h root
over 2 g h, from H 1 to H 2 from H 1 to H
2. Alright, so if you carry out this integration
it will be very simple, it is equal to C d
is constant, if we consider C d same at all
points, it is constant throughout the orifice.
B is the width geometry of the orifice root
2 g h, it simply becomes root over h d h from
H 1 to H 2. Simple, very simple school level
things, 2third C d B root over 2g into H 2
to the power, what is that half, 3 by 2minus
H1 to the power 3 by two. That is, this simple
orifice is large it may not be a rectangular
orifice, depending upon it’s geometry you
can integrate accordingly, so this is the
concept.
Now, another concept is that, what is meant
by drowned orifice, what is meant by drowned
orifice. What is meant by drowned or submerged
drowned or submerged orifice, what is meant
by that, very simple, now first let us consider
a drowned orifice or submerged orifice of
small orifice. Let us consider a very small
orifice, compared to the head which is above
the orifice, so from any point the head is
h, so better we represent it from the axis,
that means the height of the orifice or the
area of the orifice is very small, height
of the orifice is very small compared to this
height, small orifice. Drowned orifice means
that orifice does not discharge into atmosphere,
it discharge into another water vessel that
means other side, downstream side there is
a pressure, very simple, hydraulics is very
simple, this is hydraulics that means instead
of discharging into atmosphere it is discharging
into another tank where the height of the
liquid, free surface of the liquid which we
call head is h 2 from its axis. Then if we
write the Bernoulli’s equation from 1 to
2, now this is maintained that constant h
2, this is maintained at constant h1 for.
So, flow will be from this tank to this tank
always, but it has maintained constant because
of the inflow, that arrangement we make, that
is not very much concerned that how we make
it, so the height of the 2tank maintained
constant h 1 being more than h 2 the flow
takes place from this tank to this tank. So,
this orifice is discharging in this direction
the flow takes place this way, so now if we
write the Bernoulli’s equation between 1
and 2, we will get that V, the velocity of
discharge for an ideal fluid, 2 g h 1 minus
h 2, can you get it, because you write the
Bernoulli’s equation very simple one. Bernoulli’s
equation pressure head is 0 p by rho g, taking
the pressure head above the atmospheric head.
Then V square by 2g 0. Let Z1 be the elevation
head with respect to any datum, reference
datum is equal to at this point what is the
pressure head p by rho g, it is h 2 because
pressure here is h 2 rho g.
Because of this height of the liquid column,
so very simple velocity which we want to find
out V square by 2g, all in terms of the head.
That means energy per unit weight meter, unit
and then rest is it is Z 2. Therefore V 2
square is 2g okay, into h 2, Z 1 plus, so
V 2 square by 2g is minus, h 2 I am sorry
Z 1 minus Z 2 plus h 2 minus h 2 and Z 1 minus
Z 2, that means the elevation head between
this 2is nothing but h 1, that means ultimately
you get V is equal to very simple, that means
effective head causing the flow is h 1 minus
h. Where this is the back head that means
it is flowing under an effective head of h
1 minus h 2 so you will have to make it h
1.
But in this case one interesting thing is
that there is no difference between a large
and a small orifice, thing is that if we have
a large orifice for example this is a large
orifice this is a large orifice, and let this
side also, this is a large orifice let us
this H 1 and this side let this is H 1 dash
there is no difference between large and why,
how I show you. Let this is H 2, that height
of the water level from the bottom in this
side and correspondingly this side H 2 dash,
so entire orifice is under an effective head
of H 1 minus H 1 dash, can you recognize it
because this large orifice can be considered
similar as a small orifice under an effective
head H 1 minus H 1 dash. Why because here
what happens when the head varies point to
point, then whatever is added from this side
is also added from this side, for example
at this point what is the effective head causing
the flow H 1 minus H 1 dash. What is the effective
head causing the flow, H 2 minus H 2 dash
that is H 1 minus H 1 dash.
At any point the effective head causing the
flow, let this is h and if this is h dash,
then the effective head causing the flow at
this point is h minus h dash, which is nothing
but H 1 minus H1 dash, that means any increase
in head from this side favoring the flow,
a same way obstructed by the same increase
in head in this side. So therefore, for a
drowned orifice, a large orifice whose height
is comparable to the head of the liquids above
it can be treated as a small orifice under
a drowned condition or under an effective
head, which is H 1 minus H 1 dash that means
difference of the water level from it is upper
point, clear alright.
So this is a trick that if there is a large
orifice and it is a drowned or submerged orifice
it behaves as, that means the velocity through
this is theoretical case is root over 2 g
h 1 minus H1 dash, when there was no drowned
orifice then the difference in head counted
because downstream side head is uniform that
is atmospheric pressure here the same head
is added okay, it is very simple, drowned
or submerged orifice.
Now, we will come to time of emptying tank
time of emptying a tank, is an unsteady problem,
so far we have discussed the discharge with
a constant head, but if head is not maintained
constant and there is for example, consider
a orifice at the bottom of a tank centrally
and if the initial head is H 1. Let us, this
is a transient problem, unsteady problem that
means we do not have any provision for inflow
to maintain this head constant. So, what will
happen the streamlines like this, let me draw
the, that means what will happen. The water
will be drained out through the orifice coming
out so head will be decreased, so the problem
is posed like that. What time will take for
the head to come from H 1 to H 2.
Next example, what is the time taken this
type of problem so it is very simple, so these
orifice is under a varying head, therefore
the distract is maximum when the height above
the orifice is maximum, that means at initial
level. Then as the height falls the discharge
rate is reduced, let us consider an any instantaneous
height h above the orifice, this bottom bottom
plane h. So, in that case the discharge through
the orifice, that means Q, discharge through
the orifice Q is what.
Let dq small discharge through the orifice
is C d into orifice area is a naught into
root over 2 g h C d, takes care of all the
losses frictional losses and contraction and
all this thing. So, if we consider d h is
the decrease in the height, now this is the
d h is the decrease in height for a time t
or rather we can think in a various ways,
when various ways. We can consider that if
A is the cross-sectional area of the tank,
so rate at which the volume flow rate is reduced
in the tank is a d h d t and that must be
equal to C d a naught root over 2 g h with
a negative sign, why because h is decreasing
with time, that means A d h d t is the rate
at which the volume is decreasing in the tank.
The volume flow rate is decreasing, that means
it can be conceived this way that during a
given interval of time d t the head has fall,
the height has fallen by d h.
So, therefore the volume this has been decreased
is minus A d h and this must be equal to the
flow which has gone through the orifice, a
50 years back, explanation d t so ultimately
A d h d t , but you should write straight
in terms of the rate the rate of volume flow
in the tank is A d h d t in terms of the instantaneous
height of the tank and rate of the volume
flow through the orifice is this, the equal
this 2rate under continuity equation. So therefore,
it becomes simply d h d t is minus C d a naught
root over 2g, all the constant together into
root h d h d t , is equal to sorry, is equal
to minus.
So we have can found out time like this d
t is equal to, so d t is equal to this, will
be like this. A by C d a naught root over
2g into h to the power minus half d h with
a negative sign with a negative sign, but
sorry I am sorry, with a negative sign. Now,
the time taken for the height to come from
H 1 to H 2 can be written as by integrating
d t is equal to, let that time is t then you
just integrate these these are the constant
A C d a naught root over 2g, integrate h to
the power minus of d h. Now, from a initial
height H 1 to H 2, so it should be H 1 to
H 2. I take care of the minus sign so I write
H 2 to H 1, that means if you integrate it
this time taken will be simply A by C d a
naught root over 2g. Now, this integration
will be what 2minus, half plus, 1 half 2and
H 1 to the power half.
So, this is precisely very simple, now if
one has to find out the time of emptying,
which I started time of emptying means H 2
is 0, so starting from an initial height H
1 that time of emptying that time will be
this multiplied by H 1 to the power half,
but here one thing is very simple the tank
height, a tank area is constant or uniform.
It does not vary with h, so integration becomes
simple. There may be a tank of some geometry
where the area may be a function of h, so
there is no fluid mechanics, it is simply
school level integration that means there
may be an orifice in a prismatic tank, like
that difference is that here the area that
means at different head the area is changing.
That means if you start with, let this is
the H 1 the same problem if you pose the time
of emptying or reducing the height from H
1 to H 2, whatever we do so same problem if
you pose at any instantaneous height h.
From this if you consider the area of the
tank there that means, okay the through orifice
the flow is same dQ is rate of flow is C d
into the orifice area is same root over 2
g h, but the rate of volume flow rate at this
level is minus A which is a function of h,
A is changing A hd hd t which has to be made
equal to C d a naught root over 2 g h that
means you will have to take d t in one side
and d h. So, A as a function of h only you
have to incorporate there is no fluid mechanic,
these you can find depending upon the geometry
if this is a prismatic tank, say straight
surface from the inclination something will
be given in the problem. So that from the
geometry you can express the area at this
section as a function of h and can integrate
it, there are several problems, standard problems.
One standard problem is solved in my book,
there is a hemispherical there is a hemispherical
vessel and there is a, this is the hemispherical
vessel. Let, it is upper surface, let this
is the centre of curvature for this hemisphere,
hemispherical, which is the fluid is coming
out. So, it starts with some H 1 and then
it goes to another H 2 goes to another H2.
What is the time taken, very simple at any
instant you consider this as h so C d a naught
this orifice area is a naught root over 2
g h, this must be equal to the area here,
that means an elemental area here which is
a as a function of h minus A h d h dt .
Now, we have to find out h for a very hemisphere,
it is very simple if this is h already i have
done it, so therefore this is a circular cross-section.
So, this area will be if I define this as
x, this is equal to pi x square and this x
from the geometry, see this is the radius
of the hemisphere which will be given the
geometry of the tank and this part is R minus
h. This is also R because of it is a hemisphere,
so this is equal to pie into R square minus
R minus h whole square. So, you can find out
x as a function of h, R is constant so therefore
h as a function of h you put it and then take
t at the one side d t and integrate this function
of h from H 1 to H2 take care of the sign
and get the answer. That is all, so this is
the unsteady problem that means when the orifice
is discharging under a varying head, which
is the practical problem of emptying a tank;
that means the water level in the tank falls
down as the orifice is placed or orifice is
opened.
Alright now last part I will discuss the mouthpiece,
what is a mouthpiece, mouthpiece; now it has
been found that by attaching a short length
of pipes cylindrical length of pipe at the
side instead of an orifice. A short length
the discharge is enhanced as compared to an
orifice that means if there was an orifice,
sharp edged orifice, what could have been
the discharge. The discharge will be more
if instead of that orifice if we attach a
short projected part, short of a cylindrical
cross-section, very short which is known as
mouthpiece. What is the physical mechanism
for it, why the discharge is increased, what
happens is that this contraction which takes
place is again filled up. So, therefore when
the discharge is taking place then it is full
flow.
Let the area a naught is same as the orifice,
if we consider an orifice what could have
happened in case of an orifice, if there was
an orifice this area could have been just
like that. I just show you if there could
have been orifice, what could have happen
this. So therefore, the area a c gets reduced
from this a naught. So, therefore the reduction
of area contraction is not taking place, it
expands the tube and then finally flows. So,
therefore there is possibility of enhancing
the head, enhancing the discharge Q, but there
is a question. Let us consider a short height
that, that’s the head or the height of the
free water surface is same for all points
and we take a representation as the axis.
Now, the question is that okay, I understand
here contraction is there here, also contraction
takes place but this is again refilling the
tube.
A short portion is there so that when it comes
out it is fully filling the a naught, that
means the effective flow area is a naught
where, here effective flow area is a c, which
is less than the a naught, a c is less than
the a naught and the way we define the coefficient
of contraction c, c as a c by a naught, but
at the same time due to contraction there
is some loss if I consider the short section
so that the friction we neglect then at this
but we cannot neglect the losses due to contraction
which we have discussed this is the contracted
area c, c, so whether the loss counterweighs
the discharge rate or not let us see and it
has been found. It can be proved theoretically
that discharge rate is enhanced by refilling
the liquid in the tube that means to increase
or expand this stream tube, so that the effective
flow area becomes the total area of the mouthpiece
a naught even if there is a contraction loss
okay.
Let us see, now this is the section, let two,
let this is the section 1 and we write the
Bernoulli’s equation, let us write the Bernoulli’s
equation here pressure head is 0 atmospheric
pressure. We take 0, let us now consider unnecessary
Z 1, Z 2 this axis is the datum. So, velocity
is 0 h h is this now at 2pressure rate is
0 atmospheric discharge and this is the velocity
which we are finding out. Now, let us forget
about the friction, first of all friction
part we neglect and it is a short tube, when
you will compare with an orifice we will consider
frictionless orifice also, so that do not
worry comparison will be at the same basis
but loss, this loss you have to consider h
l, that loss is due to contraction. So, therefore
you see V 2 square by 2g is equal to h minus
h l alright.
So, what is h l now you tell me, what is h
l. So, 0 plus V 2 square by 2g plus h l, what
is h l if you recall h l is, it is due to
the contraction. Actually the contraction
loss is due to the expense and itself, I told
it repeatedly and this is nothing but V c
minus V 2 whole square by 2g. We have deduced
it earlier that it is the loss due to expansion,
not due to contraction. After contraction
when the stream tube expands, so it is just
the loss due to expansion of the stream tube
that means V c minus V 2 by 2g whole square.
That means if I take this V 2 square by 2g,
common. Then it is V c by V 2 minus 1 whole
square.
Now, V c by V 2, if you see here the continuity
equation V c into C c is equal to V 2 into
a naught. The continuity between this 2sections,
therefore V c by V 2 is what, a naught by
a c, that means 1 by C c C c. Definition is
a c by a naught that means, we can write h
l is equal to v 2 square by 2g 1 by C c minus
1 whole square. So therefore, if we place
this here what we get, if we place H 2 here
that means it goes this side, therefore V
2 square by 2g into 1 minus 1 by C c minus
1 whole square. I take this here h l plus,
very good plus is equal to h okay, this is
okay.
Now, I write here therefore, V 2 is equal
to, now if I take this as K, which is very
important what is K K is equal to 1 plus 1
by C c minus 1 whole square. This is k that
means this coefficient of V 2 square by 2g
that V 2 is root over 2 g h. Okay, then what
is the discharge Q area is a 0. Here area
is full a naught, there is no contraction
a naught into, oh god, divided by under root
K. I am sorry, so Q naught into root over
2 g h by under root K that means we can tell
that C d for mouthpiece is 1 by root over
K, this acts as the C d for mouthpiece into
a naught root over 2 g h.
This is the mouth piece Q, now what is the
Q for leave it ,what is the Q for orifice
Q for orifice, is C d into root over 2g, under
the same head now C d. We can write as C c
only, why because here also I want to neglect
the friction that means C d is equal to C
c into C v, we make C v is equal to 1 fluid
friction we neglect, but contraction is there
so C d is equal to C d. So, therefore here
C d is equal to C c, now if I can prove that
1 by root K is greater than C c under all
conditions, then I can prove that mouthpiece
always enhances the flow rate. This can be
proved, this as simple school level mathematics
that for values of C c less than 1 1 by root
k is greater than C c or 1 by k square is
greater than C c square, okay you see that
this can be always proved or you can prove
that other way, K square, this one K square
is less than K, sorry then we can write K
is less than 1 by C c square. Alright, so
provided I can prove 1 by root K is greater
than C c and 1 by root K is always greater
than C c for small values of C c. How can
it prove, how can you prove it, let us prove,
like it this is the K let us reform it in
this way 1 by root K is greater than C c,
I will have to prove that means, I will have
to prove K is less than 1 by C c square K
is less than 1 by C c square.
One line proof, you just break it 1 then plus
1 minus 2 by C c alright plus 1 by C c square
2 minus 2 by C c, 2 by C c is greater than
2 2 minus 2 by C c means this is a negative.
So, 1 by C c square minus something, so K
is less than 1 by C c square, very simple,
one line proof because for all values of C
c less than 1, because 2 by C c is greater
than 2 2 minus 2 by C c plus 1 by C c square.
So, therefore K is always less than 1 by C
c square, so this prove is not given in my
book, but i have written exclusively that
for all values of C c less than 1 this one.
So, this is proved that mouthpiece enhances
the, well the discharge rate. Any question
please, I think this is so simple that it.
Now, one dangerous thing is there, can you
tell me what is the dangerous thing? Now here,
you see whenever there is a contraction that
means here the velocity if you compare V 2
and V c V c is greater than V 2. Because a
c is, because a c is smaller than a a naught.
So therefore, p c is less than p 2 and what
is p 2 p 2 is atmospheric pressure that means
if there is an atmospheric pressure discharge
and if there is a area where in the upstream,
where the pressure is lower or there is an
smaller area in the upstream, where the velocity
is higher than the discharge velocity there
is a change. That the pressure will be chance
means obviously there is a pressure will be
lower than the atmospheric pressure. So, therefore
it is proved qualitatively or physically that
the pressure at the vena contract a is lower
than the atmospheric pressure, so we can find
out these vena contractor pressure by applying
the Bernoulli’s equation either between
point 1 and 2 or between this point c 1 and
let a point c, here between point 1 and c
or between point c and 2.
Alright let us write point 1 and c, you write
0 plus 0 plus h is equal to, what the point
is under question is p c by rho g plus V c
square by 2g. Alright, so p c by rho g is
equal to, but I think this will not held,
we want to find out how much it is less than
the atmospheric pressure. So therefore, you
write between this 2points, so forget it so
you write p c by rho g plus velocity head
V c square by 2g plus let this line is the
datum line 0 and you equate it with the 2at
any point in this section, this will be the
atmospheric pressure. Let I, write the p atmospheric
that means p c I express in their absolute
rho g plus V 2 square by 2g plus h l.
Well if we do so then you can write p c by
rho g is equal to p atmosphere by rho g minus
V 2 square by 2g minus V c square by 2g minus
h l. Now this can be written as p atmosphere
by in terms of C c only, so very simple you
just take h l now this I can write V 2 square
by 2g minus V 2 square by p. Atmospheric pressure
plus V 2 square very good, minus plus V 2
square minus V c square by plus h l, very
good so p p c by rho g is p atmospheric by
rho g minus V c square by 2g V 2square by
2g plus h l, very good, .
now minus V 2 square by 2g, one plus V 2 square
by 2g, 1 minus V c by V 2 okay whole square,
alright plus h l.
What is h now, this can be written as p atmosphere
by rho g minus, this is the discharge what
is V c by V 2 again, V c into a c is equal
to a naught into V 2. So, V c by V 2 is 1
by C c. So, 1 minus 1 by C c square, 1 by
C c V c by V 2 okay, plus what is h l h l,
why not we see here V 2 square by 2 g h. What
happened, very good C c minus whole square.
Now I can write p c by rho g, all plus p atmosphere
by rho g cut the minus V 2 square by 2g this
minus 1 by C c square minus 1 and plus 1 minus
1 by C c 1 by C c minus 1 whole square, this
is V 2 square by 2g minus. So, I have taken
the minus so it will be minus I am sorry minus
1 by C c whole square.
So, now this can be also told, I have also
shown that it is positive quantity, so that
this is always less than p atmospheric by
rho g, that I leave is, leave you as an exercise
that I have shown for all values C c less
than 1. This is a positive quantity now, I
tell you a representative value if you take
C c is equal to 0.62, then you get the value
of K, this one if I come to this value that
here if you see that root over 2 g h, where
I deduced the earlier one V 2 is yes. So,
root over 2h by root over K, the value of
K if you recall, the value of K this is the
value of K 1 plus 1 by C c minus 1 whole square
with the value of C c is equal to point six
2the value of K is 1.375.
So, c d which is 1 by root over k for mouthpiece
become 0.855 and C d for orifice is, C c that
is 0.62, 2very important thing and if you
put that value 0.62, here you get this is
a representative value C c usually lies between
0.6 to 0.65.
So, minus 1.225 roughly the value is like
that V 2 that means this will be lower than
the atmospheric pressure by 1.225 times the
discharge velocity, here so therefore, we
see by attaching a mouth piece that means
a short portion cylindrical, portion, short
portion. So, the discharge rate will be increased,
but thing is that why I am telling a short
portion, if instead of that we give a very
long cylindrical part projected, discharged
may not be increase, why this is because frictional
loss, frictional loss will be more that means
apart from this contraction loss there may
be more frictional loss so that the actual
velocity of discharged will be much less than
root over 2 g h plus all those. So, these
h l will increase much that is why a short
because sharped edge orifice friction is very
less, C v value usually lies between 0.9 to
0.95, so therefore we must have a very short
piece which is just sufficient to expand this
stream tube alright.
So, this is a very important statement sometimes
it is written in the short portion, what is
the meaning of the short portion because in
the field of science the words used are not
redundant there is no use ah there is no scope
of using more adjectives and all this thing,
so each and every word has got its implication.
So today, yes I conclude this talk, so we
have completed the application of Bernoulli’s
equation. If we look back the beginning, we
have first recognized the Bernoulli’s equation
that is the pressure head plus the velocity
head plus the kinetic head, first the datum
head the potential head then the 3 components
of mechanical energy remain constant everywhere
in the flow field provided the flow field
is irrotational but for a rotational flow
field it is constant only along a streamline.
Then after that we recognized a solid body
rotation that is a forced vortex, what is
a free vortex and irrotational vortex. Pressure
distributions in forced vortex and a free
vortex then we we recognized the different
losses, siphon then probably siphon, we started
siphon what is the principle of siphon, how
the siphon work then different minor losses,
losses due to expansion, contraction. Then
the concept of static pressure, stagnation
pressure, pitot tube, application of pitot
tube, then the discharge through orifice and
mouthpiece, this we discussed so today I conclude
this chapter, Tomorrow, next class I will
start the incompressible viscous flow thank you.
