Good morning, I welcome you all in this session
of fluid mechanics. In the last class, if
you recall, we were discussing about the manometers.
A manometer is a device, which measures the
pressure of a system in terms of its difference
from the atmospheric pressure or the pressure
difference between two points, whether the
fluid is at rest or fluid is in motion. At
last we discussed how the manometer registers
the pressure difference between two points,
when that fluid is flowing through a pipe.
Let us recall this again, that if we recall,
this is like that that is there is a pipe
where the fluid is flowing in this direction,
it is the flow direction. Then, if the two
points A and B are connected, if the pressure
PA is greater than PB, then the difference
of pressure is measured by connecting a manometer,
which is essentially a U tube like this, sorry
like this. We use a manometric fluid, which
is which is having a density higher than that
of the working fluid, that is the fluid flowing
through this pipe. We register this is the
fluid a deflection; that means, the interface.
That means the separation surface between
the working fluid and the manometric fluid.
The interfaces stand like that. This is known
as the let this is y.
If we take the manometric fluid density is
rho m and the density of working fluid is
rho w, we found that the delta P that is equal
to PA minus PB becomes equal to rho m minus
rho w g into y. That means, observing this
deflection manometer deflection. That means,
the difference in label of the meniscus y.
We can find out the pressure difference. How
to find it by writing the manometric equation?
That means, we take this PQ line as the same
horizontal line in this same expanse of fluid,
which is the manometric fluid. Then what we
do is we write the pressure here; equate the
pressure here from this limb. When equate
the pressure here from this limb and find
out this equation.
Now, what we start at that what is the situation?
We see a very interesting thing that when
the fluid flows through an inclined pipe,
let the flow direction is like this. The fluid
flows through an inclined by where we are
interested to measure the pressure difference
between these 2 points A and B. Let us connect
the manometer like this. Sorry, a U tube is
like this. Let us connect a U tube like this
U tube manometer.
Then, what we do? The manometric fluid will
be having a deflection definitely like this.
Let us have a deflection of the manometric
fluid. So, we just this is the working fluid
this is the working fluid. Let this deflection
as it is we call it as y. Let from B this
distance is x. A and B, there is a difference
in vertical height between the section A and
B. Let it be delta j just I give this nomenclature
delta j. Now, from simple manometric equation;
that means, if you take this line as the PQ
horizontal line, if we equate the pressure
from both the limbs.
Now, from this limb, if we come from this
limb here first, PA plus if the working fluid
density is rho w, the same nomenclature and
rho m is the density of the manometric fluid.
That means rho w delta z plus x; that means,
rho w into g into delta z plus x. That means,
this is the pressure of this column of liquid
plus rho m into y, y is the deflection. This
is y. This is y.
So, this becomes equal to what PB. That means,
the same pressure is here at the same horizontal
level within the same expanse of fluid. So,
this is PB plus this pressure due to working
fluid of height this x plus y. That means,
rho w g x plus y. See, if we equate this,
we will see PA minus PB. So, rho w g x cancels
from both the sides PA minus PB. So, plus
rho w g delta z is equal to rho m minus rho
w into g into y.
So, therefore, we see that if we compare these
2 equations, we see that from the deflection
y manometer deflection; that means the difference
in level of the meniscus. If I equate this
rho m minus rho w into g y, this gives an
additional term from the pressure difference
rho w g delta z in case of a horizontal pipe
A and B at the same horizontal level. So,
delta z is 0. So, we can put delta z 0. These
is the special case becomes this. But, what
is the physical interpretation of this term
actually?
Now, if I express the vertical elevation of
A and B, this we will come afterwards in our
studies in our course and z 2 from a reference
datum, from any reference datum, from any
reference datum. If I just specify the elevation
of A as z 1 and elevation of B as z 2, then
I can write this as PA plus rho w g z 1, this
quantity minus PA PB. PB plus rho w g z 2
is equal to this 1 rho m minus rho w. This
one will come like that.
So, therefore, we see this manometer deflection
multiplied by the difference of density into
acceleration due to gravity gives the difference
of static pressure plus some term which is
rho g z one where z 1 is the elevation of
that point from a reference datum. So, this
term is known as P star that is piezometric
pressure. That is piezometric pressure. It
is defined the static pressure plus the equivalent
pressure corresponding to its vertical elevation
from a reference datum.
So, this is very piezo important definition,
which we will come across afterwards piezometric
pressure. So, always piezometric pressure
is defined as static pressure plus the pressure
equivalent of its elevation from a fix datum
rho g. So, therefore, it registers a piezometric
pressure difference PA star minus PB star.
That means, the same equation we can use for
an inclined pipe where if we find this quantity,
this will give you the piezometric pressure
difference instead of static pressure difference.
So, if I will have to find out the static
pressure difference, of course, you will have
to deduct these things. Sometimes, directly
we want the piezometric pressure difference.
In that case, the manometer reading from the
manometer reading y, we can straight away
find out the piezometric pressure difference
in case of a horizontal pipe. The piezometric
pressure difference becomes equal to this
static pressure difference because delta z
is 0.
Now, next we come to the next part. Here,
we see that basically the pressure difference
between 2 points in a fluid flow is measured
by a manometer. Basically, by this equation,
either it is a static pressure difference
or it is a piezometric pressure difference
depending upon whether the pipe is horizontal
or inclined. Now, if we want to measure a
very small pressure difference, then what
we have, we should have we should have a substantial
value of y for a small pressure difference.
In other words, you can say, if you want to
increase the sensitivity of the manometer;
that means, for a small change in delta P.
If we want to have a substantial value of
y or readable value of y, what we should do?
We should make this part very small very small.
So, that even what small delta P we have a
large y; that means, rho m minus rho w should
be very very small nearly equal to 0. Rho
m that is the manometric fluid density will
be very close to the density of the working
fluid
In practice, it is very difficult to have
a manometric fluid, whose density is close
to the working fluid and at the same time
defining a meniscus good meniscus. That means,
a separation of surface between the 2 liquids
has to be defined. So, well defined meniscus
should be another reason why we cannot read
y. So, a well defined meniscus between the
immiscible liquids and also a close density
values are not practically feasible. So, without
going for changing the properties of the liquid
go for some other changes in the construction
of the manometer to measure small pressure
difference.
This is one, which is known as inverted tube.
This is done inverted tube manometer inverted
tube manometer inverted tube manometer. What
is inverted tube manometer? In this case,
what happens if we consider a horizontal pipe
and the fluid is flowing through it? What
do we do if we want to measure the pressure
difference between 2 points A and B? So, inverted
tube manometer is placed like that in an inverted
way.
What happened is that there is a bulb inverted
tube manometer contains air as the working
fluid, air as the manometric fluid, air as
the manometric fluid with some pressure. So,
pressure of the air is adjusted by the bulb.
That means, if you open the bulb, some air
may come out. You know the mass of the air
is reduced. The pressure in the close system
of the air is reduced. So, this way, we can
we can change the pressure of the air inside.
So, what happened?
The working fluid depending up on the pressure
will go inside. For example, the working if
the pressure PA is greater than PB that will
be adjusted. It is very simple. So, therefore,
if we measure this height, this side, let
this side is y. Then we can get simply the
PA here is equal to some pressure. Let air
pressure is P0, so P0 plus rho w into y. Let
this is x rho w into g into x plus y. Similarly,
here PB is equal to P0. The same air pressure
in the closed system, the air will impress
the same pressure. There is no difference
in the pressure because of the height of the
air.
So, you can tell say the pressure here is
P0. So, here pressure will be P0 plus height
of the air. So, that height of the air we
neglect because the density of the air is
small compared to the working fluid. If this
is a liquid is already working, fluid is a
liquid. So, we neglect for small height, the
variation due to pressure because of the height
of the air. So, the same P0 pressure is impressed
here. So, that is P0 plus rho w g into x.
So, if you just subtract 1 from other, you
get PA minus PB is equal to what will be there
rho w g x will cancel rho w g. That means,
there only appears there appears the density
of the working fluid liquid. So, therefore,
by this we can have that. So, this is low
compared to the manometric fluid. If you use
a manometric fluid in usual conventional U
tube manometer, we can have a large deflection
even for a small pressure difference.
Then, another modification is inclined tube
manometer inclined tube manometer inclined
tube manometer inclined tube manometer inclined
tube manometer. What is it? It is very simple.
Again, on principle, if 1 limb of the U tube
is made inclined, what happened? If 1 limb
of the U tube is made inclined, just imagine
what will happen? That means, it is very simple
to have amplification. That means, if a pressure
is connected to 1 pressure point, another
pressure point, let this is A, this is B with
the working fluid. Then what will happen?
The manometric fluid will be deflected like
this. So, let us have the deflection like
this. That means, for a vertical deflection
y, we can read this as it is.
So, this corresponds to for example, this
vertical deflection y corresponds to a deflection
s along the pipe with a magnification. That
is if this angle of theta, so s is equal to
y by sine theta. That means, if we measure
the displacement of the liquid mercury or
the manometric liquid is any mercury is used
as manometric fluid or manometric liquid along
the pipe, then that will be amplified by a
factor of 1 by sine theta with a y. So, this
y will be proportional to the pressure differences
PA and PB applied here.
So, the magnification is made by this inclination.
So, that this s if we measure along the pipe
along the inclined tube of this U tube manometer.
So, we measure a greater displacement than
y because of this geometrical magnification
1 by sine theta. This is an inclined tube
manometer. Usually, we have to find out the
difference of level. That means, we will have
to find out s from this pressure is applied.
That means, we will have to recognize the
difference, the deflection here and the deflection
here. To avoid this, what is done that one
part is made with larger cross section compared
to the other part?
Why? In that case, we will understand that
only the movement of the liquid in 1 limb
is sufficient to be read. That means, let
us consider a manometer like this inclined
tube manometer where this part of the limb
is made with a larger cross section let A.
This part cross section is A where A is very
very larger than greater than A. So, pressure
is being this. It is connected to a system.
This is connected to another system PA, PA
before the connection. Let the level of the
liquid was like this. This is the original
level of the liquid, manometric liquid. There
is manometric liquid.
Now, after the connection with the pressure
where PA is greater than PB, then the manometric
liquid will be depressed here. So, manometric
liquid will come here. It will go here like
this. So, now, we will have to find out this
deflection. If you write the manometric equation,
you will see we require this deflection y
to be measured. That means, in that case,
we have to measure this s. Try to understand
this s.
So, this s to measure we will have to know
this length, but we do not do that. What we
do? This part we do not make transparent.
Only this part is made transparent. We see
the initial level in the transparent inclined
part. We see the final level. Instead of measuring
this, we measure this s 1. That means, we
neglect the measurement of this. Why we can
neglect? How we can neglect?
If this area is much larger than this, so
this deflection means, this vertical deflection
is much much smaller compared to these displacements
in the tube from the continuity because the
same fluid which is depressed here will be
going up in this tube. As the cross sectional
area is much larger, the cross sectional area
is made 2 times, 2 times in the order of magnitudes
100 times larger than it. The diameter is
made some 10 times more than this is well
cases. So, therefore, this depression is much
more negligible compared to this displacement.
So, error in taking s 1 as the measurement
as instead of s; it is very, very less. So,
that we need only this inclined part of the
tube to be transparent and this can be made
metallic if the cross sectional area is larger.
So, simply what we do? We take the initial
level. We take the final level. This displacement
along this inclined tube is measured. So,
this displacement is magnified by a factor
of 1 by sine theta instead of having a vertical
displacement here that we have explained I
have explained. Now, here the one question
comes if the magnification factor is 1 by
sine theta, we can go still lower and lower
value of theta.
So, we can get more magnification that for
a small difference of delta y, very small
difference in pressure or small difference
in y. We get a larger value of s. So, there
should be a limit. Limit is like that as you
go on decreasing the theta; you will see this
meniscus free surface of the manometric fluid
will be flat. It is very difficult to find
out its exact position. We will take its end
point. We will take its middle point or here.
So, this we will call error to the system
because free surface will be horizontal.
So, if we make a higher value of theta, this
surface will be such that the difference in
s, that means, the distance along the limb
from this two points on the free surface will
be this. So, if you make and make more horizontal,
the surface you see like this, then a free
surface will be wide because it has to be
horizontal. So, this 2 point, if you measure
the displacement in this direction, this we
will call substantial error. So, that theta
should not be very very low. Theta is usually
restricted between 10 degrees. So, this is
principle of inclined tube manometer. Now,
we we will discuss the pressure.
Now, next section is the pressure exerted
by a surface submerged in a expanse of static
liquid that is hydrostatic force exerted on
surfaces submerged. So, when a surface is
sub merged in a fluid or at rest, then the
pressure will be exerted on the surface. What
should be the pressure exerted on a surface?
So, let us write that hydrostatic. This is
already done. Let us see that hydrostatic
force on a submerged surface hydrostatic force
on a submerged surface. We should first consider
plane surface. Now, let us consider a fluid
at rest with a free surface. Let this is its
free surface of the fluid.
Now, let consider this is a liquid and a surface
is submerged in a liquid within the liquid.
This is the surface at an inclination of theta.
That means the plane of the surface and the
plane of the free surface makes an angle theta.
That means, if we see the surface from this
top view, this is the surface of area A. That
means that the surface is inclined surface.
So, this is the free surface. This is the
surface, which is inclined at an angle theta,
which will be seen from the top. So, this
is the view. So, this is the surface an arbitrary
surface.
Now, you see in this case, what happens? You
can argue that the pressure force is acting
on the both the sides of the surface due to
the water. At any point, they will be equal
the pressure on this surface. Then pressure
on the other side will be equal at this point
also. They will be equal. So, the net resultant
force will be 0, but this is not the case
we are considering.
In actual case, a surface submerged will not
experience any of force. We are interested
to find out what is the force exerted on the
surface if other side is exposed to atmosphere.
That means, what is the gauge pressure exerted
or the gauge force exerted on the surface
from one side. That means we will consider
as if the other side is exposed to atmosphere.
What should be what could be the net force
exerted on this surface? You can just think
the problem in that way otherwise you can
argue. In the actual case, they balance each
other.
So, we are not considering the forces from
the other side. We are considering forces
from one side. Other side is as if open to
atmosphere. Now, as you know, the pressure
varies linearly with h in compressible fluid
if it is a liquid. So, therefore, there will
be a variation of pressure like this. That
means, the pressure force will act on the
surface like this. The pressure force will
act on the surface. Let this is the surface.
Let AB is the surface pressure force. This
is the pressure force will act because the
pressure at any point is irregularly proportional
to the height.
Now, let us consider an elemental area. Now,
first of all, before that let us consider
a co ordinate axis. This point is the point.
This line seen as a point is where the plane
of the surface and the free surface meets
the plane of the surface. The free surface
meets at a line, which is seen at a point
from this direction perpendicular to the plane
of the paper. This is taken as the origin
o and o o y y co-ordinate is taken a line
along this plane along the direction of the
plane. Perpendicular to this, ox axis is taken.
I think there is no problem.
Now, let us consider a small elemental area
d A, which is at a depth h. That means, this
one this one is h. That means, this elemental
area here, which is seen as the area from
the surface area d A is at a depth h from
the free surface. So, what is the force on
the elemental area? First of all, what is
the pressure exerted by the liquid on this
elemental area? It is at a depth h from the
free surface. Here, we are considering the
force or pressure above the atmospheric pressure.
So, this will be simply rho g h. Rho is the
density of the liquid rho. So, rho g h, h
is the depth at which this elemental area
is considered. So, therefore, the force on
this elemental area will be d F will be rho
g h times d A and this h from the simple geometry
if I define y as the co ordinate y co ordinate
of this elemental area from the x axis.
So, this h will be simply y sin theta. So,
it will be y sin theta well. So, if we integrate
this force over the entire area; that means,
double integration over the entire area. We
will get rho is constant, g is constant, sin
theta is constant. Why not we take out rho
g sin theta? h will not be there in this case.
Rho g h, h is y sin theta rho g sin theta
y d A. I think this is clear. Rho g h is the
pressure here in this area. So, the force
on the area elemental area d F is rho g d
A that is the rho g h into d A. d A is multiplied
because the pressure into area is the force
and h is substituted as y sin theta. This
is theta. This is y. This is h. So, y sin
theta becomes rho g sin theta y d A.
If you make a double integral over the area
A, this is row g sin theta y d A. Now, what
is this term? This term is an implication
you see that y d A over the entire area represents
what is this moment of the area about x axis.
So, this term is d y d A, this term is the
moment of the area about the x axis. So, now,
if we define a center of area, you know what
is center of area very well. So, if you define
c as the center of area whose y co-ordinate
from the reference x axis if it is y c, then
by definition y c is what? It is the moment,
first moment of area over the area A divided
by the total area d A. That means, the total
area A; that means, y d A is equal to y c
into A.
So, if we replace this here, we get a very
interesting result that F becomes is equal
to rho g y c into A instead of integral y
d A. That means it is rho g y c sine theta
into A. What is y c sine theta? y c sine theta
is similarly, the vertical depth from the
free surface. So, this is y c sine theta.
If I denote this as h c; that means, h c represents
the vertical depth of the center of area.
That means if it is y c, y c sine theta is
this vertical depth. That means, if the center
of the area is at a vertical depth from the
free surface, h c F is equal to rho g h c.
So, this is a very interesting result. That
means, if there is an inclined surface, there
is a inclined surface where we see the pressures
at each and every point is varying because
of their depth from the free surface. The
total pressure due to this total force due
to this hydrostatic pressure is simply the
pressure at the centroid.
If one can find out what is the pressure value
of the centroid times the total area, will
give you the force. That means, it is it can
be equivalently told that if this surface
AB could have been placed horizontally, this
will be at this end. So, if the surface AB,
this is wrong surface AB could have been placed
horizontally at a depth h c below the free
surface, then what could have happened?
The uniform pressure in a horizontal surface
pressure at each point is uniform. This is
the pressure P. So, this P is equal to rho
g h c because of this depth. That means, placing
an inclined surface in a fluid submerged in
a fluid is equivalent to place the same surface
at the depth of its center of area. That means,
if we know the center of area depth at the
center of area, we can find out the pressure
and times the area is the total force. That
means, it is equivalent to it will experience
the same force like that. We say experience
the same force if it could have been placed
in a horizontal position at a depth h c, which
is the depth at which the center of area of
this area lies.
So, this is one important conclusion. Now,
next is to find out the point of application
of this force. Now, it is true that the point
of application of this force well I can put
it here. Here itself, you can see the point
and how to find out the point? Now, you see
that the resultant force magnitude is known
that is equal to rho g h c into A. So, if
I know the value of x c, the centroid depth,
depth of the centroid from the free surface,
I can find out the total force. The direction
of the force is always known that this is
perpendicular because for all elemental surfaces,
the pressure force is perpendicular to the
surface.
Since, it is a plane surface, the perpendicular
at all points are parallel. So, a scale of
summation will do. So, the direction will
not change. Direction is perpendicular to
the surface. Where is the point of application?
It is also very simple. The point of application
definitely will not coincide with the center
of area y. This is because the forces are
not proportional to the area. As we go along
surface, we see that depth is increasing.
So, even for the same area here, if we have
another same area here that 2 pressure forces
and the 2 areas are not same.
This is because the pressure intensity is
changing, because of the depth. So, it is
common sense that as we go below the surface
free surface, the pressure is more and more.
So, therefore, center of pressure will lie.
For example, here we denote this c p as the
center of pressure which will lie beyond c
center of area towards the depth of the fluid.
So, let us consider this is the center of
pressure. Let us define the co ordinate of
the center of pressure as y p from the x axis.
Let us define this as the x p from the y axis.
Now, common sense will tell us to make this
c p here, not here c p would have coincided
with c. The center of area of the forces could
have been proportional to the area element,
but it is not. So, since the pressure is increasing;
that means, with the depth, that means, the
lower portion of the area is having more force.
So, that it will be this side from this center
of area. Now, if we apply simple mathematics
to find out the y p and x p what is that?
That means that the resultant force multiplied
with y p; it will be the sum of the moment
of the component forces above the o x axis.
That means if I tell write this F into y p
will be the sum of the moment; that means,
integral of the component forces moment of
the component forces on the x axis. What is
this component force rho g d a y sine theta?
Its moment will be about y axis. x axis is
another y; that means, it will be simply rho
g y square sine theta d A rho g y square sine
theta d a. That means, y p will be integral
rho g y square sine theta d A divided by F.
F is rho g y sine theta d A. That means, it
is integral rho g sine theta y d A.
So, this is already found as integral y d
A as h c into A. So, therefore, I write is
at h c into A. Here, I will write as integral
y square sine theta will cancel y square d
A y square d A into h c by A. What is y square
d A in this case? y square d A, you see is
the second moment of area of the surface above
the x axis. Second moment about x axis, sometimes
it is second moment of area or moment of inertia
of the area about this x axis. So, x, x divided
by h c into A. Is there any problem? No problem.
Now, what we get? What we do that this can
be replaced by a parallel axis theorem. Before
that, I let me do the x 1. So, what will be
the x component? Now, x component, if x p
we want to find out, we can take the moment
in the similar way about the o y axis. That
means, this will be simply integral of rho
g x y sine theta d A divided by what is this
thing? This is rho g sine theta y d A. now,
if we come here, we see that this becomes
x p is equal to integral x y d A divided by
what is that? This we can write as h c into
A.
So, therefore, we see that this is known as
the product of area I x y. What is this x
y d A? That is product of area. That is known
as product of area. That means, if this be
the area, this x co-ordinate of at any point
or at any elemental area, this is the y co
ordinate. So, the integration of x y d A over
the entire area is known as the product of
area, I just simply write it is as x y this
already you know h c into. So, we know x p
by a parallel axis theorem. Now, if you know,
any question please you ask, you can ask at
this moment.
So, before we proceed to the next one, is
there any problem?
This will be y c.
Yes yes yes yes. I am sorry.
This will be y c. This will be y c. I am sorry
I am sorry. This is because sine theta I am
cancelling, this will be y c. This is because
sine theta I am cancelling from these sides,
numerator and denominator. So, I cannot take
y sine theta as x c. So, it will be y c because
sine theta has such as cancelling from the
numerator and denominator. It should be y
c. I think there is no other problem.
Now, by a parallel axis theorem, you know
that if I define the moment of inertia that
means, I take the parallel shifting of the
axis as you have read earlier. If I define
x dash and y dash are the co-ordinate axis
parallel to x and y through the centroid c,
if I now define. You understand this x dash
and y dash through the centroid c.
If I define that I x dash is the moment, second
moment of area or moment of inertia about
o x dash axis, then the relationship between
these and the moment of inertia of the old
x dash is I x plus A into y c square. That
means the moment of inertia about this axis
is the moment of inertia or the second moment
of area above this axis plus area into y c
square. That is the distance of this axis
from the earlier axis y c square. Similarly,
for the product of area, if I define with
respect to new x dash y dash axis, this will
be at the relationship I x y plus A.
The two co-ordinate will come, which will
define the co-ordinate of the new axis from
the old one. That means this is y c in that
case. This is x c in our case. Our new axis
is to center of area. In that case, y c, x
c are the y and x co ordinates of the center
of area. So, now, if you substitute see this
interesting results come y p is equal to if
I substitute this here. That means, I x, it
will be minus. It will be minus. I x will
be I x dash plus this will be minus. I am
sorry. So, I x will be I x dash plus this
I x y. That means, in the old coordinate will
be in the new co ordinate plus this.
So, if I substitute here, what I will get?
I will get I x dash divided by y c A plus
cancels y c. So, I get y p minus y c is I
x dash divided by y c A, y c into A. Similarly,
if I substitute this one here, what I will
get? This is I x y I x dash y dash. So, we
will get x p minus x c is I x dash y dash
divided by y c A plus. What we will get x
c y c. That means that if I substitute here,
y c will cancel A simply. x c has come here.
Nothing will be there. So, I get this and
this one. So, therefore, we see that y p minus
y c is this one. This is a positive quantity
because I x dash contain an integral with
the square of the ordinate.
So, therefore, the second moment of area always
a positive quantity y p minus y c is greater
than 0. That means, this quantity gives you
the displacement in this direction. This is
y p minus y c. Sometimes, this is told as
the eccentricity. That means how far the center
of pressure is shifted from centre of area
in the direction along the plane. So, this
is y p minus y c. So, this distance is given
by a simple expression I x dash divided by
y c A. I x dash is the moment of second moment
of area about this o x dash axis. y c is the
center of area in respect to this co-ordinate.
A is the area
The most interesting fact is that x p minus
x c becomes 0 because the product of area
about the centroid axis is 0. This means that
x p is equal to x c because if I take the
axis x dash y dash to centroid plane, the
product of area is 0. So, therefore, x p minus
x c is 0. This one can make from a common
sense that if this c is the center of area.
So, whatever may be the arbitrary area, the
center of pressure cannot be shifted in a
direction perpendicular to the plan. This
is because any area in this side and another
area will be always at the same tip. So, therefore,
the pressure forces are proportional to the
area element only.
So, therefore, the center of pressure and
center of area is not displaced in a direction
perpendicular to the direction of the plane.
So, it cannot be disturbed. So, therefore,
this is x p minus x c; that means, this is
the x p and this is the x p. So, x co-ordinate
of the center of pressure coincide where the
y co-ordinate. That means, along the plane
is shifted towards the depth of the fluid
by the amount, which is given by y p minus
y c is this. You have understood this plane
figure, plane surface. So, next class, we will discuss the pressure forces 
on curved surface.
Thank you.
