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Welcome to this forth lecture on Fluid Mechanics
for Chemical Engineering under graduate students.
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In the previous lecture, we described the
continuum hypothesis and explained when it
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is valid and when it can break down potentially,
and we also said that in most engineering
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applications, it is it is possible to use
the continuum hypothesis without any problem.
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And in the continue hypothesis, we treat the
variables in the fluid such as pressure, velocity,
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temperature, density and so on as smooth and
continuous functions of position coordinates
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like x, y, z and as well as time.
Then, we started our discussion on what a
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fluid is. In order to do this, it is useful
to contrast the mechanical behavior of a fluid
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with that of a solid.
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So, that is what we will do. We will start
again with the notion of an elastic solid,
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and then, will contrast the behavior of fluid
immediately after this discussion. So, what
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we did in the last lecture was to take a slab
of a solid like a rubber, elastic solid light
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rubber, imagine you have a rectangular slab
and this width of this slab in this third
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direction; let us put a coordinate system
x, y and the third direction that is pointing
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out of the board is z.
So, this is w, and the thickness of the slab
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in the y direction is h, and we are going
to consider a slab such that w is very very
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large compared to h. So, when w is very very
large compared to h, you need not worry about
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the variations in the z-direction. So, all
we will do is to consider a plane, the planar
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cross section in the x y plane.
So, we will take, we will assume that will
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take the x y coordinate system like this and
we will assume that the slab is like here,
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just a cross section in the x y plane, and
what we did was, we imagine that this slab,
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this piece of elastic material is kept between
two plates, and the bottom plate is stationary,
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and the top plate; we want to apply a stress
in the x direction which I will denote as
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f x.
So, essentially what we are doing is, let
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us consider this area of the top surface which
I am going to shade, this force in the x direction
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is being acted upon the entire area. f x acts
on the top surface and let the area of the
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top surface be a. So, f x is acted upon f
x axis on the top surface which is at y equals
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h. And this at y equals zero, and this thickness
is h.
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This is the system. One can think of doing
this experiment in lab, and I am going to
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discuss this as if we are doing this experiment
in a mind and then, so we are doing a thought
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experiment, and then we are going to discuss
how this solid is going to behave under the
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influence of applied forces.
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So, let us imagine, at in the… This is the
unstressed, suppose at time, at some initial
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time, the solid is unstressed
state before the force has been applied. No
force has been applied in the unstressed state,
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and in the unstressed state, imagine that
we are going to mark a line in the solid;
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a vertical line at a given location, and we
are going to follow the position of this line
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as we are going to apply a force on the top
surface. So, when there is no force, no stress
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in the solid. So, this line will be a vertical
line.
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Now imagine that we are going to consider
the same rectangular slab in between two plates,
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and this is x and y as usual, and we are going
to follow, this is in the undeformed state
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or the unstressed state, but now we imagine
that you are applying a force. So, we start
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applying the force at some time t equals zero,
and then we allow an interval delta t, a time
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interval delta t.
And then watch the evolution or the motion
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of this line. When you apply a force to the
solid on the top surface, the solid responds
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by undergoing a deformation. So, what this
solid will do is in fact, this line; force
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is being applied only on the top surface.
So, this point that was here will move here,
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whereas, this point at the bottom surface
is stationary, there is no force is applied;
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the solid is stuck to the bottom surface.
So, this point will remain here. It would
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not deform. So, this line will move in general
in the deformed. Upon application on force,
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this line will move like this.
So, this green line is no force in the unstressed
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state, whereas, this red line is when you
apply a force, and then you record the motion
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or location of this line at a time delta t.
Now, imagine applying this force for continuously
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on the top surface, and what will and let
us try to understand what will happened to
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this red line as you continuously apply the
force.
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So, even if you where to wait for more times;
2 delta t, 3 delta t, 4 delta t, a solid merely
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undergoes a deformation in response to an
applied force, and once it undergoes, once
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it has once it has underwent a deformation,
it will stop deforming due to resistant resisting
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elastic forces that are built in the solid.
So, this let us call this displacement of
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the point on top close to the top plate as
delta l. This delta l will remain the same
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even at higher times, even suppose you have
to measure the displacement of the top point
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versus a function of time. So, at time delta
t, 2 delta t, 3 delta t and so on, in a solid,
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you will observe experimentally that the displacement
of the top point remains delta l. And you
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can also characterize this deformation by
this angle delta alpha. So, and this thickness
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is of course h.
So, a solid responds to an applied stress
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by undergoing a deformation, and this deformation,
the solid if deformation stops after you know,
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it does not continue to increase.
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It undergoes a finite deformation and if you
were to do experiments at different values
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of stresses, so, this is for a given value
of stress if you want to do experiments at
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different values of stress or force sorry,
so, if you have to do that experiment for
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f one, let us call it just f; a force f x,
the displacement will be delta l, then if
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you were to do 2 f x, the displacement will
be 2 delta l, and 3 f x.
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If it is a purely a elastic solid, you find
that the force that you applied will be directly
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proportional to the displacement the solid
undergoes or the displacement which is a response
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to the forces is directly proportional to
the force in a solid.
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So, while this is how experimentally you will
characterize the deformation in a solid, in
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order to make the information from experiments
more general, it is useful to talk in terms
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of a stress rather than a force. Stress is
force per unit area.
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Now this force, in this example, in this example,
the force is applied in the x direction; plus
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x direction and the area of the surface on
which force is exerted is A.
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So, this is simply f x divided by A. So, that
is what we said in the beginning that... So,
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the area is A. So, the area of surface is
A and this is called as stress.
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Now the stress is given denoted by the symbol
usually tau in fluid mechanics; in mechanics
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in general, mechanics in general. And it is
described with two subscripts. One is the
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subscript which is let me write down the subscripts
and explain what those two subscripts mean.
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The subscript x is the direction of the force.
The stress is force acting on a surface per
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unit area. So, force itself has a direction.
This y denotes the unit normal to the surface
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direction of the unit normal; unit vector
if you want to the surface on which force
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is acting, force is acting. Now in our example,
the surface; we took this slab and the force
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was in the x direction, and the direction
of the unit normal in the plus y direction
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which is traditionally denoted by unit vector
j.
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So this y denotes a direction of the unit
vector which is divide that unit vector along
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the y direction. So, it is a direction of
the unit vector perpendicular to the surface.
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That is called unit normal, and x denotes
the direction of the force. So, this is called
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stress. Stress is force by the area, but we
have to specify the direction of the force
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on the surface as well as the orientation
of the surface by specifying the unit normal
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or unit vector perpendicular to the surface.
So, for a solid, you will find that tau y
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x in our example, tau y x is simply f x divided
by A.
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And you will find that this stress if you
do experiments is directly proportional to
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the deformation which is characterized by
the angle delta alpha. We call that you had
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this line, original line in stressed case
and then deformed case.
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The deformation can be characterized by the
angle by which this line tilts upon application
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of force in the x direction. So, this is x,
this is y. So, this angle can be taken as
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a suitable measure of the deformation or the
strain in the solid.
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Once a solid undergoes a deformation, we said
that it is strained. So, the strained; a measure
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of the strain in the solid, deformation of
solid is the angle delta alpha. So, you will
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find that delta alpha; the stress that you
apply is directly proportional to the inclination
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of the this tilt of this line upon the application
of stress and the stress will not change in
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a purely elastic solid as you wait long enough,
sorry, the angle will not change if you wait
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long enough, even if you apply a stress continuously,
the angle will still remain the same. That
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is because the nature of the solid to resist
deformation. So, it resists deformation. It
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under undergoes some deformation that it does
not continue to deform under the application
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of a force.
So, that is the definition of a purely elastic
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solid. So, let us do some simple geometry.
So, this high course, h. So, from this and
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this, displacement of this line from here
to here was delta l at the top plate. So,
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tan delta alpha from this figure is delta
l by h, but when you apply small enough forces,
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delta alpha will be small.
So, when delta alpha is small, tan delta alpha
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is roughly proportional to delta alpha. So,
this equation tells you the delta alpha is
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approximately delta l by h or I can write…
So, that is a delta l by h, that is fine.
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So, this stress; tau y x is proportional delta
alpha. So, I can write tau y x is proportional
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to delta l by h, because delta alpha this
is, please do not confuse this, this is a
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proportionality sign. This alpha is the angle.
So, let me try to write it like this. The
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proportionality sign slightly different from
alpha. So, tau y x is proportional to delta
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l by h.
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And you can replace the proportionality constant,
the proportionality sign with a constant of
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proportionality. That is called a modulus
of elasticity. This is called the modulus
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of elasticity. So, and we can work out the
dimensions or units of this quantity tau y
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x is of dimensions of stress. Stress is force
per unit area. Force is mass times acceleration
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divided by area.
So, stress becomes m l to the minus 1 t to
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the minus 2, and this group is a ratio of
two lengths is dimensionless. It has no dimensions
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because it is length divided by length. So,
strain in a solid is a dimensionless. So,
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the modulus of elasticity will have the same
dimensions of stress. So, the mod normally
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the dimensions of a quantity are denoted by
this square bracket.
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So, m l to the minus 1 t to the minus 2. This
is the dimension; these are the dimensions
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of modulus of elasticity and or stress for
that matter.
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And in SI units, g is, so if you put m as
k g, per meter for l to the power minus 1
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and t to the minus 2 second square, this is
called one pascal. So, stress and modulus
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of elasticity everything is measured in pascal.
So, now this stress is therefore, directly
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proportional to the strain. Now previously,
in this simple thought experiment, we considered
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a slab of the thickness h, but we could take
the tiny thickness delta y in the y direction,
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within the slab itself, we will take a tiny
thin slice, sorry take just to illustrate,
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this is your slab and even within this slab,
you can take a tiny slice of the slab.
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And then worry about what is the stress with
respect, what is a how does the deformation
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change with respect to stress within this
slice. So, we will do the same thing. So,
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we will find the tau y x is proportional to…
So, this within this thin slice, line that
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was original like this would have moved like
this. Lets called as delta l. So, again it
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will be proportional to delta l by delta y.
So, in the limit, now if you take the limit
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when the thickness of the slice goes to zero,
this becomes a derivative in calculus. So,
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tau y x is proportional to d l by d y. And
we can write this as a constant proportionality
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as d l by d y. This is called a strain in
the solid.
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So, while the previous discussion where you
took a finite piece of material h is the valid
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for the particular experiment alone. This
is this expression is valid in general because
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you can take a solid of any thickness and
look at the deformation at the point or within
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the continuum approximation, a tiny slice
of volume around a point, and then you will
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find this equation is valid.
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So, now let us contrast this behavior with
a fluid. I am going to do the same thought
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experiment have a fluid between a two slabs
like a viscous liquid; lets imagine a viscous
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liquid like honey. Take two slabs; let us
mark the coordinate systems. Again I am going
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to assume as before that in principle, it
is two slabs, and on the top slab, you apply
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a force f x. This width is very very large
compared to the thickness edge in which the
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fluid is present. So, we need not worry about
the variation in the z direction. So, in our
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scheme of thing, this is x, this is y and
this is z. So, we will just consider the x
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y plane and we will put a colored dye in the
fluid at time t equal 0, when there is no
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force that is exerted.
Now a time a time t is equal to 0 plus, we
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start applying a constant force f x. Now what
we are going to do in this discussion is look
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at the evolution of this colored line as a
function of time, when the force is being
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applied on a top plate and as before, the
area of the top plate is A.
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Now, so, let us imagine that after a time
delta t, we look at this line. So, the bottom
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plate is stationary. There is no force. So,
the fluid here close to the bottom plate,
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just adjacent to the bottom plate will not
move, while the top plate; fluid process a
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top plate since you are exerting a force,
this line would have deformed like this.
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So, this is at a time let say delta t, but
a fluid continues to deform under the application
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of stresses, especially shear stresses. So,
here we are applying a force on a surface
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and the force is parallel to the surface.
So, the force is parallel. This is the surface
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on which the force is being applied. And these
are called shear or tangential stresses. Tangential
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forces or stresses are force per unit area
is a stress; shear or tangential stresses
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because it is tangential to the surface on
which the force is acting. The force is tangential
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to the surface itself.
In contrast to normal stresses which are perpendicular;
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as the name suggest, so, here we are applying
a shear stress or a shear force on a tiny
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slice of a fluid with between two plates and
we are finding that you will find that, if
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you do this experiment, if you watch this
evolution of this colored line as a function
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of this line, it will continue to move at
later and later times. A fluid continues to
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deform under the application of shear stress
in contrast to solid which deforms to some
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extent and then stop deforming.
So, the fluid continues to deform. So, what
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you will find in experiment is that, at 2
delta t, this line will become like this,
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at 3 delta t, this may become like this. So,
if you think of this distance as delta l,
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and this is a 2 delta t, this line is a 3
delta t
and so on. So, this line continues to deform;
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that means, the fluid continuous to move under
the application of shear stresses. We say
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that a fluid flows under the application of
shear stress.
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So, you will find that if you do this experiment,
that if you were to plot at t equal to 0,
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this line will be here. So, I am going to
draw different snap shots, t equals delta
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t, this line which was originally here would
have moved like this. It would have moved
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by an angle delta alpha. And the top point
would have moved by the length delta l and
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at a later time; t equal 3 delta t, it would
have moved, I am sorry just 2 delta t; just
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00:24:53,669 --> 00:25:00,669
keep 2 for simplicity. It would have more
than two alpha. The angle would have increased
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to 2 delta alpha and the length of this point
from which original position is known to be
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2 delta l.
So, a fluid cannot resist any shear stress.
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So, it continues to move upon application
of shear stress. So, clearly, this thought
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experiment suggest; this can be done really
in a lab also, that one can do it, but here
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for the sake of illustration I am just doing
a thought experiment.
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00:25:30,669 --> 00:25:36,059
So, the stress in a fluid, unlike a solid
cannot be a proportional to deformation. The
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00:25:36,059 --> 00:25:42,629
reason is you can get any amount of deformation
if you are prepared to wait a long enough.
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00:25:42,629 --> 00:25:48,840
This slice which was originally here underwent
a deformation of delta l at time delta t,
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2 delta l at time two delta t and so on. So,
it would keep deforming as you wait long enough.
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So, stress really cannot be proportional to
deformation. So, what is then stress a function
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00:26:01,609 --> 00:26:06,070
of?
Let us slightly change the experiment. Instead
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of keeping... So, here I am applying a force
f x; same force. So, here I mean a time t
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equal to 0, there is no force. So, there is
no force, but at later times, we are applying
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the same force.
But now let us take this experiment. Instead
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of applying f x, I am applying 2 f x. You
will find that even a time delta t, this line
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00:26:34,369 --> 00:26:41,369
would have moved by 2 delta l. So, if you
increase the force, for the same amount of
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00:26:45,409 --> 00:26:51,919
time that your waiting, the deformation will
increase for the same amount. So, for example,
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00:26:51,919 --> 00:26:57,739
here f x was the force here, the force is
2 f x. For the same delta t, we are finding
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00:26:57,739 --> 00:27:04,039
that the deformation has doubled.
So, the stress is not proportional to deformation
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00:27:04,039 --> 00:27:09,830
for say because the same amount of deformation
can be obtained in a fluid at you know, if
215
00:27:09,830 --> 00:27:14,330
you are prepared to wait long enough. So,
even if you apply a very very tiny amount
216
00:27:14,330 --> 00:27:20,100
of force, you can get the same amount of a
deformation delta l, if you wait sufficiently
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00:27:20,100 --> 00:27:23,279
long enough.
So, clearly stress is not directly proportional
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00:27:23,279 --> 00:27:29,409
to deformation. Indeed, it is proportional
to how fast the fluid deforms or we say more
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00:27:29,409 --> 00:27:36,409
clearly or more specifically; the rate at
which the fluid deforms. So the stress is
220
00:27:36,590 --> 00:27:40,309
proportional in a fluid as this experiment
suggest.
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00:27:40,309 --> 00:27:47,309
So, we have that for force f x, the deformation;
this is applied force, this is the deformation
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00:27:50,929 --> 00:27:57,929
that one gets as we measure through the angle
was delta alpha at time delta t, 2 delta alpha
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00:27:59,759 --> 00:28:06,759
at time 2 delta t. And similarly, if you apply
2 f x, you get 2 delta alpha at delta t itself.
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00:28:10,809 --> 00:28:17,809
So, the stress tau y x which is the same which
has the same meaning as what we had in the
225
00:28:23,659 --> 00:28:30,659
previous illustration from elastic solid,
this is the force in the x direction; on a
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00:28:30,669 --> 00:28:37,669
surface whose perpendicular is in the y direction
to the surface.
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00:28:43,249 --> 00:28:50,249
So, tau y x is the stress that you are exerting
on the top plate. This is simply equal to
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00:28:54,029 --> 00:29:01,029
f x by A divided by A. It cannot be a proportional
to delta alpha, but infact, it proportional
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00:29:01,210 --> 00:29:07,539
to the rate at which alpha changes with time;
delta alpha by delta t
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00:29:07,539 --> 00:29:13,889
Then this, because this what the experimental
result would suggest, if you wait for delta
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00:29:13,889 --> 00:29:19,239
t for the same f of x, you get delta alpha.
If you wait 2 delta t, you have 2 delta alpha.
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00:29:19,239 --> 00:29:26,239
But if you double this stress, for the same
delta t, the deformation doubles. So, all
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00:29:26,779 --> 00:29:33,779
this is captured by simple hypothesis or propotion
that the stress must be proportional to the
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00:29:33,869 --> 00:29:40,179
rate of deformation.
For example, in this proportion, if I double
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00:29:40,179 --> 00:29:47,179
the stress, so, let say tau 1 was the stress,
the delta alpha 1 was the deformation at time
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00:29:47,690 --> 00:29:54,690
delta t 1. If aware to wait for 2 delta t
1, for the same stress, then proportional
237
00:29:54,710 --> 00:30:01,710
to, then I would get 2 delta alpha 1; because
I am keeping stress constant. If I wait 2
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00:30:02,509 --> 00:30:06,669
delta t 1; since it directly proportional,
it will be 2 delta alpha 1.
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00:30:06,669 --> 00:30:13,669
But if I keep 2 tau 1 and I keep delta t 1
the same; since is directly proportional,
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00:30:14,159 --> 00:30:18,399
the angular displacement has to be increased
by 2 delta alpha 1. So, all this is captured
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00:30:18,399 --> 00:30:24,279
by this simple relation that tau y x must
be directly proportional to delta alpha by
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00:30:24,279 --> 00:30:28,239
delta t in a fluid.
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00:30:28,239 --> 00:30:35,239
Now, let us look at the geometry again, once
again. So, this was the original line. This
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00:30:38,820 --> 00:30:45,820
was the deformed line, this is delta alpha,
this is delta l, but in a fluid, the top plate
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00:30:47,539 --> 00:30:50,729
continues to move because the fluid is also
moving. The top plate will also; if you exert
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00:30:50,729 --> 00:30:56,820
a force, top plate will continue to move.
So, this delta l, so, the top plate will acquire
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00:30:56,820 --> 00:31:03,820
a velocity, if you apply a force to the fluid.
So, it is u times; delta l will be u times
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00:31:05,029 --> 00:31:12,029
delta t, and this is h. So, from geometry,
we know that tan alpha is delta l by h, tan
249
00:31:16,679 --> 00:31:23,460
of delta alpha, sorry and for small delta
alpha, tan delta alpha is approximately delta
250
00:31:23,460 --> 00:31:29,979
alpha; small angles. This we discussed few
a minutes back for the case of elastic solid
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00:31:29,979 --> 00:31:36,979
also. So, delta alpha is delta l by h, delta
l is delta u times delta t divided by h. So,
252
00:31:47,239 --> 00:31:54,239
delta alpha by delta t is equal to u by h.
So, u is a velocity of the plate; top plate,
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00:31:55,639 --> 00:32:02,639
and h is the thickness. So, tau y x is proportional
to u by h, where u is the velocity. Suppose
254
00:32:03,019 --> 00:32:08,450
you have exerted a force in the x direction
on the top plate, this plate will start moving
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00:32:08,450 --> 00:32:14,739
if whatever is the material that present between
the two plates is a fluid, and you can characterize
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00:32:14,739 --> 00:32:21,460
that motion with a velocity u of the top plate.
So, this is the velocity of top plate, and
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00:32:21,460 --> 00:32:28,460
in the x direction, h is the thickness of
the fluid.
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00:32:33,899 --> 00:32:40,899
So, instead of taking finite thickness, we
can also consider an infinitesimal thickness
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00:32:45,190 --> 00:32:52,190
delta y, then tau y x will be proportional
to… So, infinitesimal slice of fluid will
260
00:33:01,039 --> 00:33:08,039
be proportional to delta u by delta y, which
is essentially the velocity of this top layer
261
00:33:15,259 --> 00:33:20,279
with respect to the bottom layer.
And when all the things when in the limit
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00:33:20,279 --> 00:33:26,179
delta y tending to 0, tau y x will be proportional
to; this will become a derivative, will be
263
00:33:26,179 --> 00:33:32,929
proportional to d u d y. This is called the
velocity gradient. It is the derivative of
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00:33:32,929 --> 00:33:39,929
the x velocity u with respect to the y coordinate.
It is called the velocity gradient.
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00:33:41,259 --> 00:33:48,259
So you can replace the proportionalities six
symbol with a constant of proportionality.
266
00:33:49,330 --> 00:33:54,830
And that constant proportionality is denoted
by the greek symbol mu. It is called the viscosity
267
00:33:54,830 --> 00:34:01,830
of the fluid.
So, a fluid; in a fluid, the stress is not
268
00:34:03,469 --> 00:34:08,909
proportional to deformation. It is proportional
to rate of deformation and through this simple
269
00:34:08,909 --> 00:34:15,840
geometric analysis or argument, they have
shown that the rate of deformation is proportional
270
00:34:15,840 --> 00:34:19,340
to the velocity gradient, that is equal to
the velocity gradient.
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00:34:19,340 --> 00:34:24,859
Therefore tau y x which is proportional to
the velocity gradient, can be replaced by
272
00:34:24,859 --> 00:34:30,049
a constant mu, the proportionality sign can
be replaced with a constant which is called
273
00:34:30,049 --> 00:34:34,129
a viscosity of the fluid.
So, what this says is that suppose we were
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00:34:34,129 --> 00:34:41,129
to do this experiment of two fluids; one with
higher viscosity, and the other with lower
275
00:34:44,210 --> 00:34:51,210
viscosity, and you apply the same force of
f x on the top layer, and you watch the motion
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00:34:53,799 --> 00:35:00,760
of these lines at a time, after a time delta
t.
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00:35:00,760 --> 00:35:07,760
So, if you do that, you will find that the
lower viscosity fluid would have deformed
278
00:35:08,200 --> 00:35:13,510
more compared to the higher viscosity fluid
at the same time, but if you are prepared
279
00:35:13,510 --> 00:35:19,660
to wait long enough at higher values of delta
t, even the higher viscosity fluid will achieve
280
00:35:19,660 --> 00:35:23,270
the same amount of deformation as low viscosity
fluid.
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00:35:23,270 --> 00:35:29,089
So, in simple terms, a solid cares how much
you deform. The stress is proportional to
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00:35:29,089 --> 00:35:35,359
the deformation, the amount of deformation
while liquid cares how fast you deform. So,
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00:35:35,359 --> 00:35:42,359
a fluid resists deformation not in the sense,
a fluid resist deformation by the rate at
284
00:35:43,119 --> 00:35:45,579
which it is deforming, not by the deformation
itself.
285
00:35:45,579 --> 00:35:52,579
If you take a solid like steel which is very
very rigid, and if you take a soft material
286
00:35:53,119 --> 00:35:59,240
like a rubber, then if you apply the same
amount of stress, both of these materials
287
00:35:59,240 --> 00:36:06,240
will deform, but the extent of deformation
will be more in rubber than in steel.
288
00:36:06,380 --> 00:36:11,470
Whereas, here in a fluid, if you take two
different fluids; one with very high viscosity
289
00:36:11,470 --> 00:36:18,440
and other with low viscosity, if you apply
the same amount of stress, the amount at which,
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00:36:18,440 --> 00:36:23,160
the rate at which the fluid deforms will be
different which is given by suppose you keep
291
00:36:23,160 --> 00:36:28,059
the time interval constant, then that is given
by angle delta alpha.
292
00:36:28,059 --> 00:36:32,760
So, this delta alpha will be small for a higher
viscosity fluid, while it will be larger for
293
00:36:32,760 --> 00:36:38,339
low lower viscosity fluid. But this delta
alpha will continue to increase in both the
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00:36:38,339 --> 00:36:45,339
cases. So, it is a matter of this how fast
a fluid deforms, and a higher viscosity implies
295
00:36:45,700 --> 00:36:52,700
that the fluid is going to resist deformation
more compared to the case where when the fluid
296
00:36:55,970 --> 00:37:00,910
has lower viscosity, where it deforms very
quickly. So, it is the rate of deformation;
297
00:37:00,910 --> 00:37:06,440
that is the crucial factor you know fluid.
And the stress is directly proportional to
298
00:37:06,440 --> 00:37:13,440
rate of deformation.
Now tau y x is mu d u d y, where the stress
299
00:37:15,740 --> 00:37:22,740
is equal to the viscosity times rate of deformation
is sometimes called Newton’s law of viscosity,
300
00:37:27,849 --> 00:37:34,849
but you should understand that this is merely
an observed material behavior, observed behavior
301
00:37:37,660 --> 00:37:42,619
of a class of fluids.
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00:37:42,619 --> 00:37:49,619
So, a Newtonian fluid is merely or the Newton’s
law of viscosity; let me write this as the
303
00:38:01,950 --> 00:38:08,950
Newton’s law of viscosity which is merely
a behavior of a class of fluids materials.
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00:38:18,779 --> 00:38:23,930
It is not a fundamental law and there are
of course, there are fluids need not obey
305
00:38:23,930 --> 00:38:30,930
Newton’s law of viscosity. So, but, it turn
out that many simple fluids like air, water,
306
00:38:32,619 --> 00:38:39,619
honey, then glycerin and several oils; they
all observe, they all obey this behavior.
307
00:38:46,349 --> 00:38:53,349
So, it is a pattern of behavior that is followed
by wide class of fluids, but there are always,
308
00:38:53,670 --> 00:38:59,539
there are lots of exceptions to this behavior,
but this one of the simplest possible relations
309
00:38:59,539 --> 00:39:05,819
between the stress and the rate of deformation
in a fluid. So, and fluids which obey this
310
00:39:05,819 --> 00:39:12,819
behavior, they are called Newtonian fluids.
They are called Newtonian fluids.
311
00:39:16,099 --> 00:39:22,410
Now, let us workout the dimensions of viscosity
because we are encountering this for the first
312
00:39:22,410 --> 00:39:29,410
time in this course. So, tau y x is mu d u
d y. tau y x is stress and stress is force
313
00:39:30,609 --> 00:39:37,609
per unit area. So, this has dimensions. We
already saw this few minutes back in the context
314
00:39:39,680 --> 00:39:46,680
of modulus of elasticity, and velocity gradient
is basically l t to the minus 1 divided by
315
00:39:48,700 --> 00:39:54,559
l. So, this is simply t to the minus… It
has dimensions of 1 over time. So, if you
316
00:39:54,559 --> 00:40:01,559
work this out, the dimensions of the mu is
m l to the minus 1 t to the minus 1.
317
00:40:03,339 --> 00:40:10,339
Just compare these two. In order for this
equation to be dimensionally consistent, then
318
00:40:12,079 --> 00:40:19,079
mu has to have this dimensions; viscosity
as to have this dimensions. In SI units, mass
319
00:40:19,539 --> 00:40:25,720
is measured in kilogram, length in meter,
and time in seconds. So, the unit of viscosity
320
00:40:25,720 --> 00:40:32,720
is k g per meter second. So, this also equal
to one pascal second.
321
00:40:32,880 --> 00:40:39,880
Now, just to give you some example of various
viscosity values that one sees in common fluids,
322
00:40:43,240 --> 00:40:50,240
suppose you have air, all the viscosity values
are in pascal seconds. Viscosity value of,
323
00:40:51,769 --> 00:40:58,769
viscosity of air is about 10 to the minus
5 in pascal second units. Water is about 10
324
00:41:00,710 --> 00:41:07,710
to the minus 3 and castor oil 100 times more
viscous than water; 0.1, and blood is which
325
00:41:13,079 --> 00:41:18,220
is a bodily fluid has viscosity of 8 times
to the minus 3, eight times to viscosity of
326
00:41:18,220 --> 00:41:21,650
water.
But interestingly, if you consider mercury
327
00:41:21,650 --> 00:41:27,039
which is a liquid metal, the viscosity is
very close to that of water. It is only 1.55
328
00:41:27,039 --> 00:41:34,039
times larger than water. So, mercury has large
density we know that, but the viscosity is
329
00:41:34,579 --> 00:41:41,319
a completely different property. It is not
correlated with density in any direct way
330
00:41:41,319 --> 00:41:45,779
as you can see here. Even though mercury is
very very dense, the viscosity of mercury
331
00:41:45,779 --> 00:41:51,819
is not very different from that of the viscosity
of water.
332
00:41:51,819 --> 00:41:58,819
So, in this course, which is about fluid mechanics
applied to chemical process industries, we
333
00:42:03,200 --> 00:42:10,200
will largely restrict ourselves to Newtonian
fluids mostly. But, at the end of the course…,
334
00:42:16,160 --> 00:42:22,440
so, when we say Newtonian fluids, we mean
that the stress is proportional to or is equal
335
00:42:22,440 --> 00:42:28,660
to viscosity times the rate of deformation,
which is the velocity gradient.
336
00:42:28,660 --> 00:42:35,510
Mostly, but at a later point of time, at the
end of the course, we will have opportunities
337
00:42:35,510 --> 00:42:42,510
to talk about fluids which do not obey this
behavior. They are called Non-Newtonian fluids.
338
00:42:43,960 --> 00:42:50,960
So, we will have an opportunity to discuss
the behavior of fluid that do not follow the
339
00:42:52,150 --> 00:42:56,930
Newton’s law of viscosity, but for the initial
part, we will certainly restrict ourselves
340
00:42:56,930 --> 00:42:58,950
to Newtonian fluids.
341
00:42:58,950 --> 00:43:05,950
Now, what this is saying is if tau y x is
nu d u d y, if you have to do this experiment
342
00:43:08,500 --> 00:43:15,500
in a lab and plot the data that you get for
tau y x versus d u d y, this is variously
343
00:43:18,799 --> 00:43:23,960
called as rate of deformation because that
is what it is. It is rate at which the fluid
344
00:43:23,960 --> 00:43:28,500
deforms.
But since the rate of deformation we showed
345
00:43:28,500 --> 00:43:35,500
is equal to velocity gradient, it is also
called as velocity gradient, and since the
346
00:43:36,980 --> 00:43:40,869
deformation; the way in which the fluid is
deforming by shear; that is, you are applying
347
00:43:40,869 --> 00:43:47,869
a tangential force. This also called shear
rate or rate of shear or shear rate. You should
348
00:43:48,260 --> 00:43:53,539
be… sometimes it is also called the strain
rate because delta alpha is a measure of strain,
349
00:43:53,539 --> 00:44:00,539
and this is delta alpha divided by delta t.
So, these are all various descriptors that
350
00:44:00,900 --> 00:44:05,500
are used to signify the same quantity which
is mathematically d u by d y, which is a velocity
351
00:44:05,500 --> 00:44:09,029
gradient.
So, if you plot the shear stress versus shear
352
00:44:09,029 --> 00:44:16,029
rate, so this is the shear rate; the rate
of the deformation for Newtonian fluid, you
353
00:44:16,240 --> 00:44:22,769
will get a straight line that passes through
the origin, and slope of this line will be
354
00:44:22,769 --> 00:44:28,650
the viscosity mu.
This is not to say that all fluids will have
355
00:44:28,650 --> 00:44:34,829
the same behavior. I have told you that there
are many fluids which do not follow this behavior.
356
00:44:34,829 --> 00:44:41,809
Let me tell you the experimental behavior
that is commonly seen. Suppose you take…,
357
00:44:41,809 --> 00:44:48,809
so these are, I am going to… the blue line
is Newtonian. So, let me write this in blue.
358
00:44:49,700 --> 00:44:56,700
So, let us suppose you take a solution of
water and a polymer like poly ethylene oxide.
359
00:44:59,339 --> 00:45:05,039
You dissolve very very small quantities less
than one weight percent polymer in water;
360
00:45:05,039 --> 00:45:12,039
polymer such as poly ethylene oxide, so, polymer
solution, and if you plot tau y x; if you
361
00:45:12,049 --> 00:45:17,289
do this experiment in a lab, you will find
that it is not linear. It is going to behave
362
00:45:17,289 --> 00:45:24,289
like this, if you take a polymer solution.
This is a polymer solution.
363
00:45:28,200 --> 00:45:35,200
Now, for a Newtonian fluid, if you have to
plot the viscosity, so, let me just go a little
364
00:45:36,760 --> 00:45:37,319
below.
365
00:45:37,319 --> 00:45:44,319
If you plot the viscosity as a function of
shear rate, it is a constant because tau is
366
00:45:50,390 --> 00:45:55,010
directly proportional to d u d y, and it is
a straight line. So, the slope is a constant,
367
00:45:55,010 --> 00:46:02,010
but if you take a polymer solution, the viscosity
will decrease with shear rate. So, such fluids
368
00:46:04,079 --> 00:46:08,740
are called shear thinning because the viscosity
decreases with shear rate.
369
00:46:08,740 --> 00:46:15,740
But there are also fluids that shear thicken.
These are colloidal dispersions. By shear
370
00:46:18,029 --> 00:46:23,839
thickening, we mean that the viscosity increases
with shear rate. So, you could also have this
371
00:46:23,839 --> 00:46:30,839
behavior. Both are non-newtonian; that is,
the viscosity is not a constant, that is,
372
00:46:32,410 --> 00:46:39,410
our tau is not linear in d u d y, but there
are different classes of non-newtonian behavior
373
00:46:40,369 --> 00:46:45,170
and there is one more type a non-newtonian
behavior which I will draw in a separate graph.
374
00:46:45,170 --> 00:46:52,170
So, you can have a material like tar, where
if you plot tau y x versus shear rate, the
375
00:46:55,410 --> 00:47:01,069
material does not flow up to a critical value
of shear rate, and it flows after that like
376
00:47:01,069 --> 00:47:08,069
a Newtonian fluid. Such fluid such materials
are called Bingham plastics. Example is tar.
377
00:47:11,029 --> 00:47:17,019
So, appears like a solid up to a critical
shear stress, after that, it starts flowing
378
00:47:17,019 --> 00:47:21,430
like fluid. So, such materials are called
Bingham plastics.
379
00:47:21,430 --> 00:47:28,430
So, this discussion is to just tell you what
a fluid is, what is a fluid, why is it, how
380
00:47:30,970 --> 00:47:37,779
does it differ from a solid in terms of its
deformation nature, and we saw that a fluid
381
00:47:37,779 --> 00:47:43,190
fundamentally differs from a solid in the
way it responds to shearing stresses. Fluids;
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we say it cannot resist any shear stresses
unlike a solid because if you apply a shear
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stress to a solid, it undergoes a deformation
and it stops deformation after some time.
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00:47:54,869 --> 00:48:00,000
It does not continue to deform unlike a fluid
which keeps on deforming as long as the force
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is applied, as long as the shear stress is
applied.
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So, a solid cares how much you deform, while
liquid cares about a how fast you deform.
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So, if viscous if the constant of proportionality
between the stress and rate of deformation
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00:48:13,890 --> 00:48:20,630
is called viscosity and fluids with different
viscosity offer varying resistance to rate
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at which they deform. Just as solid with different
moduli offer varying amounts of resistant
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00:48:26,529 --> 00:48:32,730
to how much they deform. For example, if you
apply a stress of 100 pascals to a steel bar,
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00:48:32,730 --> 00:48:38,799
it will deform very very little in contrast
to let say a piece of soft rubber.
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So there, these two materials; steel and rubber
are characterized by different elastic moduli
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00:48:45,680 --> 00:48:52,680
and material with lower elastic moduli deforms
much more compared to material with larger
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elastic moduli, whereas in a fluid, a fluid
with higher viscosity deforms at a much smaller
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00:48:59,779 --> 00:49:06,539
rate than fluid with a much lower viscosity
for the same stress that you apply because
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in a fluid, you can get the same amount of
deformation regardless of the stress applied
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because you can always wait long enough.
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Now, the next topic that we are going to worry
about is fluid statics. Now so, this is the
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00:49:31,200 --> 00:49:36,630
first series of topics we are going to cover
in fluid mechanics. So, far we are been introducing
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the subject and introducing the notions of
continuum hypothesis, what a fluids is and
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00:49:43,269 --> 00:49:46,279
so on.
The first topic that we are going to discuss
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is fluid statics and what are the forces that
are… How force distribution happens when
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00:49:53,599 --> 00:50:00,599
a fluid is completely static. Static means
there is no flow, there is no motion.
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So, this is the topic that we are going to
first discuss, and we will start from next
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the lecture. So, we will see you in the next
lecture to discuss this new topic on the force
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distributions and static fluids. Good bye.