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Welcome to this fifth lecture on this n p-tel
course on fluid mechanics for undergraduate
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students in chemical engineering in the last
lecture, we discussed the basic difference
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between newtonian fluids viscous fluids and
elastic solids and we wrote down we said that
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suppose.
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You take two plates and place a fluid layer
in between them, a layer of fluid in between
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them, so we just remove so this is the layer
this is fluid in between and suppose you exert
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a force on the top plate, f x so the top plate
itself if you recall is a plate that is extending
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in the z direction, where the co-ordinates
are as follows x y and z so this has an width
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w in the z direction.
And this is the area the top area a or which
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the force f is been applied and if the bottom
plate is stationary, no force is being applied
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and its stationary then we said that if you
follow line a line if you follow colored line
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in the fluid, which was a vertical before
the application of force and if you follow
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the line after the application of force, this
line will continue to tilt.
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As a function of times so here time increases,
increasing time, in this way. T as time t
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increases this line will keep on tilting.
So, a fluid continues to deform on upon application
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of tangential or shear stress, shear force
in contrast to a solid which deforms to some
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extent and then stops deforming upon application
of force.
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So, we said that the stress, stress is force
divided by unit area force divided by the
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area upon which the force is acting. The stress
is actually denoted by the symbol tau and
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we said that there are two suffixes for the
stress, x is the direction
of the force and y is the direction perpendicular
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to the surface upon which the force is acting.
So, it is a direction of the unit normal to
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the surface unit vector perpendicular to the
surface is called unit normal. So, it is the
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direction of the unit normal to the surface,
upon which the force is acting.
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So, this is the stress this is the definition
of a stress. So, tau y x we said must be proportional
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to the rate of deformation, which if you recall
we call this angle as delta alpha since delta
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alpha continues to change with time, we said
that the stress must be proportional to delta
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alpha by delta t and this was also found to
be equal to based on geometry considerations,
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if you call this was the line before deformation
this is lets say delta alpha and this is the
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small displacement delta l and this is the
thickness of the fluid h and delta alpha tan
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of delta alpha is delta l by h but, for small
enough angels this is approximately equal
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to delta alpha is delta l by h.
But as a fluid deforms continuously the top
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plate will move with a velocity v lets say
all. Let us say u so delta l is u delta t
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so delta alpha is u delta t divided by h or
delta alpha by delta t is u divided by h.
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So, tau y x must be proportional to u which
is the velocity at which the plate is moving
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and divided by the gap thickness in which
the fluid is moving and the constant of the
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proportionality is the viscosity of the fluid.
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This is the viscosity of the fluid
and we also gave you some examples of viscosity
values, viscosity of the fluid is measured
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in pascal second in s I units and we also
gave you some examples of the values of viscosities
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of various commonly encountered fluids like
air water oils and so on.
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Now what is the use of this formula well we
will return to this formula in detail later,
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when we do differential balances but, at the
outset this is an introduction to what how
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a viscous fluid behaves upon application of
shear stress you can use this formula to find
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for example, suppose you have.
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Fluid layer, a viscous liquid and let us say
you have to push this viscous liquid and the
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bottom you have the bottom surface is stationary.
And let say you are moving the top surface
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with a belt with a velocity. So, this is rotating
so this is a velocity u at which the belt
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is moving, so you are trying to drag the fluid
at a constant velocity suppose you ask the
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question. What is the force that must be exerted
to move the fluid at a velocity u?
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Suppose you ask this question, well the answer
is obtained in the following way we know that
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tau y x is the force f x divided by a, that
area of the belt so remember this is a these
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are all the belt is a finite. So, this whole
sheet is so this is a role cylindrical role
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in the third direction and this whole role
is rotating a the constant velocity.
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And the area of the top plate is or the belt
surface is a so that is the area so this is
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the viscosity of the fluid times u divided
by the gap thickness. So, force is therefore,
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mu a u divided by h suppose we know the values.
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Suppose you are interested in dragging a viscous
liquid using a belt, let say example viscosity
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is let us say 0.01 pascal seconds, let us
say the gap thickness is 1 centimeter. And
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the area is so we should convert this in s
I units which is 10 to the minus 2 meters,
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let say the area is 10 centimeters by 10 centimeters
so this is 10 to the minus 2 meters times
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10 to the minus 2 meters, which is 10 to the
minus 4 meter square. And let us say the velocity
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is 1 centimeter per second which is 10 to
the minus 2 meter per second so if you substitute
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everything in that expression f x is mu which
is 10 to the minus 2 0.01 pascal second is
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10 to the minus 2 times.
So, let me just write this is 10 to the minus
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2 pascal second, so 10 to the minus 2 times
area is 10 to the minus 4.Times velocity is
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10 to the minus 2 divided by h is 10 to the
minus 2 so this is about 10 to the minus 6
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newtons, if the area is just the small amount
then this is the force that must be exerted
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to move the belt at a constant velocity what
is the rate at which work must be done. Suppose
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you want to move this conveyer belt using,
some rotating machinery, so the rate at which
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work must be done by the machinery in order
to keep moving this is work must be done by
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the external agency to move this belt is work
done is force times distance rate at which
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works done is force times distance divided
by time.
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Distance by time is velocity, so it is the
force times velocity its 10 to the minus 6
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times velocity is 10 to the minus 2, so this
is 10 to the minus 8 so this is newton then
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meter per second. So this is the rate at which
work is done in order to move this fluid.
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So, this is 1 simple example of this equation
that, we just wrote down for a simple fluid.
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So we also said that this, this relation mu
u by h, this describes what are called newtonian
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fluids, because this is only an expectation
that the stress must be directly proportional
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to the rate of deformation. But, it turns
out that commonly encountered liquids, respect
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this expectation. So, this is a material behavior
and we will see that such relations are called
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constitutive relations.
Later
these describes specific behavior of how a
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given fluid flows or responds to external
applied stresses and there are fluids, which
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do not obey this for example, slurries and
solutions of polymers, molten polymers, molten
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plastics and so on. They such fluids are called
non newtonian fluids. Anything that does not
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obey if this is not obeyed they are called
non newtonian fluids. So things that obey
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this they are called newtonian fluids. So,
we also we will discuss much later in this
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series of lectures, as to how to understand
with the behavior of non newtonian fluids
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a little later.
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So, this expression for the, the shear stress
as a function of velocity of the top plate
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divided by thickness was applied to a case
of a finite fluid layer of thickness h, which
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is exerted upon which a force is exerted in
the x direction and so on. But you can also
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think of a flow suppose you have, the same
example a flow is here is the x y co-ordinate,
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this is fluid in between you can also think
of considering a tiny slice of thickness delta
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y in the y direction.
So, essentially if you recall we are extending
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everything in the z direction so this is a
tiny slice like this of thickness delta y
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in the y direction and you can try to apply
so when a the fluid is stationary here fluid
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is moving with some velocity here. So, if
you were to so if you were to plot the velocity,
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so the velocity will go from zero to some
velocity u at the top plate in this example
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so fluid layers are sliding past each other
in this simple example so if you take a tiny
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slice like, so the fluid that is flowing past
here will have a greater velocity than the
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fluid that is flowing below this tiny slice,
so if you are moving with the velocity of
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this lower fluid, the flow about this tiny
slice will again appear. So, I am just expanding
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this tiny slice will appear light a linear
flow like this.
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So the thickness is delta y so because these
two slices are fluid above the slices moving
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with the greater velocity then the fluid below
the slice. So, if you move with the velocity
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of the lower side, then this will appear stationary
the top will appear to move at a constant
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velocity. So the this expression can be generalized
in the limit delta t going to zero delta alpha
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by delta t, this was the fundamental definition
and delta alpha is by delta t is nothing but,
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limit delta t tending to 0 delta u by delta
y, where delta u is the difference in velocity
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between these two points.
So we know the two if you take a slice, this
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magnitude of the velocity differs between
these two slices in this case, so that magnitude
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of the velocity difference is delta u so as
delta t goes to 0, this is the shear stress
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this is mu times this. So, tau y x is mu d
u d y in the limit of delta t goes to 0, the
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velocity difference, so this translates to
delta y goes to 0 becomes this. So, this equation
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is not merely applicable for thicknesses of
macroscopic dimensions, Even if you take a
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tiny slice volume slice and then we try to
look at how flow is going to happen how the
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shear stresses are related we will find that
this is true even for a tiny slice.
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So this is called the newton’s law of viscosity.
now the next topic that we are going to discuss
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is fluid statics but, before I do that we
will have to first understand what are the
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various forces that are exerted on a given
fluid element.
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So, let us continue along the same lines and
then try to classify the types of forces,
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that can that can be experience by a given
fluid element volume element of fluid and
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then we will proceed to fluid statics, so,
let us imagine having continuing with the
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same example.
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You have two plates, so the thickness is h
this is x this is y this is z. And the top
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plate is moving with a velocity, let us say
capital v bottom plate is stationary, so if
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you consider a tiny volume element remember
that z is going outside into the board.
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So essentially this volume element will look
like this, into the z direction into the so
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that the z direction is going into the board.
So, but this is the volume element this is
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the volume element but, if I want to write
it in a so let me draw it separately, let
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me draw it separately here.
So, the volume element that we are going to
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analyze is somewhere something like this but,
in a 2 d plane because this diagram has been
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drawn in a plane, that is cutting across any
z location the volume element will appear
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like a slice. So, this volume element what
are the forces that are acted upon on this
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volume element suppose you are doing this
experiment in a lab and let us say gravity
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is acting downwards, now if you take this
let us for simplicity assume this to be a
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cubic volume, that simplifies our discussion
so take a cubic volume element instead of
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a rectangular volume element. So, this will
appear like a square in the two d diagram.
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What are the forces on this volume element
well if the volume of this cubic volume element
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is delta v because there is gravity, there
will be a force see whenever gravity acts
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on a mass there is a force acting in the direction
of gravity m g. So, mass times acceleration
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due to gravity is a force this force acts
through the entire body entire volume, such
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forces which act through the entire volume
are called body forces or sometimes they are
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also called as volume forces because if you
take a tiny volume this force due to acceleration
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due to gravity acts through the entire volume
not just on a particular surface or a given
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location, so it acts on the entire volume.
So, this body force acting on this volume
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element and the body force is acting in the
y direction only because, we know that the
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gravity is only in the we have align the gravity
in the direction of negative y. So, this is
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negative because negative because gravity
is acting so this is the direction of positive
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y gravity is pointing the direction on negative
y so that must be negative mass is density
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of the fluid times the volume infantilism
volume we are considering times, acceleration
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due to gravity the magnitude of acceleration
due to gravity which is 9.8 meter per second
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square in s I units.
So, this is the force that is acting on this
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volume element due to acceleration due to
gravity such forces are called body forces
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or volume forces, now apart from this because
of the fact that the top plate is moving with
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the constant velocity. So, I told you that
so, mething like this should happen.
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So, there will be fluid flow layer by layer
in this simple case and if you take a tiny
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volume element I am exaggerating this volume
element for illustration sake but, this is
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a very tiny volume element if you take a tiny
volume element, let us again draw this square
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separately here.
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So, there is fluid flow above the square with
some velocity and below the square with some
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velocity and this difference in velocity means
at there is a shear stress exerted by the
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fluid on the top on this surface, such forces
which act suppose you consider this, remember
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that it is a cubic volume it has a surface
on the top I am drawing the 2 d version of
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it so there is a surface, so along this surface
the top surface there is a force that is acting
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in the x direction due to the fact that fluid
layers are flowing pass this surface and due
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to viscous action there is a stress or force
per unit area or a stress, such forces are
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called surface forces.
They act only on the surface, of a volume
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element
of a fluid the element when I say fluid element
we mean a volume element in the fluid. So,
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if you take this volume element there is a
tangential force on the top surface due to
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the fluid flow above in this direction and
this will constitute a surface force this
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is essentially due to the tangential force
is due to viscous action. But, there could
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also be force due to in the normal direction
due to pressure those are also surface forces.
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So, surface forces can be acting normal to
a surface and tangential to the surface, the
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normal force to the surface is usually pressure
dictated by a pressure also we will clarify
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a little later when we do differential balances
this statement but, you can think of at this
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introductory level normal force on a element
on a surface largely due to pressure this
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is acting perpendicular to the surface, so
its normal therefore, it is a normal force
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where as the tangential force is acting parallel
to the surface that is why is called tangential
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or shear force.
So pressure forces and shear forces can act
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on a surface and these are acting only on
a surface and not to the entire volume. So,
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these are called surface forces, so this is
in the context of a simple example like this
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like flow between two plates but, in general
you could have a fluid that is flowing in
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a arbitrary way and you can construct any
volume element.
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So, any arbitrary volume element and this
volume element will have a surface and the
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surface is characterized. So, this is the
volume element so this is a volume element
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in 3 d its arbitrary shape, but since I can
draw only a 2 d version in this board, so
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I have drawn a 2 d version and the unit normal
to the surface is n so the n suppose you consider
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this let me draw this slightly better. So,
this is the arbitrary surface at each and
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every point this n will keep varying.
So, n will be like this here like this here
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like this here and like this here and so on.
So, n keeps changing this is an arbitrary
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volume element in the fluid, in a fluid flow.
So, this arbitrary volume element in a fluid
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flow is surrounded by fluid also, this is
surrounded by fluid
and if you look at a given point on this volume
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element and you worry about, let us say you
take a tiny patch of area delta a, this has
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00:25:22,580 --> 00:25:29,170
a unit normal n at this location and if you
ask what is the force the force can in general
195
00:25:29,170 --> 00:25:35,060
be in any direction or the force per unit
area which is denoted by the letter r stress
196
00:25:35,060 --> 00:25:42,060
vector it can point in any direction. So,
let me just illustrate this again separately.
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00:25:42,440 --> 00:25:49,440
So you take any surface on a volume element,
let the so take a point p on a volume element
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00:25:49,930 --> 00:25:56,930
on the surface of a volume element and the
surface has a unit output normal n and the
199
00:25:58,370 --> 00:26:02,520
force that is exerted by the fluid, that surrounding
on this the surface force that exerted by
200
00:26:02,520 --> 00:26:07,840
the fluid is surrounding, this area on this
surface can in general point in any direction
201
00:26:07,840 --> 00:26:14,840
so let us so that force is denoted by r.
This is the area now this r is of the force
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00:26:17,510 --> 00:26:22,190
and that need not in general be in the direction
of the normal but, it will have a component
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00:26:22,190 --> 00:26:27,810
perpendicular in the direction of the normal.
So, let us call that so I am going to draw
204
00:26:27,810 --> 00:26:34,810
this again here. So, this is r the direct
the component of r and the direction of the
205
00:26:40,510 --> 00:26:47,510
normal is called the normal force per unit
area so r is the surface force per unit area.
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00:26:51,460 --> 00:26:58,460
So, there is a normal component r n and along
the surface there are two tangential components
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00:27:02,230 --> 00:27:09,230
r t 1 and r t 2. So, these are the this is
the normal stress these are the tangential
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00:27:10,640 --> 00:27:14,060
stresses.
To any surface you can construct a normal
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00:27:14,060 --> 00:27:19,860
unit outward normal and you can for this normal
perpendicular to this normal there are two
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00:27:19,860 --> 00:27:23,010
tangential vectors, this can be constructed
in any way that you want I am just showing
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00:27:23,010 --> 00:27:28,490
you for illustration in this manner, so this
force that is exerted by the fluid on the
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00:27:28,490 --> 00:27:32,650
outside on this surface can be resolved in
the direction of the normal, as well as in
213
00:27:32,650 --> 00:27:37,330
the direction perpendicular to the normal
that is parallel to the surface itself so
214
00:27:37,330 --> 00:27:42,120
these are. So, the surface force can have
in general a normal component.
215
00:27:42,120 --> 00:27:49,120
This is the surface force remember can have
a normal component, as well as tangential
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00:27:49,520 --> 00:27:56,520
component, now and I given you a simple example
that the surface force is the surface forces
217
00:28:02,000 --> 00:28:09,000
normal as well as tangential components. So,
the surface forces can, surface force is can
218
00:28:17,640 --> 00:28:24,640
have normal as well as tangential components,
now I given you some simple illustration in
219
00:28:25,110 --> 00:28:29,640
which the tangential component of the surface
force, why it occurs due to viscous action
220
00:28:29,640 --> 00:28:33,070
but, we will discuss more about this in detail
a little later.
221
00:28:33,070 --> 00:28:39,130
A clear understanding of what are the forces
that can exert that that can be exerted on
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00:28:39,130 --> 00:28:46,130
a volume element of fluid is necessary to
understand, how to analyze a fluid a fluid
223
00:28:46,410 --> 00:28:49,520
mechanics because ultimately.
224
00:28:49,520 --> 00:28:55,660
As I told you fluid mechanics is a branch
of mechanics and if you take any tiny volume
225
00:28:55,660 --> 00:29:00,190
element and if you know, what are the forces
that are acting the body force, which is rho
226
00:29:00,190 --> 00:29:07,190
g delta v and the surface forces, if you have
a knowledge of body and surface forces then
227
00:29:07,230 --> 00:29:13,400
one can apply newton’s second law of motion
that is which says us mass times acceleration
228
00:29:13,400 --> 00:29:20,400
is sum of all forces.
Which can be used to compute the motion of
229
00:29:22,210 --> 00:29:28,770
the fluid so this is the program, rough program
in any branch of mechanics but, in fluid mechanics
230
00:29:28,770 --> 00:29:33,580
the forces are not of various types one is
the body force, which is acting entirely through
231
00:29:33,580 --> 00:29:38,810
the volume and the other is the surface force,
which acts only on the surface the surface
232
00:29:38,810 --> 00:29:42,710
forces can be due to viscous action and as
well as due to pressure.
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00:29:42,710 --> 00:29:49,710
So, the first topic that we are going to consider
is fluid statics. Now, we are going to consider
234
00:29:53,400 --> 00:30:00,400
fluid statics, deals with a fluid that has
no motion, that is no flow. So, it is no flow
235
00:30:11,060 --> 00:30:16,180
this only a fluid is stationary.
Still there are some interesting aspects to
236
00:30:16,180 --> 00:30:21,840
the force distributions even in a fluid, that
is even in a fluid that is not moving it is
237
00:30:21,840 --> 00:30:28,840
a stationary. So, let us understand what forces
can act on at static fluid you take a fluid
238
00:30:30,800 --> 00:30:37,800
volume. Construct a surface at any point you
take a point in the volume, construct a surface
239
00:30:39,320 --> 00:30:44,660
of area delta a this surface is denoted by
unit normal n.
240
00:30:44,660 --> 00:30:50,950
So, in general the surface forces, are in
the direction of n which is what I called
241
00:30:50,950 --> 00:30:57,950
r n and in the direction perpendicular to
n which is what I call r t 1 the tangential
242
00:30:58,880 --> 00:31:05,880
forces. These are the shear forces.
Now, we are now considering fluids that are
243
00:31:09,080 --> 00:31:15,530
not moving a fluid this static, the fluid
is static, now what can we say about the forces
244
00:31:15,530 --> 00:31:19,730
the nature of surface forces, when the fluid
is static by definition a fluid undergoes
245
00:31:19,730 --> 00:31:25,380
a flow, when there is a non zero shear stress,
we just saw this in detail in the last lecture,
246
00:31:25,380 --> 00:31:29,960
where we constructed simple thought experiments,
where we showed that a fluid continues to
247
00:31:29,960 --> 00:31:34,340
deform upon application of a shear or tangential
stresses.
248
00:31:34,340 --> 00:31:41,340
So, if you consider fluid that has no motion
no velocity and no flow, then it implies that
249
00:31:41,680 --> 00:31:47,120
the fluid the tangential stresses in the static
fluid must be 0. So, the tangential stresses
250
00:31:47,120 --> 00:31:54,120
must be 0 in a static fluid.
So, you cannot have tangential stresses, so
251
00:31:55,330 --> 00:32:02,330
the only type of surface forces that can act
on a volume element, so the only surface force
252
00:32:03,100 --> 00:32:09,230
is the normal force because, if there is a
tangential force on any point in a fluid that
253
00:32:09,230 --> 00:32:13,830
means it the fluid will start flowing.
Because, a fluid cannot resist any non zero
254
00:32:13,830 --> 00:32:19,940
tangential stress, even at a slightest application
of tangential stress the fluid will start
255
00:32:19,940 --> 00:32:26,940
moving therefore, we can conclude safely that
in a static fluid the tangential stresses
256
00:32:28,580 --> 00:32:35,580
are zero. So, zero tangential stresses and
the only surface forces must be purely normal
257
00:32:40,640 --> 00:32:47,640
to any surface. Now, this is true, for suppose.
258
00:32:50,940 --> 00:32:57,940
You take a static fluid and consider a volume,
let us take a simple volume, like a cube
so if you take any face on this volume, if
259
00:33:07,620 --> 00:33:13,270
you take the top face the forces will be purely
normal, if the fluid is static because there
260
00:33:13,270 --> 00:33:15,920
is no other way no way the fluid can support
a tangential stress.
261
00:33:15,920 --> 00:33:20,930
If you take the side way, side face, here
again the forces will be purely normal and
262
00:33:20,930 --> 00:33:26,370
same with bottom and so on. So, the forces
will be purely normal in a static fluid but,
263
00:33:26,370 --> 00:33:30,740
this is true for any volume element because,
you can construct a volume element the shape
264
00:33:30,740 --> 00:33:32,440
of the volume can be as arbitrary.
265
00:33:32,440 --> 00:33:39,440
As you please, even in this complicated volume
element the forces will be purely acting at
266
00:33:41,390 --> 00:33:46,770
each and every point on the surface of this
volume element the forces has to be purely
267
00:33:46,770 --> 00:33:51,960
normal.
Because, by definition in a static fluid in
268
00:33:51,960 --> 00:33:57,960
a static fluid, if you construct a volume
arbitrary volume element, the forces have
269
00:33:57,960 --> 00:34:02,870
to be purely normal because at each and every
point in the fluid if there cannot be any
270
00:34:02,870 --> 00:34:09,870
tangential stress. So, the normal in a static
fluid this so that can in general be normal
271
00:34:11,500 --> 00:34:14,720
force a static fluid can support normal forces.
272
00:34:14,720 --> 00:34:21,720
The normal force, exerted per unit area is
usually compressive in nature in a fluid,
273
00:34:32,539 --> 00:34:39,539
that is if you consider this is the fluid
inside the volume element, this is the fluid
274
00:34:42,029 --> 00:34:49,029
outside. So, this fluid outside exerts a force
that is compressive the tenses compress the
275
00:34:50,159 --> 00:34:54,769
fluid inside the volume element.
And but, the unit outward normal is pointing
276
00:34:54,769 --> 00:35:00,220
like this, to the to this volume element at
each and every point on the surface the unit
277
00:35:00,220 --> 00:35:04,740
outward normal will act like this but, the
normal force will act in the direction, this
278
00:35:04,740 --> 00:35:10,180
is the direction of the force in the direction
opposite to the unit outward normal.
279
00:35:10,180 --> 00:35:15,180
This normal force exerted per unit area, this
compressive normal force, exerted by unit
280
00:35:15,180 --> 00:35:22,180
area in a fluid is called the pressure. Since,
its force per unit area it is dimensions of
281
00:35:23,029 --> 00:35:30,029
newton’s per meter square.
The normal force that is experienced by a
282
00:35:32,130 --> 00:35:39,130
fluid, a static fluid element volume element
the fluid is static. So, if you take a volume
283
00:35:39,289 --> 00:35:46,289
element I just told you that this volume element
in a static fluid will experience only a surface
284
00:35:46,460 --> 00:35:51,420
force, that is purely normal to each and every
point on the surface and this is compressive
285
00:35:51,420 --> 00:35:58,420
in nature that is called the pressure.
So, if you think if you say this is n and
286
00:36:02,279 --> 00:36:09,279
if you say r vector is the force per unit
area, exerted by the fluid outside. So, this
287
00:36:24,299 --> 00:36:31,299
is in, this is out on the surface, since the
fluid is static this is, this has to be in
288
00:36:39,140 --> 00:36:44,249
the direction I mean since its purely normal
it has to be either in the direction of n
289
00:36:44,249 --> 00:36:48,660
or opposite to n.
But the pressure is normally compressive so
290
00:36:48,660 --> 00:36:54,900
this is in the direction of minus n because
n is pointing from in to out, where as the
291
00:36:54,900 --> 00:36:59,819
force is from out to in and the magnitude
of the force compressive force is pressure,
292
00:36:59,819 --> 00:37:02,499
force per unit area is pressure p.
293
00:37:02,499 --> 00:37:09,499
So R that is force per unit area is given
by minus p n in a static fluid because, a
294
00:37:11,650 --> 00:37:18,650
static fluid cannot support any shear stress.
So, and a typically the units for pressure,
295
00:37:25,609 --> 00:37:32,609
is in SI units it is 1 newton a force per
area newton per meter square, this is also
296
00:37:38,400 --> 00:37:45,400
equal to a pascal 1 pascal but, it turns out
that the pressure that is experienced by us
297
00:37:48,880 --> 00:37:55,880
in the free atmosphere is called the atmospheric
pressure. So, the atmospheric pressure is
298
00:37:56,730 --> 00:38:02,200
approximately 10 to the 5 pascals it is a
large number, so pressure is normally expressed
299
00:38:02,200 --> 00:38:07,690
not just in pascals but, also in kilo pascals,
mega pascals. Kilo pascal is 10 to the 3 pascal,
300
00:38:07,690 --> 00:38:14,690
1 mega pascal is 10 to the 6 pascal and so
on. So, it should be getting used to this
301
00:38:16,410 --> 00:38:21,599
prefixes to this units which tell you multiples
of thousands.
302
00:38:21,599 --> 00:38:28,599
And sometimes pressure is expressed in terms
of 1 atmosphere that is approximately 10 to
303
00:38:28,789 --> 00:38:35,789
the 5 pascals. Sometimes, it is also expressed,
we will show little later in terms of columns
304
00:38:36,319 --> 00:38:42,930
of mercury in a mano meter its 760 m m Hg.
We will see this a little later once, we finish
305
00:38:42,930 --> 00:38:49,930
the basics of a fluid statics, now let us
first worry about what is the nature of pressure.
306
00:38:50,710 --> 00:38:57,710
So if you take any macroscopic volume element
in a static fluid, if you take a macroscopic
307
00:38:58,670 --> 00:39:05,670
volume element I have just told you, that
the force will be purely compressive its force
308
00:39:07,039 --> 00:39:12,490
exerted by the surrounding fluid on the surface
of this volume element purely normal to the
309
00:39:12,490 --> 00:39:15,359
surface and it is compressive.
310
00:39:15,359 --> 00:39:22,359
This is for a macroscopic volume in a static
fluid. Suppose, I consider a point, in a volume
311
00:39:28,130 --> 00:39:34,759
and construct a surface, so construct a surface,
this is a point and I can construct a surface
312
00:39:34,759 --> 00:39:39,849
with some orientation, whenever you construct
an area there is orientation associated with
313
00:39:39,849 --> 00:39:45,400
it because I can orient this area in many
ways. Now, what is the nature so this is really
314
00:39:45,400 --> 00:39:52,400
a point and about this point I construct a
tiny area delta a and I ask the question,
315
00:39:53,249 --> 00:40:00,249
what is the pressure measured at this point.
So, I can do this thought experiment I can
316
00:40:09,799 --> 00:40:14,240
orient this surface in one direction at a
point. So, I take a point construct a very
317
00:40:14,240 --> 00:40:20,930
tiny small area and I oriented in a particular
way I measured the pressure I change the orientation
318
00:40:20,930 --> 00:40:25,970
the fluid is static and this is really a very
very small area. So, that we are shrinking
319
00:40:25,970 --> 00:40:30,410
it eventually to a point.
And as I change the orientation what is going
320
00:40:30,410 --> 00:40:36,809
to happen, what is the value of pressure,
is if does the value of pressure. So, we will
321
00:40:36,809 --> 00:40:43,809
ask this question, if you do this experiment
does the value of pressure at a point in a
322
00:40:47,329 --> 00:40:54,329
static fluid, does it depend on n the orientation
of the surface essentially you are considering
323
00:41:04,309 --> 00:41:07,989
a point but, about a point you can construct
a infantilism area.
324
00:41:07,989 --> 00:41:13,869
So as I told you repeatedly in the continuum
approximation, when you say at a point any
325
00:41:13,869 --> 00:41:17,450
property like density at a point you have
to construct a tiny volume about that point,
326
00:41:17,450 --> 00:41:21,650
when you take consider a pressure you have
to construct a tiny area but, you can construct
327
00:41:21,650 --> 00:41:27,700
this area with many infinitely many orientations
of out of point, So, is there does the is
328
00:41:27,700 --> 00:41:34,700
the does the pressure that one obtains about
a point as the area shrinks to zero in the
329
00:41:34,849 --> 00:41:39,420
continuum sense.
Does it still does it depend on n the orientation
330
00:41:39,420 --> 00:41:46,420
we will show that it will not, so the pressure
is purely a scalar that is at a point pressure
331
00:41:53,390 --> 00:41:58,220
is force per unit area but, at a given point
it.
332
00:41:58,220 --> 00:42:05,190
If you construct a tiny volume about a point
at a of arbitrary shape orientation at the
333
00:42:05,190 --> 00:42:11,130
surface its always compressive its always
in the direction of the in the direction opposite
334
00:42:11,130 --> 00:42:15,160
to the unit outward normal. So, it pressure
does not have a sense of direction at a point
335
00:42:15,160 --> 00:42:22,160
it is the force per normal force per area.
At any surface, of a surface of any orientation
336
00:42:22,549 --> 00:42:27,410
at that you would construct about a point.
So, how do we show this, we show this in a
337
00:42:27,410 --> 00:42:34,410
very simple way, so you construct consider
so this proof, that the pressure is a scalar
338
00:42:34,769 --> 00:42:39,829
does not depend on the orientations about
a many orientations, that you can have when
339
00:42:39,829 --> 00:42:46,829
you construct an area. So, let us take a volume
that is wedge shape like this
and this third direction is so large, that
340
00:42:57,180 --> 00:43:04,180
we will neglect variations in the direction.
So, consider a only the cross section triangular
341
00:43:04,309 --> 00:43:11,309
cross section.
Now, this angle let it be theta, now let us
342
00:43:18,480 --> 00:43:25,480
put a co-ordinate system x z and y is going
into the board that is not relevant. So, if
343
00:43:27,019 --> 00:43:34,019
you consider this x y into the board and z
vertical, it is a gravity is acting with this.
344
00:43:38,700 --> 00:43:45,249
So, this is a static fluid the fluid is still
static we are instead in fluid statics right
345
00:43:45,249 --> 00:43:52,249
now. So, let us not draw y here it is a purely
a plane so this is x and z.
346
00:43:52,470 --> 00:43:59,470
So, let us call this surface the let this
distance be delta x, let this distance be
347
00:44:05,549 --> 00:44:12,549
delta z, let this distance be delta s, now
there is also gravity that is acting down,
348
00:44:21,019 --> 00:44:24,950
the pressure at this surface and these are
all infantilism links.
349
00:44:24,950 --> 00:44:29,450
So, the pressure is approximately a constant
the pressure on this bottom surface of this
350
00:44:29,450 --> 00:44:36,450
wedge like volume element, let us call it
p z because this is p at this location z equal
351
00:44:37,119 --> 00:44:43,619
to zero and let us call the pressure here
at this face p x.
352
00:44:43,619 --> 00:44:48,359
And let us call the pressure, pressure is
acting normally and its compressive, so it
353
00:44:48,359 --> 00:44:53,999
will act like this on this volume element,
let us call this p n and by geometry you can
354
00:44:53,999 --> 00:45:00,180
verify that this angle is also theta. So,
pressure will act on this incline plane incline
355
00:45:00,180 --> 00:45:04,049
surface that is the stop surface in a purely
normal way.
356
00:45:04,049 --> 00:45:11,049
So, what are various forces acting the body
force due to gravity is the mass times acceleration
357
00:45:13,739 --> 00:45:20,079
due to gravity, the body force acts only in
the minus z direction remember z is going
358
00:45:20,079 --> 00:45:27,079
up so body force in the minus z direction
in minus negative z direction is equal to
359
00:45:29,509 --> 00:45:36,509
mass is rho times volumes, rho times volume
of this wedge is basically half times delta
360
00:45:36,599 --> 00:45:43,599
x delta z times, let us call this width b,
b is very very large.
361
00:45:51,230 --> 00:45:58,230
So, W times g if this were a complete cube
a cuboids you would say that the volume is
362
00:46:01,089 --> 00:46:07,299
delta x delta z times b but, you are cutting
it into half by from across the diagonals.
363
00:46:07,299 --> 00:46:13,140
So, it is half delta x delta z w g that is
the body force that is acting in the negative
364
00:46:13,140 --> 00:46:19,249
z direction.
Now, we will do a force balance, so and along
365
00:46:19,249 --> 00:46:26,249
the other shear a surface force is are acting
the way we have shown here in the three faces,
366
00:46:27,769 --> 00:46:34,769
the surface force are acting along the three
faces like this. So, now let us do a force
367
00:46:41,230 --> 00:46:46,420
balance, force is a vector, so you have to
do the balance in component wise.
368
00:46:46,420 --> 00:46:53,420
So, x component of the force, so first of
all what is the force balance it is a static
369
00:46:57,670 --> 00:47:04,670
fluid, so by newton’s second law sum of
all forces acting on this volume element is
370
00:47:04,950 --> 00:47:09,539
0 because the fluid is static this is a from
mechanics.
371
00:47:09,539 --> 00:47:14,900
So, sum of all the x forces must be 0, if
it is, if you are worrying about the x component,
372
00:47:14,900 --> 00:47:21,900
so this is the wedge along this direction
it is so this is plus x this is y, so along
373
00:47:26,220 --> 00:47:33,220
the plus x direction you have the force p
x times the area, the area is b which is in
374
00:47:37,160 --> 00:47:44,160
this into the board times delta z I am sorry
this is not y this is z that is way we drew
375
00:47:46,019 --> 00:47:51,029
times delta z this is the force in the plus
x direction.
376
00:47:51,029 --> 00:47:58,029
In this face and then you have this normal
force p n, this normal force can be resolved
377
00:47:58,369 --> 00:48:05,369
into two and this angle is theta into two
components, this is theta along the vertical,
378
00:48:05,710 --> 00:48:12,710
it is p n cosine of theta and along the horizontal
p n sin theta, if such is the case then and
379
00:48:16,329 --> 00:48:21,599
it is acting in the minus, so this force is
acting in a minus x direction, so I will put
380
00:48:21,599 --> 00:48:28,599
a minus sign times p n sign of theta, times
b times delta s is 0. So, let us put 0.
381
00:48:39,549 --> 00:48:46,549
This is the x component, now the b factor
were cancelled right through, now from geometry
382
00:48:50,799 --> 00:48:57,799
delta s sin theta is nothing but, delta z.
383
00:48:58,130 --> 00:49:05,130
So, remember that this is theta, this is delta
z, this is delta x, this is delta s. So, sin
384
00:49:12,519 --> 00:49:19,519
theta is delta z by delta s therefore, delta
s sin theta is delta z, so in the above equation
385
00:49:23,680 --> 00:49:30,680
in this equation I write the first term as
p x delta z minus p n sin theta delta s is
386
00:49:30,769 --> 00:49:37,769
nothing but, delta z is 0. So, we have shown
that by using the x momentum balance.
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00:49:37,809 --> 00:49:43,200
That the magnitude of this force p x must
be the same as the magnitude of the force
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00:49:43,200 --> 00:49:50,200
per unit area here, so p x is p n, now let
us do the vertical balance which is the z
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00:49:53,940 --> 00:50:00,940
component, so here the force is the sum of
all the force as in the z direction must be
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00:50:03,099 --> 00:50:10,099
0 along the z direction the force is so remember
this is the z direction, so the forces are
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00:50:11,390 --> 00:50:18,390
in this plane acting in the positive z direction
p z times delta x times b and then you have
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00:50:22,470 --> 00:50:29,470
in the negative z direction minus p n cos
theta, that is the component of the force
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00:50:31,210 --> 00:50:34,970
p n in the vertical direction.
But it is negative because it is acting in
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00:50:34,970 --> 00:50:40,910
the minus z direction and then you have to
multiply by the area of the triangular sorry
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00:50:40,910 --> 00:50:47,910
the wedge area of this surface because, that
is where this force is acting that is delta
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00:50:49,299 --> 00:50:56,299
s times b.
And then in the vertical direction this is
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00:50:56,690 --> 00:51:03,690
z there is gravity acting, so gravity is you
have already found this is rho times g times
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00:51:05,650 --> 00:51:12,619
half b delta x delta z is 0.
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00:51:12,619 --> 00:51:17,940
Sum of all force is gravity is acting in the
negative z direction, so is this force which
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00:51:17,940 --> 00:51:24,940
is the component of the force on this triangular
surface on this face. So, we can sub cancel
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00:51:25,239 --> 00:51:32,239
b right through and by geometry delta s cos
theta is delta x, so we can cancel this, this
402
00:51:38,059 --> 00:51:45,059
and this to give p is a minus p n minus rho
g delta z by 2 is equal to 0.
403
00:51:51,700 --> 00:51:58,700
Or p z is p n plus rho g delta z by 2 but,
in the limit as the volume shrinks to a point
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00:52:06,349 --> 00:52:13,140
delta z will tends 0, so this term will be
negligible so p z is approximately equal to
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00:52:13,140 --> 00:52:19,960
p n in the limit and that is also equal to
p x.
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00:52:19,960 --> 00:52:26,960
So, what we ended showing finally, is that
if you take a point in a static fluid and
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00:52:27,380 --> 00:52:33,349
if you want to measure the pressure, so you
measure the pressure by taking a tiny area
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00:52:33,349 --> 00:52:37,799
and the area can be oriented in any way.
What we have shown with this example is that
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00:52:37,799 --> 00:52:44,130
p x is p n is p z is the same and it is independent
of orientation of the surface, so we took
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00:52:44,130 --> 00:52:51,130
this wedge, this is the wedge and we looked
at the side face which is p x the bottom face
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00:52:55,069 --> 00:53:01,789
which is p z, the pressure acting is p z and
then incline surface the pressure acting is
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00:53:01,789 --> 00:53:06,440
p n and we find that the pressure must be
the same in a static fluid.
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00:53:06,440 --> 00:53:13,440
Regardless of the orientation of the surface
at which it is measured. Now, we will stop
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00:53:15,430 --> 00:53:20,890
here and then we will continue in the next
lecture and we will worry about force distribution
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00:53:20,890 --> 00:53:25,559
in a static fluid remember that in a static
fluid there are pressure forces which are
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00:53:25,559 --> 00:53:30,819
due to the surface forces and body forces
balance between a pressure and body forces
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00:53:30,819 --> 00:53:35,369
gives raise to force distributions of pressure
distributions. So, we will continue with this
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00:53:35,369 --> 00:53:42,369
in the next lecture and we will see you in
the next lecture. Thank you.