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Welcome to this sixth lecture, in this course
on fluid mechanics for chemical engineering
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undergraduate students.
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In the last lecture we were discussing f luid
statics that is the forces that are present
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in a fluid that is not under any kind of motion
that is the fluid static. And we will see
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that there are some interesting features that
come about even in a static fluid and that
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will be the subject of our discussion for
this lecture. If you recall, forces and fluids
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can be classified into two types.
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We divided them into broadly as body forces
and surface forces. So, remember we are carrying
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on all our analysis within the continuum frame
work, where the fluid is assumed to be a continuous
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medium in which various quantities such as
pressure, density, temperature, velocity.
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They are all assumed to be smoothly varying
functions of both spatial and time co-ordinates.
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So, within the continuum frame work, if you
consider an element of fluid for simplicity,
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I take a volume element, which is cubic in
shape this is a volume element in a fluid.
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We can ask the question why is this volume
element is in general going to move, if the
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fluid is not under static conditions. Well,
it will move, because of the fact that if
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you apply Newton’s laws of motion. So, mass
time’s acceleration of this volume element
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is equal to sum of all the forces in the volume
element.
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If there is an imbalance of all the kinds
of forces that are present in the volume element
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then the fluid element will accelerate. So,
it is important to know, what are the forces
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that are acting so, one force that is known
to us is called the body force. The body force
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acts entirely through the entire volume of
the fluid.
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Well known example is that of gravitational
force. Suppose, you have this volume element
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the volume to be delta V then, the mass of
this volume element is rho times delta V times
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the acceleration due to gravity g, will give
you acceleration as a vector. So, will give
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you a body force that acts through the entire
element of the fluid, entire volume of the
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fluid. So, these are called body forces or
sometimes these are also referred to us volume
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forces. In contrast, surface forces are forces
that are acted upon only on the surfaces for
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example, if you consider the top surface of
this cubic volume element.
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The fluid that is present just immediately
outside will exert a force on this surface.
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And this is called a surface force; the nature
of the surface force is ultimately due to
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molecular interactions. For example, if you
consider a very dilute gas, if you consider
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a tiny volume element within the gas, then
the gas molecules present outside will be
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colliding with these molecules present inside
and that will result in this force. And that
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force will just be present only on the surface
it would not extend in to into the interior
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of the volume element. So, since these forces
act only on the surface of the given volume
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element they are called surface forces and
the surface forces in general. So, let us
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discuss a little bit more about surface forces
before going to static fluids.
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If I take a very simple surface element, this
surface element there is fluid present outside.
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So, the surface element forms part of a volume
element, which is the cubic volume element
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that we have chosen for simplicity. But let
us, consider one of the top one of the surfaces
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namely the top surface. There is fluid present
outside and there is fluid present inside
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and the every surfaces marked by a normal
the unit outward normal let us call it n.
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So, n is the unit outward normal to the surface.
It is a unit normal perpendicular to the surface
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and it is pointing from inside to outside
therefore, it is called unit outward normal.
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Now, if you consider the forces in general
that are exerted by the fluid that is present
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outside on the fluid that is present inside
on this surface.
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This force, which we will call R can point
out in general in any direction. This R is
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a function in general; it is a function of
the surface on which we are focusing that
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is a top surface. So, it is a function of
the unit outward normal, but the direction
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of this force can need not be always in the
direction of n. It can be in any arbitrary
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direction. So, R is the force, but per unit
area, R is a function of n. So, let us write
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this as R of n, R of n is the force per unit
area acted upon by the fluid outside
on the fluid inside at the surface. The inside
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and outside are demarcated by a surface. So,
R is the force that is x per unit area exerted
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by the fluids as present outside on the fluid
that is present inside.
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So, if you look at this surface of interest
and let us say this is the unit normal. This
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R will can be resolved into a component that
is parallel to the normal. Let us, call it
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R n and then it can also be resolved in the
direction perpendicular to the normal. Let
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us, call it R t 1, R t 2.
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Now, R n is called the normal stress, it is
a normal component of the force that is being
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exerted by the fluid outside on the fluid
inside R n is called the normal stress while,
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R t 1 and R t 2 are called tangential or shear
stresses. So, in general the fluid that is
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present outside will exert a force on the
fluid that is present inside only on the surface
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and this is because of viscous action. And
this force can have a component along the
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normal as well as a component perpendicular
to the normal.
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But when, we look at a static fluid, a static
fluid by definition cannot bear any shear
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stress because if there is any finite non
zero faces the fluid will start moving. So,
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the shear stress is in a static fluid must
be 0. So, the static fluid is a fluid where,
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the shear stresses or the tangential component
to the stresses are 0 so, the only, because
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otherwise, a fluid will start moving here.
Now, considering a fluid that is completely
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stationary or static. So, if a fluid is stationary
or static, if you take any volume element
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and enquire what is the force is on the surface.
They cannot be any component that is parallel
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to the surface because if there were to be
a force that is parallel to the surface then
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that will constitute to shear force or shear
stress and the fluid will start moving.
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So, we since, we are considering only static
fluids where there is no motion, there cannot
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be any shear stresses. So, if you take a static
fluid and if you take a volume element in
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a static fluid on each and every surface the
force can exert only purely normally. If you
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take a static fluid and you take a volume
element that is big then, regardless of the
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nature of the volume element. The force will
act purely normally to the surface of the
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volume element. There cannot be any force
that is tangential to the volume element.
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So, shear stresses are 0 in a static fluid.
Now, this need not the result that shear stresses
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are 0 and the force is acting purely normal
to the surfaces holds for any arbitrary volume.
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Suppose, I take an very arbitrary volume,
this is a three dimensional volume, but I
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can draw only a 2 d sketch of it. The force
is always in the direction of the normal for
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any arbitrary volume element in a static fluid.
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So, R of n in a static fluid must be; since,
R is a vector and it is purely in the direction
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of the normal must be some constant times
n, because r is purely in the direction of
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n and so, it must be some constant times n.
So, R is a vector, n is a vector this constant
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is a scalar constant. So, in general quantities
of physical interest can be classified as
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scalar or vectors. Scalars or quantities,
which are described by a single value numerical
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value at a given point in space. For example,
if you see temperature at a point, it has
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only when one value and it has no sense of
direction at a given point in space in a fluid
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likewise, if you consider density. You can
ascribe a value of density to a point in a
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fluid with in continuum hypothesis, but there
is no sense of direction associated with density.
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So, where as quantities as such as forces
and velocities or vectors, which have a magnitude,
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that is a numerical value as well as a sense
of direction at a given point in fluid so,
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this constant C is a scalar. So, if you take
a surface area delta a in a static fluid and
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you ask what is R the force exerted by a fluid
that is present outside on the fluid that
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is present inside it has to be in the direction
of the n, it has to be purely normal to the
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area element surface element. Usually, it
turns out that this normal force in a static
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fluid is compressive. That is the force is
acting inwards while, the unit normal n is
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pointing outwards the force is pointing in
the direction of minus n. So, R of n is normally,
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written has minus constant a scalar p times
n where this p is called the pressure.
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So, pressure is the magnitude of the compressive
force that is exerted by the fluid that is
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outside on the fluid that is inside. And since,
R of n is proportional to n and n is pointing
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outside and p has is compressive this negative
sign accounts for this compressive nature
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of this force or stress in a static fluid.
So, this is called the pressure.
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So, the pressure, pressure is denoted by the
letter p is the normal force, normal compressive
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force per unit area
acting in a static fluid. We can also; well,
before I precede further the units of pressure
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is Newton’s per meter square is one Pascal
S I. In S I system units is p is one Newton
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per meter squared is one Pascal it is that
is the standard units. We also showed in the
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last lecture that at a suppose, I will take
a point in a fluid and I construct a tiny
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volume element about that point infinitesimal.
Suppose, I take an infinitesimal volume element
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about a point and then, I asked the question,
what is the value of pressure? So, this for
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simplicity this volume element is again, a
cubic volume element.
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So, there are eight faces one two or there
are six faces to this cube. There are four
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side faces on top and bottom there are six
faces to this cube. So, you can ask the question
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the pressure will act purely normally in all
this six faces, but what is the value of the
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pressure p as the volume element shrinks to
zero. So, the question is we know that pressure
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will act purely normally at each and every
face in a static fluid. But what is the value
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of p is it same or it is different along the
different faces we are going to show that
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it is in fact the same. So, in order to this
we take a wedge like volume element and let
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us put a co-ordinate system, this is x and
y is going into the board and z is so, y is
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the direction that is in the plane perpendicular
to the board.
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So, since and let us call the dimension perpendicular
to the board as b and let us call this dimension
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as delta z and let us call this dimension
as delta x. And let us say there is a body
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force of gravity, acceleration due to gravity
g acting and let us call this top dimension
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as delta s. So, since this b is very very
large compared to the thicknesses delta z
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and delta x. We need not consider any variation
in the z direction, in the y direction in
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the plane perpendicular to the board. So,
now first we will ask, what are the various
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forces in order to prove that pressure so,
let us give some notation.
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Let us call this p z; this is the pressure
that is acting on a surface which is; whose
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perpendicular is in the z direction let us,
call this p x this is the pressure acting
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on a surface whose, perpendicular unit normal
is in the x direction. And let us call this
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on the pressure on the incline surface as
p n. So, let us call this as theta, angle
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theta this is also angle theta by geometry.
So, what are the various forces? Well, firstly
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what is the body force acting on this wedge
like volume element? The body force is the
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mass times acceleration due to gravity mass
is nothing but the density times volume. Volume
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of this wedge suppose, you consider a rectangle
a cubic volume and then you divide it into
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half through the diagonal and then it is a
volume element in the third direction like
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this.
So, the volume of this is half the volume
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of these rectangles so, I will put half delta
x delta z time’s b times g. This is the
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mass times acceleration due to gravity g.
This is the body force that is acting on this
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wedge like volume element. What are the surface
forces?
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So, in order to this I will just consider
a cross section of this wedge like volume
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element. I will take a cut in the y direction
and this is p x, this is p z, this is p n
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and this angle is theta so, is this angle.
So, this p n will have component in the horizontal
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and vertical direction. So, what are the various
forces acting on this static fluid element,
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this fluid element is static. So, what are
the various forces? Let us, take the x component
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of the force. So, just to remind you we had
x like this and z like this. The x component
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of the force is the acceleration due to gravity
is acting purely in the minus z directions.
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So, the x component of the force balance will
tell us that on this face the force is the
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pressure times the area of the face, which
is nothing but delta x sorry delta z times
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b and minus this acting in the plus x direction.
But there is also a force due to this force,
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which is acting normally to this incline surface.
It will have a component in the horizontal
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direction or x direction, which is nothing
but p n. So, the area of this face is delta
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s time’s b and the force has a component
p n sin theta, if this surface is completely
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vertical that is theta is phi by 2. The normal
force will be purely p n, but this is surface
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is incline. So, there will be a component
of this surface of this force on this in the
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x direction, which is p n sin theta.
So, the sum of forces must be 0, because by
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Newton’s second law the fluid is static.
So, sum of all the forces must be 0. So, likewise,
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the z component but before I do that I can
simplify this further. So, just from geometry
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delta s sin theta so, we have this, this is
delta s, this is theta, this is delta z, this
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is delta x. Sin theta is nothing but delta
z by delta s. So, which implies delta s sin
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theta, is delta z. So, instead of this delta
s sin theta, I will put delta z. Then this
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equation will tell me that delta z b will
cancel with delta z and b. So, this will implies
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that p x minus p n is 0 or p x is equal to
p n. So, likewise, we can do a similar thing
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for the y direction.
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In the y direction again, we look at forces
so, the forces will be; so, the force on this
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face is p z p z times delta x times b that
is the force in the plus y direction. And
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there is a force the component of the force
on the incline surface that will be p n cosine
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theta that will be acting in the minus z direction.
So, will be minus p n cos theta time’s delta
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s time’s b also we have the acceleration
due to gravity the boy force acting in this
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direction, which is minus rho half b delta
x delta z time’s g and
sum of all these must be 0. Now, just by geometry
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delta s cos theta is nothing but delta x so,
I can cancel delta x right through the equation.
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So, I get and then I can cancel b right through
the equation.
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So, I get p z minus p n is equal to half rho
g delta z and in the limit as delta z shrinks
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to 0, we get p z is p n, but we also showed
p x is p n so, p z is p n is p x. So, if I
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take a surface like this, if I take a point
and then I construct a volume element like
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this, which is extending in the third direction
which we not worry. The pressure is acting
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in a static fluid as this volume element shrinks
to 0 the pressure acting on this faces with
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different orientations. They are this in the
magnitude of this force per unit area is exactly
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the same. So, the pressure; so, what we find
is that you take a point, if you take a large
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enough volume macroscopic volume in a static
fluid the pressure will be purely normal to
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this arbitrary volume element.
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But as I shrink this volume element to a tiny
tiny tiny poin then, the value of the pressure
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will be the same and purely normal to the
surface of this tiny volume element. But as
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this volume element gets shrunk to a point
even the magnitude of the pressure the numerical
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magnitude will be the same. And it is independent
of the direction of the orientation of the
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surface. Now, the next task for us is to find,
what is the pressure force on a fluid element?
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00:24:28,659 --> 00:24:35,659
In order to this, we will take a simple volume
element again, take a cubic volume element.
191
00:24:42,799 --> 00:24:49,799
And I will put co-ordinate system x y and
z along the three directions and I will call
192
00:24:55,960 --> 00:25:02,960
this as delta y or and this as delta x and
this width as delta z.
193
00:25:08,450 --> 00:25:15,450
Now, let us also call put some labels for
the faces let us, call this face A B C D;
194
00:25:16,980 --> 00:25:23,980
let us call this face E F G H. So, we want
to find out, what is the pressure force, if
195
00:25:30,600 --> 00:25:34,679
this volume element is surrounded by a fluid
and that is exerting a pressure under static
196
00:25:34,679 --> 00:25:41,679
conditions. What is the pressure force net
pressure force acting on this fluid element?
197
00:25:41,690 --> 00:25:48,690
To answer this pressure on the face or force
to be precised, force on the face A B C D.
198
00:25:58,970 --> 00:26:05,970
A B C D is this face is nothing but p of y.
So, this pressure in general will be different
199
00:26:09,799 --> 00:26:14,380
from this point to this point. Since this
is a tiny volume element the pressure at a
200
00:26:14,380 --> 00:26:20,799
given face is essentially the same, but it
is value will change from one point to another
201
00:26:20,799 --> 00:26:26,470
times the area of this element. The area of
this element is nothing but delta z times
202
00:26:26,470 --> 00:26:33,470
delta x. So, this is the force acting on the
face A B C D, the force on the face E F G
203
00:26:41,960 --> 00:26:48,960
H is nothing but in general the pressure will
vary from this point from to this point.
204
00:26:49,809 --> 00:26:56,809
So, we write this as p of y plus delta y times
delta z times delta x. Now, we are going to
205
00:27:01,820 --> 00:27:08,820
consider infinitesimal tiny volume elements,
for infinitesimal volume elements that is
206
00:27:17,150 --> 00:27:24,150
as delta y delta x delta z tends to 0. We
can always write by what is called a Taylor’s
207
00:27:27,630 --> 00:27:34,630
series. p of y plus delta y is nothing but
p at y plus partial p by partial y times delta
208
00:27:40,250 --> 00:27:47,250
y plus half partial square p partial y square
delta y square plus so on. Typically, this
209
00:27:50,320 --> 00:27:55,850
higher order sums are smaller so, we can neglect
and we can keep our expansion up t only up
210
00:27:55,850 --> 00:27:59,070
to first order.
211
00:27:59,070 --> 00:28:06,070
So, if such is the case, then the force the
net pressure force in the x direction on the
212
00:28:17,039 --> 00:28:24,039
volume element is given by nothing but we
have to worry about the force on the face
213
00:28:28,770 --> 00:28:35,770
A B C da nd the force on the face E F G H.
So, this is p; so, the net pressure force
214
00:28:40,570 --> 00:28:47,570
acting in the y direction, because remember
this is x, this is y, this is z, dF p y this
215
00:28:49,840 --> 00:28:55,020
is the net pressure force the y component
of the net pressure force. So, this is the
216
00:28:55,020 --> 00:29:02,020
y direction is nothing but p at y times delta
x delta z minus p at y plus delta y times
217
00:29:07,399 --> 00:29:14,399
delta x delta z. But I can use a Taylor series
expansion to write this as p at y minus p
218
00:29:14,710 --> 00:29:21,710
at y minus partial p partial y time’s delta
y time’s delta x delta z. So, these will
219
00:29:23,419 --> 00:29:30,419
cancelled to give dF p y is nothing but minus
partial p partial y times delta x delta y
220
00:29:33,429 --> 00:29:40,429
delta z. This is nothing but the volume of
the cubic element. So, I can write is minus
221
00:29:42,649 --> 00:29:46,340
partial p partial y times delta v.
222
00:29:46,340 --> 00:29:53,340
So, similarly, now what we have done is just
to recapitulate, if you take a volume element
223
00:29:58,529 --> 00:30:05,529
a cubic volume element with x y and z like
this. The net y component so, let me go back
224
00:30:16,750 --> 00:30:20,799
and correct any problems with notation.
225
00:30:20,799 --> 00:30:27,799
So, we are this still doing it in
so that is perfectly fine.
226
00:30:32,809 --> 00:30:39,809
So, the net force in the y direction on this
fluid element is due to the pressure acting
227
00:30:43,580 --> 00:30:48,559
acted upon by the fluid that is present outside
and the fluid inside is this. This is the
228
00:30:48,559 --> 00:30:55,559
y component, but just by analogy similarly,
you can do it as an exercise at home. You
229
00:30:56,149 --> 00:31:03,070
can show that d F p x, the net x component
of the force due to pressure alone is minus
230
00:31:03,070 --> 00:31:10,070
partial p partial x delta v and dF p z is
minus partial p partial z time’s delta v.
231
00:31:13,649 --> 00:31:20,649
So, we can construct the vector force. So,
if you have three components of a vector;
232
00:31:21,480 --> 00:31:28,480
suppose, I have a vector f and I have three
components f x f y f z. I can simply reconstruct
233
00:31:29,600 --> 00:31:36,090
this is in the cartesian coordinate system
as f x times i is f y times j is f z times
234
00:31:36,090 --> 00:31:41,860
k, where i j k are the unit vectors in the
x y and z direction.
235
00:31:41,860 --> 00:31:48,860
So, d F p can be simply written as dF p x
i plus d F p y j plus dF p z times k, where
236
00:31:59,309 --> 00:32:03,779
i j k are the three unit vectors along the
x y and z co-ordinates of the cartesian co-ordinate
237
00:32:03,779 --> 00:32:10,779
system. So, I can substitute for d F p x minus
partial p partial x i minus partial p partial
238
00:32:15,269 --> 00:32:22,269
y j minus partial p partial z k times delta
v. So, for those of you have taken elementary
239
00:32:27,740 --> 00:32:34,100
courses in mathematics, engineering mathematics,
where you would be dealing with vector calculus.
240
00:32:34,100 --> 00:32:41,100
You would recognize the this is nothing but
what is called the gradient of p. I will just
241
00:32:42,320 --> 00:32:47,899
take couple of minutes for those who are not
familiar with these concepts.
242
00:32:47,899 --> 00:32:53,970
So, gradient of a scalar quantity, what does
it mean physically? So, let me just take two
243
00:32:53,970 --> 00:32:59,399
minutes to explain this before I discuss the
physical significance of that result that
244
00:32:59,399 --> 00:33:06,399
we are just derived. So, let me just rewrite
this d F p is minus del p times del v. So,
245
00:33:07,950 --> 00:33:13,450
what do you mean by gradient of a scalar?
Suppose, I was single a function that is,
246
00:33:13,450 --> 00:33:20,450
where function of only one variable f of x.
Now, if I want to know, if I change the value
247
00:33:24,000 --> 00:33:31,000
of x by an amount dx. Then the function f
will change the delta f is the change that
248
00:33:32,490 --> 00:33:37,279
is obtain by the function that is incurred
in the function as you change x by delta x
249
00:33:37,279 --> 00:33:44,279
is from definitions of calculus dF dx times
delta x as the value of delta x becomes small,
250
00:33:45,399 --> 00:33:48,380
small and small.
The change in the value of the function suppose,
251
00:33:48,380 --> 00:33:55,380
I have function f verses x and I take a tiny
element x of length dx. This change that is
252
00:34:02,350 --> 00:34:09,350
incurred by the function delta f this change
is nothing but the slope at that point times
253
00:34:09,770 --> 00:34:16,129
the value of the displacement we are doing.
So, this is true for normal functions, but
254
00:34:16,129 --> 00:34:21,399
suppose, you have a function that is not a
function of single variable. It is a function
255
00:34:21,399 --> 00:34:28,399
of three variables not just one variable three,
where the three variables are the three co-ordinates
256
00:34:30,639 --> 00:34:36,230
directions in cartesian co-ordinate system.
Then how do you think of delta f, what is
257
00:34:36,230 --> 00:34:43,060
the change in the function f? So, if you consider
a function not just of one variable, but three
258
00:34:43,060 --> 00:34:48,530
variables. And you ask the question, what
is the change incurred in the function as
259
00:34:48,530 --> 00:34:55,530
you proceed by a small distance d x, but x
is no longer a scalar it is a vector, because
260
00:34:56,079 --> 00:35:01,780
you can undergo a displacement in the three
co-ordinate directions dx times i times d
261
00:35:01,780 --> 00:35:08,780
y times j plus dz times k. Then what is the
value, the change in the value of the function
262
00:35:09,130 --> 00:35:09,540
f?
263
00:35:09,540 --> 00:35:16,540
It turns out that gradient of the function
has that information so, gradient of a function.
264
00:35:21,500 --> 00:35:28,500
So, if f is a function of x y z then, if x
changes by a small amount delta x, y changes
265
00:35:33,400 --> 00:35:40,400
by a small amount delta y, z changes by a
small amount delta z. Then d f is nothing
266
00:35:40,400 --> 00:35:47,400
but partial f by partial x time’s delta
x keeping y z constant, partial f by partial
267
00:35:49,990 --> 00:35:56,990
y constant x z times d y there by plus partial
f by partial z x y delta z. So, this can also
268
00:36:04,599 --> 00:36:11,599
be thought of as or d f the change in the
value of the function. This is the displacement
269
00:36:11,640 --> 00:36:18,640
vector that we are under gone is delta x i
plus delta y j plus delta z k. So, I can also
270
00:36:20,650 --> 00:36:27,650
re write this as a quantity, which is a vector,
which is nothing but partial f partial x i
271
00:36:29,450 --> 00:36:36,450
plus partial f partial y j plus partial f
partial z k dotted, this is the dot product
272
00:36:38,780 --> 00:36:45,780
in vectors with delta x i plus delta y j plus
delta z k. This is nothing but the displacement
273
00:36:50,800 --> 00:36:57,800
vector that we have undergone dx. This quantity
is identified as grad f dotted with dx, this
274
00:37:01,960 --> 00:37:03,109
is df.
275
00:37:03,109 --> 00:37:10,109
So, this quantity is called the gradient of
the function f. So, the gradient of the function
276
00:37:18,970 --> 00:37:25,579
gives you information about the changes a
function encounters as you suppose, you have
277
00:37:25,579 --> 00:37:32,579
x y z co-ordinate system and you are traversing
any arbitrary distance delta dx. The gradient
278
00:37:34,670 --> 00:37:39,450
of the function has the information as to
how much this function f will change that
279
00:37:39,450 --> 00:37:45,440
is d f as undergo a vectorial displacement
d x about a point.
280
00:37:45,440 --> 00:37:51,220
So, the quantity the expression that we derive.
Now, going back to static fluids, the expression
281
00:37:51,220 --> 00:37:58,220
that we derived is that the force acting on
a fluid element is minus partial p partial
282
00:37:59,540 --> 00:38:06,540
x i plus partial p partial y j plus partial
p partial z k times delta v. This is nothing
283
00:38:12,770 --> 00:38:19,770
but minus grad p times delta v d F p. So,
this force that is acting on a tiny volume
284
00:38:24,510 --> 00:38:30,589
element recall that our volume element is
cubic in shape is proportional to that volume
285
00:38:30,589 --> 00:38:37,589
of the volume element. And therefore, I can
define a force density that is small f p that
286
00:38:38,470 --> 00:38:43,060
is force differential force acting on this
volume element divided by the volume is nothing
287
00:38:43,060 --> 00:38:47,740
but grad p.
So, there will be a net force on this volume
288
00:38:47,740 --> 00:38:54,650
element only, if there is a change or variation
in pressure. So, if each face has the same
289
00:38:54,650 --> 00:39:01,140
pressure can obviously all these forces will
cancel out. There would not be a net force
290
00:39:01,140 --> 00:39:04,960
due to pressure variation, because there is
no variation in pressure. Therefore, the forces
291
00:39:04,960 --> 00:39:10,510
will cancel out giving raise to no net force.
But if for some reason there is a pressure
292
00:39:10,510 --> 00:39:17,440
variation in a volume element, then that will
give raise to a net force on this volume element.
293
00:39:17,440 --> 00:39:24,440
So, this is called this is the net force acting
on a volume element due to pressure.
294
00:39:24,520 --> 00:39:31,520
So, one can make a very simple corollary so,
if the net force acting on a volume element
295
00:39:33,660 --> 00:39:39,230
is proportional to grad of p or the gradient
of pressure. That means that if there is constant
296
00:39:39,230 --> 00:39:46,230
pressure acting everywhere on a volume element
that will not give raise to any net force.
297
00:39:46,910 --> 00:39:53,910
A constant pressure acting all around a volume
element will not result in a net force. Why
298
00:40:10,500 --> 00:40:14,720
is this important? This is important, because
in many practical applications in terrestrial
299
00:40:14,720 --> 00:40:20,010
applications there is a constant atmospheric
pressure that is acted upon by the air that
300
00:40:20,010 --> 00:40:25,200
is present. And this constant atmospheric
pressure will not cause a net force on many
301
00:40:25,200 --> 00:40:29,290
problems of interest in fluid statics and
fluid flow.
302
00:40:29,290 --> 00:40:36,290
So, this force this pressure is called the
atmospheric pressure. This is the pressure
303
00:40:37,960 --> 00:40:44,960
due to the air that is present in the atmosphere.
So, this is called the atmospheric pressure
304
00:40:58,869 --> 00:41:05,869
commonly denoted as p atmosphere. So, if you
measure the pressure p of a fluid then that
305
00:41:06,240 --> 00:41:13,079
is called the absolute pressure. This is the
actual pressure that is present in the fluid,
306
00:41:13,079 --> 00:41:19,220
but since, if you in many problems you have
atmospheric pressure acting all around. It
307
00:41:19,220 --> 00:41:22,760
is only the difference between the atmospheric
absolute pressure and atmospheric pressure
308
00:41:22,760 --> 00:41:29,760
that matters, this is called the gage pressure.
So, this is if the pressure in the fluid is
309
00:41:31,500 --> 00:41:37,790
greater than atmospheric pressure. If it is
less than the reverse of it is called the
310
00:41:37,790 --> 00:41:44,790
vacuum pressure. If p is less than p, this
definition just ensures that vacuum pressure
311
00:41:49,660 --> 00:41:51,319
is a positive quantity.
312
00:41:51,319 --> 00:41:58,319
So, an uniform pressure acting all through
all around our body will not result in a net
313
00:41:58,660 --> 00:42:05,660
force. Now, let us worry about; if you consider
a volume element like this cubic volume element.
314
00:42:08,640 --> 00:42:15,640
We know that so, let us put co-ordinate system
x y and z. We know that the force is due to
315
00:42:15,920 --> 00:42:21,170
pressure, the force per unit volume due to
a pressure is a net force by the unit volume
316
00:42:21,170 --> 00:42:27,800
due to pressure is minus delta p and the net
force per unit volume due to gravity is simply
317
00:42:27,800 --> 00:42:28,099
pro g.
318
00:42:28,099 --> 00:42:35,099
So, if this fluid element is static
is static. Then the sum of all forces on this
volume element must act to 0, which implies
319
00:42:49,940 --> 00:42:56,940
that minus delta p time’s rho delta v plus
delta v rho g must also be equal to 0. So,
320
00:43:00,819 --> 00:43:07,819
if I cancel delta v right through we get minus
delta p plus pro g equal to 0. This is the
321
00:43:08,849 --> 00:43:15,849
fundamental equation of hydro statics minus
delta p plus rho g is 0.
322
00:43:29,619 --> 00:43:34,750
This is a vectorial equation any vectorial
equation can be written in a three component
323
00:43:34,750 --> 00:43:41,750
form. So, this will have three forms, a three
components minus dp dx plus rho gx is the
324
00:43:42,750 --> 00:43:47,990
component of the acceleration due to gravity
vector in the x direction minus dp d y plus
325
00:43:47,990 --> 00:43:54,990
rho g y is 0 minus dp dz plus rho g z is 0.
So, it is traditional or conventional in fluid
326
00:44:01,010 --> 00:44:06,300
flow to align the gravity vector vertically
along with one of the co-ordinate axis. So,
327
00:44:06,300 --> 00:44:13,300
suppose, I call this x y and z and the three
unit vectors are i j and k. And gravity is
328
00:44:17,550 --> 00:44:24,550
acting downwards so, g the vector g is written
as minus or g times minus k. So, that means
329
00:44:31,900 --> 00:44:38,900
that g x is 0, g y is 0 g z is minus g.
So, one can simplify this equations to write
330
00:44:44,460 --> 00:44:50,859
as partial p partial x is 0. All these are
partial derivatives, because we are looking
331
00:44:50,859 --> 00:44:57,859
at variations in all the three directions.
So, you have partial derivatives
332
00:45:08,069 --> 00:45:12,099
partial p partial y is 0. There is no variation
in the pressure in the x direction, no variation
333
00:45:12,099 --> 00:45:17,829
in the pressure in the y direction, but in
the z direction partial p partial z will go
334
00:45:17,829 --> 00:45:24,829
as minus rho g, because g is z is minus g.
So, if I take this rather side I will get
335
00:45:25,569 --> 00:45:27,640
minus rho g.
336
00:45:27,640 --> 00:45:33,000
This is a fundamental equation; this is a
fundamental result in hydrostatics. If you
337
00:45:33,000 --> 00:45:40,000
place the gravity vector in the minus k direction,
then pressure is independent of x and y direction.
338
00:45:44,940 --> 00:45:49,569
And it is a function only of the z direction
and the vary in which is going to vary is
339
00:45:49,569 --> 00:45:55,480
this. So, what are the assumptions of this,
when this equation is valid? This is a very
340
00:45:55,480 --> 00:46:02,480
important equation. This is valid for static
fluid and the gravity is aligned along the
341
00:46:03,030 --> 00:46:10,030
minus k direction and the gravity is the only
body force. There can be other body forces
342
00:46:14,050 --> 00:46:19,640
in general, but just in this introductory
discussion we keep think simple so, we take
343
00:46:19,640 --> 00:46:22,599
gravity to be the only body force.
344
00:46:22,599 --> 00:46:29,599
Now, can we solve this or can we simplify
this further? So, you have minus partial p
345
00:46:30,800 --> 00:46:37,140
partial z is rho g in order to integrate this.
We need to know whether rho and g are functions
346
00:46:37,140 --> 00:46:43,849
of z or not, g is practically a constant in
terrestrial applications and most engineering
347
00:46:43,849 --> 00:46:50,849
and situations so, g is a constant. So, next
thing is this rho a constant or not? Is it
348
00:46:51,089 --> 00:46:58,089
a constant? Well, it depends on context to
context in many fluid flow problems, the fluid
349
00:46:59,310 --> 00:47:06,310
can be consider to be in compressible that
is rho is independent of the pressure. There
350
00:47:10,109 --> 00:47:14,970
is no change in the density, because of the
fact that the pressure in the fluid is changing.
351
00:47:14,970 --> 00:47:21,359
So, rho is considered to be independent of
pressure then the fluid is in compressible
352
00:47:21,359 --> 00:47:28,359
and rho is independent of and rho is a constant.
So, if you consider the density to be constant
353
00:47:30,300 --> 00:47:35,069
then this equation can be integrated in a
straight forward way. So, you write minus
354
00:47:35,069 --> 00:47:42,069
integral z 1 or rather if I integrate, if
I take two points in the fluid where the elevation
355
00:47:45,190 --> 00:47:51,250
is or the height is z naught and z. And where
z is z naught p is p naught and where the
356
00:47:51,250 --> 00:47:58,250
vertical co-ordinate z pressure is p. Then
I write minus integral p naught p dp is minus
357
00:47:59,980 --> 00:48:04,230
rho g. Since, rho is considered to be a constant
for an in compressible fluid I will integrate
358
00:48:04,230 --> 00:48:07,670
this from z naught to z dz.
359
00:48:07,670 --> 00:48:14,670
So, p minus p naught is minus rho g times
z minus z naught or is equal to rho g times
360
00:48:17,619 --> 00:48:20,530
z naught minus z.
361
00:48:20,530 --> 00:48:27,530
Now usually, what happens is we can; so, the
way it is we have done the problem is z is
362
00:48:29,540 --> 00:48:36,160
pointing like this and gravity is pointing
in the opposite direction. Suppose, you have
363
00:48:36,160 --> 00:48:43,160
trough of water and there is a free surface.
This is let us say water and here it is air
364
00:48:43,250 --> 00:48:50,250
the pressure is atmospheric. Suppose, you
take z naught z equals z naught as the interface,
365
00:48:57,250 --> 00:49:03,460
where the pressure is atmospheric. So, p naught
is p atmosphere and let us take z naught minus
366
00:49:03,460 --> 00:49:10,460
z as h. So, h is the depth from the free surface,
where the pressure is atmospheric. So, we
367
00:49:12,510 --> 00:49:19,510
can write p minus p atmosphere is essentially
rho g h, where z naught is the location, where
368
00:49:23,670 --> 00:49:30,540
the pressure is atmospheric and z is any arbitrary
location. So, this is z, z naught minus z
369
00:49:30,540 --> 00:49:37,540
is denoted by the symbol h.
So, which implies p is p atmosphere plus rho
370
00:49:40,609 --> 00:49:47,609
g h. This is a fundamentally equation in hydrostatics.
The pressure varies so, if you consider the
371
00:49:47,839 --> 00:49:52,140
pressure here in the air to be atmospheric
and if you look at each and every point in
372
00:49:52,140 --> 00:49:59,140
the fluid as you go down in h. So, this is
z, as z decreases then the depth of the water
373
00:50:00,550 --> 00:50:05,650
increases from the free surface. Then the
pressure will increase compare to an atmospheric
374
00:50:05,650 --> 00:50:10,780
pressure due to the weight of the liquid column
that is present above. This is so, we take
375
00:50:10,780 --> 00:50:16,140
any h the weight of this liquid column will
act to increase the pressure.
376
00:50:16,140 --> 00:50:23,140
So, this is the fundamental equation of hydrostatics
in incompressible fluids. And the pressure
377
00:50:23,180 --> 00:50:28,849
increases in an incompressible fluid, because
of the fact that the pressure at any point
378
00:50:28,849 --> 00:50:35,849
in the fluid has to; suppose, if you take
a force, if you take a cylindrical volume
379
00:50:36,859 --> 00:50:41,059
element. If you do a force balance the pressure
here is acting normally like this, the pressure
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00:50:41,059 --> 00:50:45,099
here is the atmospheric. The pressure has
to balance not just the atmospheric pressure
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00:50:45,099 --> 00:50:51,440
but also the weight of the fluid element that
is present in between the two levels. So,
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00:50:51,440 --> 00:50:57,160
this is a fundamental result in hydrostatics
for incompressible fluids.
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00:50:57,160 --> 00:51:03,230
Now, you must have seen or learnt of what
is called the Pascal’s law or Pascal’s
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00:51:03,230 --> 00:51:10,230
principle for incompressible fluid, which
says that pressure applied to an enclosed
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00:51:14,059 --> 00:51:21,059
liquid
is transmitted and diminished to every portion
in the liquid and to the walls and to the
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00:51:43,150 --> 00:51:50,150
walls of the container. So, where does this
principle come from? This principle merely
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00:51:52,270 --> 00:51:59,270
is a consequence of this relation, this equation.
Suppose, you consider a container in which
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00:52:00,000 --> 00:52:05,839
you are imposing a pressure p naught and this
is let us say water it is incompressible,
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00:52:05,839 --> 00:52:12,839
there is liquid here. So, p at any; so, this
is gravity. So, p at any location is p naught
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00:52:13,800 --> 00:52:19,980
plus rho g h suppose, you change p naught
to some other value.
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00:52:19,980 --> 00:52:26,980
The change in pressure is because of change
in p naught plus change the change in pressure
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00:52:27,780 --> 00:52:34,780
at any depth h from the surface is delta p.
And there is change in p naught and then there
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00:52:36,220 --> 00:52:43,079
is change in rho g h, but we are looking at
the same h. So, this change in rho g h will
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00:52:43,079 --> 00:52:48,109
amount to only change in h, because there
is rho is constant for an incompressible fluid
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00:52:48,109 --> 00:52:53,800
g is constant. So, rho g delta h, but we are
looking at the same h so, delta h is 0. So,
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00:52:53,800 --> 00:53:00,650
this implies that the pressure that you apply.
The change in pressure that you make at the
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00:53:00,650 --> 00:53:05,690
surface here will transmit undiminished to
any in each and every point in a static fluid
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00:53:05,690 --> 00:53:12,609
this is the Pascal’s principle. Now, there
are some other ways of integrating this equation.
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00:53:12,609 --> 00:53:19,609
For example, you have dp dz is minus rho g.
Now, we assumed initially the fluid to be
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00:53:21,270 --> 00:53:26,250
incompressible. Now, let us say the fluid
is not in compressible, but it is compressible
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00:53:26,250 --> 00:53:30,480
and when the fluid is compressible the pressure
and density are related through an equation
402
00:53:30,480 --> 00:53:37,480
of state. Let us assume the equation of state
to be that of an ideal gas, where p v is given
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00:53:37,640 --> 00:53:41,550
as number of mole times universal gas constant
times t.
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00:53:41,550 --> 00:53:48,550
Usually, this is written has p is number of
moles is mass divided by the molecular weight,
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00:53:49,230 --> 00:53:56,230
the molecular weight of the gas and bringing
V down R T. So, m by V is the density and
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00:53:58,359 --> 00:54:03,940
R by molecular weight is called as specific
gas constant so, let us call it R g. This
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00:54:03,940 --> 00:54:10,050
not the universal gas constant it is specific
gas constant. So, p is rho R g T; so, instead
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00:54:10,050 --> 00:54:17,050
of writing rho g I am going to from this equation.
I am going to write dp dz is instead of rho,
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00:54:18,180 --> 00:54:25,180
I am going to write p by Rg T is minus p by
Rg T times g so, this can be integrated.
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00:54:32,140 --> 00:54:39,140
So, dp by p is nothing but minus g over Rg
T dz. This is law of p is minus g by Rg T
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00:54:49,280 --> 00:54:55,030
and the temperature is assume to be a constant.
So, the here we are looking at an isothermal
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00:54:55,030 --> 00:55:01,559
condition, but the temperature is fixed, the
gas is a constant temperature so, t is constant.
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00:55:01,559 --> 00:55:07,359
So, you can integrate this just to get t is
a constant. So, you can integrate this trivially
414
00:55:07,359 --> 00:55:14,359
to give z plus c. So, if you assume p is p
naught at z is z naught you can fix this constant.
415
00:55:15,809 --> 00:55:22,809
So, you get log p over p naught is minus g
by R T naught so, t is t naught a constant
416
00:55:24,819 --> 00:55:31,819
it is isothermal. So, times z or p is p naught
exponential minus g by R T naught times z.
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00:55:39,020 --> 00:55:46,020
So, we can integrate this equation for static
fluids not just with the incompressibility
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00:55:52,240 --> 00:55:52,890
assumption.
419
00:55:52,890 --> 00:55:58,579
So, even if you assume the flow the fluid
to be compressible, but by supplying the ideal
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00:55:58,579 --> 00:56:05,579
gas law. We got a simple relation for the
variation of the pressure with the elavish.
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00:56:06,619 --> 00:56:13,619
Now, this is for ideal gas and p is p naught
plus rho g h, this is for incompressible.
422
00:56:20,750 --> 00:56:27,750
Now, this expression can be simplified, when
g z by R T naught is much less than 1. If
423
00:56:31,530 --> 00:56:38,400
this is the case, when the elevations are
not so high then I can; when n suppose, you
424
00:56:38,400 --> 00:56:45,170
have e to the minus x when x is small then
I can write it as one minus x approximately
425
00:56:45,170 --> 00:56:52,170
so in the limit gz by R T 0 is small. So,
this is the universal stress specific gas
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00:56:52,430 --> 00:56:59,430
constant, which I denote by R g. So, p by
p naught is written as 1 minus; so, since
427
00:57:00,760 --> 00:57:07,660
there is 1 minus gz by Rg T0.
428
00:57:07,660 --> 00:57:14,660
Now, p by or p is p naught minus p by Rg T0
g is a from the ideal gas flow p is rho Rg
429
00:57:23,369 --> 00:57:30,369
T0 so, p by Rg T0 is nothing but rho so, p
is p naught minus rho g z. So, this is the
430
00:57:35,420 --> 00:57:41,059
same equation that we obtained for incompressible
fluids, but this is obtained as a special
431
00:57:41,059 --> 00:57:48,059
case, when gz by Rg T0 is very small compare
to 1. So, even if you assume compressibility
432
00:57:49,069 --> 00:57:53,780
when the elevations are not so large. This
essentially balls down to the same linear
433
00:57:53,780 --> 00:57:59,140
variation of pressure with respect to depth.
So, when is this considered to be small? Well,
434
00:57:59,140 --> 00:58:06,140
when z is less than 800 meters roughly then,
if plug in values of g and R g for air and
435
00:58:06,799 --> 00:58:12,160
T naught to be 300. You will find that this
is small when z is less than 800 meters.
436
00:58:12,160 --> 00:58:17,220
So, when you are looking at applications even,
if you assume and have to be compressible
437
00:58:17,220 --> 00:58:21,559
when the elevations are not large. Then this
linear variation in pressure with respect
438
00:58:21,559 --> 00:58:27,670
to the vertical distance is a very good one.
So, we will stop here and we will continue
439
00:58:27,670 --> 00:58:34,670
in the next lecture. So, we will see you in
the next lecture.