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Welcome to this third lecture in this NPTEL
course on fluid mechanics for under graduate
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students in chemical engineering. In the first
two lectures, we introduced this course by
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telling you how chemical engineering is important,
how fluid mechanics is important in chemical
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engineering in chemical process industries,
and why chemical engineers have to have a
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thorough understanding of fluid mechanics
in order to design many process many chemical
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processes better.
And we also told you the approaches that one
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normally takes in understanding fluid flows
for chemical engineering applications. We
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said that there are macroscopic approaches
where in one writes integral balances of mass,
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momentum and energy, and then there are microscopic
approaches where one derives differential
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equations that are valid at each and every
point in the flow.
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And then when these two approaches are not
workable in a very very complicated industrial
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setting, there is also the need for experimentation
and there we said that dimension also will
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play an important role in using in properly
designing experiments and using those experimental
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results for scaling up of results to from
a lab scale to industrial scale.
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So, after introducing the course, we also
gave a brief outline of what are all the topics
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that we are going to cover in this course,
and after that we introduced the system of
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units that we are going to follow; SI units
that we are going to follow the SI units in
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this course which is the conventionally accepted
units everywhere in this scientific world.
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And we also told that if there are other systems
of units that are being used in some context.
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Then, there are ways of converting from SI
units to the other units if you need to convert
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them.
Finally, in the last lecture, we started a
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discussion on the frame work that we are going
to use to analyze fluid flows in chemical
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process industries and that frame work is
called the continuum approximation.
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It is called the continuum approximation.
What this approximation says is that in a
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flow for example, if you are interested in
flow in a very long pipe just schematically
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shown. So, there is flow. You have flow in
very long pipe. In principle, that continuum
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approximation says that you can define velocity
at each and every point in the flow.
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The continuum approximation says that you
can define velocity, density, pressure or
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any fluid property, temperature as a smooth
and continuous function of spatial co-ordinates.
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We will see that in order to analyze any problem
in fluid mechanics, you have to first set
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up a proper co-ordinate system. For example
here, a simple rectangular co-ordinate system.
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So, here for example, if you have density,
density is a smooth continuous functions of
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x, y, z that the three spatial co-ordinates
in the rectangular co-ordinate system and
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time.
So, the key thing is, it is a smooth function,
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continuous function that is defined at each
and every point. So, all these quantities;
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velocity is a vector. It has both magnitude
and the direction. Velocity is also a function
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of all the three spatial co-ordinates and
t is time.
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So, what the continuum approximation says
is that you can plot all these properties
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if you take density as a function of x which
says that you can plot as a function; it is
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a smooth function. So, the question is first
of all, why is this approximation being made?
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We saw that a fully molecular approach
that one can conceive because we said that
a fluid ultimately beat a liquid or a gas
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is ultimately comprised of molecules. So,
one is tempted to first go for a fully molecular
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approach.
So, even if you consider any reasonable macroscopic
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volumes that are encountered in any practical
application, you will have the number of molecules
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to be a very large number. It is of the order
of Avogadro numbers 10 to the 23 in a very
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reasonable volume in any practical applications.
So, this is a huge number. So, it is not possible
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to solve the motion of all these huge number
of molecules to obtain for example, quantities
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of practical interest towards, such as forces
that are experienced by a the walls of the
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pipe and so on.
For example, if you want to pump a fluid in
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this pipe, the reason why you need a pumping
cost is because of the drag force exerted
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by the surrounding walls on the fluid flow.
So, if you want to estimate or predict the
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pumping cause, you have to know you have to
have an idea of these forces. So, ultimately
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these forces are due to the molecular interactions
that are there, but it is simply not possible
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to compute the forces exerted by each and
every molecule on the surface of the wall
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because simply because of the huge number
that is present.
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Even if you have that information ready, even
if you have to solve for all these forces,
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such an information is not really required
because all we require is the total force
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and we are not overly concerned in engineering
studies has to a which molecule is exerting
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a force. All we are interested in design is
the force that is being experienced by a solid
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surface for example.
So, a fully molecular approach we said in
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the last lecture is not feasible generally
and even if it were there to be feasible,
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it is not necessary in most practical applications,
in engineering applications.
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So, the fully molecular approach is not necessary
of feasible. So, we rule that out. Then the
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only other option is this continuum approximation
where we completely ignore molecular details
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and treat the fluid as a continuous medium
wherein with respect to a co-ordinate system,
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at each and every point in space, various
quantities such as density, pressure, velocity,
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temperature, concentration; all these are
defined at each and every point. And therefore,
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you can and these are assuming to this smooth
and continuous functions of all these spatial
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co-ordinates and time. And this is the continuum
approximation.
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Now whenever we make an approximation on hypothesis,
it is first useful to see the domains of validity
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of the hypothesis. So, how do we go about
doing that?
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Let us take again this case of a pipe flow.
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We have this pipe which is very commonly encountered
in chemical industry, and let us say that
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we are interested in suppose we are interested
in defining the density of the fluid at each
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and every point in the pipe. So, we first
put a co-ordinate system. This is called a
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rectangular co-ordinate system. There are
three mutually perpendicular axis.
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So, you want to define in the continuum approximation
within the continuum approximation; quantities
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like density at each and every point in the
pipe. What do you mean by density at each
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and every point in the pipe?
Well, let us take a point. This point a is
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labeled by the co-ordinate x y z for example,
if you have to construct a position vector.
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So, it will have a label x y z. So, you want
to tell you want to see what is what do you
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mean by, we want to understand what do we
mean by density at this point. So, how do
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we calculate density at a point? Well, a point
really has no volume. So, we cannot dens…
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First before I say that, first what is density?
A fluid density, as we all know is denoted
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by the greek symbol rho is mass per unit volume
of the… So, you take a particular volume,
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and measure the mass of the fluid present
in the volume, divide the mass with the volume.
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So, this is the density, and in SI units,
it is k g per meter cube. Those are the units.
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So, here we are worried about density at a
point. A point has zero volume. If there is
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zero volume, then there is zero mass. So,
that does not make sense. So, what we mean
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the density at a point is that, you take a
point, let us say a point here just for clarity,
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you take a point and construct a tiny volume
around it. You construct a tiny volume. For
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simplicity, I am constructing a cubic volume.
So, this cube has side delta l cube. So, that
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is the volume; delta v. So, you take any point
in the pipe. So, that is fluid flow in the
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pipe.
You want to understand for simplicity let
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us say, density at a point. So, what is density
at point a? So, let us call this a just because
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we are changed at. Let us call this point
as a. So, density at point a, what is the
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density? Well what you will do is you will
count the number of molecules in that volume
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delta v. So, we will take the delta v which
is basically delta a whole cube in the denominator.
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You will count the number of molecules. Let
us call it as n; capital n times the mass
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of each molecule. So, this is the total mass
of, this is molecular mass. This is the total
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number of molecules in that volume. So, this
will be the density at point a.
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So, but, since we are saying the density associated
with the point, we have to be careful in saying
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that this is in the limit as the volume delta
v shrinks to a point; that is, tends to zero.
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So, this is the way in which density is defined
formally in a continuum approximation the
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to take a point, construct a tiny volume,
count the mass or count number of molecules
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and find the mass, divided by the volume
Now, clearly we can up take. So, firstly,
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in order for this to make sense, this delta
v; the definition a not for this definition
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to make sense, this density that we obtain,
the numerical value should be independent
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of delta v. That is what you should we should
expect because if we find if you calculate
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density at a point by considering one delta
v, and if I calculate the density by considering
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a slightly different delta v, and if we come
up a different answers, then density is not
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uniquely defined at that point.
So, in order for the continuum approximation
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to make sense, the density; the numerical
value of density that we obtain should be
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independent of delta v. Now if that is the
case, let us take the extreme cases. If that
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is the expectation that we have, then let
us take delta v to be very small.
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Now what do we mean by very small? Well, we
know that a fluid is ultimately comprised
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of molecules. So, let us take, let a be the
diameter of the molecules; that compresses
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the fluid. So, first is, consider delta v
to be very small. If a be the diameter of
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the molecules, and delta v will be roughly
proportional to a cube, there will be some
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numerical pre-factors. Those are not of importance
to us in our argument.
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So, if delta v is proportional to a cube,
then you a have a very tiny volume that is
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proportional to, tiny volume that is comparable
to molecular scales. So, if you consider a
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such a tiny volume, this is a tiny volume
proportional to molecular volume, then at
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a molecular level, the fluid molecules are
not stationary or static. They are at any
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non zero temperature, owing through thermal
motion, they will be rapidly moving at very
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large velocities.
So, the presence or absence of a molecule
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is a purely probabilistic event. So, if you
consider a volume at a if you take a point
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and construct a volume about that point and
let if that volume is comparable to molecular
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dimensions, then whether there are molecules
are not itself is a probabilistic event. So,
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this is true, even if the molecules, sorry,
if the volume size is slightly larger, even
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if it is let us say, five times the molecular
diameter. I mean just for the sake of argument.
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If delta v is proportional to five times a
cube, even so, then, the number of molecules
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that are present; so, here there could be
either one molecule or no molecule. Here there
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could be let us say few molecules; three.
If you have to consider the same volume element
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around that point at some other later time,
this could have no molecules at all. So, the
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number of molecules in a very tiny volume
that is proportional to the molecular dimensions
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is a highly fluctuating quantity.
Therefore, the density that you will get,
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which is the number of molecules times the
molecular mass divided by delta v, in the
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limit delta v goes to 0, this will also be
highly fluctuating, if delta v is comparable
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to molecular dimensions. So, clearly there
is a lower limit as to how small your delta
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v can be because if your delta v is too small
density, then t is not uniquely defined because
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it will fluctuate very wildly.
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So, in the next in the next slide, what I
am going to do is to plot, if you have to
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do this as a thought experiment, where you
plot density which is defined as limit delta
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v going to 0 divided by delta v, now, since
I am going to discuss wide variations of delta
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v, it is useful to plot density as a function
of a logarithm of delta v in the x axis, and
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I am plotting the density in the y axis. Now
what will I get? Now when delta v is proportional
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to a cube, very small, where a is the molecular
diameter, then density will be a widely fluctuating
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quantity for the reasons I just mentioned.
Now as delta v increases in this axis, now;
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that means that your volume about that point
is increasing larger and larger. So, the density
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that is defined as number and small molecular
mass divided by the delta v, we will slowly
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settle to a constant value because if the
volume becomes larger, a number of molecules
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that are present, which I am going to indicate
now by several red dots, even though it is
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still a fluctuating quantity, whether you
add one more or one less, should not largely
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affect the numerical value of density. So,
even though n is still fluctuating, the fluctuations
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cannot play a big role in defining the numerical
value of the density.
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So, as you increase delta v, you will find
that the density will eventually settle to
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a constant value. Now this is precisely because
of the fact that fluctuations in n about the
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mean, about the average, they are proportional
to 1 over root n. This is a fundamental result
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and statistics and statistical mechanics.
This goes by the name of central limit theorem.
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So, if the fluctuations in the number of molecules
in a volume, if they decrease as the volume
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becomes larger and larger and it decreases
in the following way, decreases 1 over square
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root of n. So, if n is of the order of 10
to the 8, then the fluctuations will be of
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the order of 10 to the minus 4.
So, it means that it is so small. This is
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the fluctuations in the average relative to
the average. So, divided by the average itself.
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So, if n is 10 to the 8, and let us define
the fluctuations with delta n, then delta
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n by n will be proportional to the 10 to the
minus 4 or delta n itself will be of the order
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of 10 to the plus 4 because n is 10 to the
8, example.
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So, fluctuations are small when the size of
the volume becomes larger and larger, but
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can we take the volume to be arbitrarily large.
The answer is no because eventually, we remember
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the context was we to a pipe in the last slide,
and we took a small point, construct a tiny
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volume about that point, and we are interested
in obtaining a point wise value for the density.
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Now clearly and trivially speaking, if this
is the diameter of the pipe, your delta v
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cannot be proportional to or comparable to
d cube because in which case, then there is
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no point. We cannot speak sensibly of a point
wise variation of density. So, clearly there
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are restrictions as to how long, how large
the volumes can be and how small the volumes
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can be.
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So, the continuum approximation works when
we say… So, let us say validity of continuum
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approximation. It works well when there is
a, we say separation of length scales. So,
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if you are if you take a point and construct
a cube of size delta l, the volume is proportional
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to delta l cube.
Then, continuum approximation where in you
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can propose point wise variations of, you
can define each physical properties at a point
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wise is valid, when delta l which is the side
of this cube which makes the volume is large
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compared to the molecular dimension. This
is the molecular dimension, molecular diameter.
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This is the side of the cube which forms the
volume; the test volume.
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And this should again be further; this delta
l must be small compared to macroscopic lengths.
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For example, the pipe diameter in our illustration.
So, this is when the continuum approximation
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is valid.
Now, this is not true only for density, even
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if you were to consider… so, we did this
very briefly in the last lecture, let me remind
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you of that example. Even if you have to consider
pressure, the pressure ultimately if you consider
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a flat surface, pressure is the normal force
exerted by the fluid on a surface per area
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of the surface divided by the area of the
surface.
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So, if you consider pressure on a solid surface;
this is a solid wall of area a, then why is
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there pressure suppose this solid wall is
exposed to air, the pressure is that because
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of collisions of molecules that are present
in the air, the gas with respect to the wall
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and the collisions means there is rate of
change of momentum that whenever there is
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a collision of a molecule with a solid surface;
that is, rate of change of momentum, that
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rate of change of momentum will manifest as
a force on the surface and that force divided
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00:23:40,860 --> 00:23:47,860
by area a is the pressure.
Now again the continuum approximation means
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00:23:48,190 --> 00:23:55,190
that first of all, a continuum approximation
works because even if you consider a area
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which is made of a square, with a side 10
to the minus 6 meter; that is, one micron,
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it is a very small; this is one micro meter.
Even if you have to consider a tiny area,
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so, suppose you have to use a pressure transducer
or pressure sensor, it will have a tiny area
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associated with it.
Even if you were to consider a tiny area which
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you would think so small that you can think
of it as a point, within that area, some from
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elementary kinetic theory, one learns in physical
chemistry classes, one can show that the number
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of collisions; the average number of molecular
collisions is proportional to is of the order
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of rather can be estimated to be about ten
to the seven collisions per second.
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00:24:47,720 --> 00:24:53,929
So, even in a given second, there are so many
large, such a large number of collisions that
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the surface of the sensor which is trying
to measure the pressure will sense only an
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average of all these collisions and it cannot
resolve individual collisions. So, when you
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00:25:09,049 --> 00:25:16,049
define pressure as force by this delta area
in the limit as delta a goes to 0, so let
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00:25:21,309 --> 00:25:28,309
me grace here. What does means is that as
you decrease delta a from this area to a smaller
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00:25:30,429 --> 00:25:35,070
area, you will still find the value of the
pressure to be uniquely defined.
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00:25:35,070 --> 00:25:40,850
So, even if you change the probe dimension,
as long as you are not encroaching on the
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00:25:40,850 --> 00:25:45,830
molecular scales, you will still find that
the pressure is uniquely defined and it is
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independent of the test area.
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00:25:47,760 --> 00:25:53,850
This is the same with density as well. In
the previous example, we to different volumes
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00:25:53,850 --> 00:26:00,409
and then we plotted; if you remember what
we plotted was that we to different test volumes
220
00:26:00,409 --> 00:26:04,789
and we tried to plot the density that one
would compute in this test volume as the function
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00:26:04,789 --> 00:26:09,090
of the size of the volume.
Now for molecular, when the volume is proportional
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00:26:09,090 --> 00:26:16,090
to the molecular dimensions, then you found
that in this regime, the density is fluctuating,
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but when the volumes are such that they are
large compared to the molecular dimensions,
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00:26:23,330 --> 00:26:28,950
but small compared to macroscopic dimensions,
then you will find that the density is a unique
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00:26:28,950 --> 00:26:35,950
value; that is, density is mass present in
that tiny volume divided by the volume.
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00:26:40,850 --> 00:26:47,850
So, it is defined as limit delta v goes to
0, sorry delta v goes to 0, mass divided by
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00:26:49,390 --> 00:26:54,309
volume. So, as you shrink the volume such
that still large compared to the molecular
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00:26:54,309 --> 00:26:58,669
scales, the mass that is present in that volume
will also decrease, but the ratio of these
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00:26:58,669 --> 00:27:05,669
two will approach a unique value. That is
the continuum density. This is the density
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00:27:05,950 --> 00:27:12,950
as defined in the continuum approximation.
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00:27:14,000 --> 00:27:21,000
And the same goes for the pressure which we
discussed in the next. So, the pressure will
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00:27:23,020 --> 00:27:30,020
approach a unique value irrespective of the
area, if the area is large compared to the
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00:27:33,149 --> 00:27:38,390
molecular dimensions. So, this is the sum
and substance of the continuum hypothesis
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00:27:38,390 --> 00:27:40,269
or approximation.
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00:27:40,269 --> 00:27:47,090
Now once we define the continuum approximation,
we can express density as a function of as
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00:27:47,090 --> 00:27:53,690
a smooth function of x, y, z and time. So,
as an illustration, I am just showing that
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00:27:53,690 --> 00:27:59,309
density can be plotted as a function of x
with one spatial co-ordinate as a smooth function
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00:27:59,309 --> 00:28:06,309
and so is pressure as a function of let say
z and so on. So, this is how once you define
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00:28:10,130 --> 00:28:15,419
the continuum approximation, once we understand
what we mean by values of density and pressure
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00:28:15,419 --> 00:28:21,100
at each point, we can sensibly define these
quantities at each and every point in the
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00:28:21,100 --> 00:28:25,669
fluid.
Now, the discussion on the meaning of continuum
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00:28:25,669 --> 00:28:32,669
hypothesis also tells us when continuum hypothesis
can fail or continuum approximation
can fail. Now the molecular length scale in
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00:28:42,909 --> 00:28:49,909
a liquid, the molecular length scale is simply
the diameter of the molecules that are present
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00:28:54,640 --> 00:29:00,140
in the liquid because if you remember from
your elementary physics or chemistry classes,
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00:29:00,140 --> 00:29:07,140
a liquid is very dense. So, all the molecules
are; so, this is diameter a, are fairly close
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00:29:07,440 --> 00:29:11,230
by with very less, very little gap between
them.
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00:29:11,230 --> 00:29:15,929
So, the molecular dense scale is that molecular
length scale in a liquid is the diameter of
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00:29:15,929 --> 00:29:22,929
the molecules, but if you consider a gas,
the molecular length scale is not really the
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00:29:26,260 --> 00:29:30,830
diameter of the molecule because a gas if
you remember is very dilute.
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00:29:30,830 --> 00:29:35,330
So, the molecules are far well separated.
Even though the diameter of the molecule is
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00:29:35,330 --> 00:29:40,110
still a, the relevant length scale here in
a gas is not the dimension though molecular
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00:29:40,110 --> 00:29:46,500
itself, but its called the mean free path.
So, let us denote it as l m. So, this is the
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00:29:46,500 --> 00:29:53,500
average distance between two molecules before
they collide. This is called the mean free
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00:30:02,330 --> 00:30:07,200
path.
So, this is the average distance a molecule
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00:30:07,200 --> 00:30:12,250
travels before it collides with some other
molecule. So, this becomes is denoted as l
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00:30:12,250 --> 00:30:19,250
m, this becomes the relevant molecular scale.
Now, this mean free path can become substantially
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00:30:21,730 --> 00:30:28,730
large for what are called rarified gases.
Rarified gases are gases with very low densities.
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00:30:33,750 --> 00:30:39,850
Rho is very very small.
So, this can be achieved if you know what
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00:30:39,850 --> 00:30:46,850
the ideal gas law is; for example, p v is
n r t, where n is the number of moles. So,
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00:30:47,590 --> 00:30:54,590
this is the mass divided by the molecular
mass r t, and I bring the volume in the denominator,
261
00:30:55,630 --> 00:31:02,630
mass by volume is the density. So, p is rho
by m r t, where r is the gas constant and
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00:31:03,159 --> 00:31:07,210
t is the temperature.
So, when can you get low densities? When the
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00:31:07,210 --> 00:31:12,279
pressures are low; that means, the density
is true because pressure is directly proportional
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00:31:12,279 --> 00:31:13,860
to density in a gas.
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00:31:13,860 --> 00:31:20,070
So, densities; these are not… These are
encountered in some applications when the
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00:31:20,070 --> 00:31:27,070
pressure is so low, the density becomes very
small, then the mean free path becomes large.
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00:31:29,779 --> 00:31:34,840
If the mean free path becomes very large,
by very large we mean that, the mean free
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00:31:34,840 --> 00:31:41,840
path; if mean free path becomes comparable
to the macroscopic dimensions.
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00:31:46,279 --> 00:31:53,279
For example, you have flow of a rarified gas
in a channel with diameter or length h, and
270
00:31:53,880 --> 00:32:00,880
let us say the mean free path is of the order
of h, then you will find that continuum approximation
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00:32:01,490 --> 00:32:08,490
fails or it becomes questionable
because you can no longer define sensibly
point wise quantities in this case. So, in
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00:32:20,940 --> 00:32:27,940
rarified gases, one has to be careful we should
I mean this is a warning. So, beware of using
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00:32:35,139 --> 00:32:42,139
the continuum approximation.
Another context is in applications, there
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00:32:44,279 --> 00:32:51,279
are in recent years, there are what are called
micro-fluidic devices which are used in several
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00:32:53,309 --> 00:33:00,309
biological and bio-technological applications.
These are made of channels whose the gap in
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00:33:04,450 --> 00:33:11,450
which fluid is flowing, is itself of the order
of let us say 100 microns or 10 microns.
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00:33:12,559 --> 00:33:18,460
If the fluid which is flowing is comprised
of molecules which are not small, which are
278
00:33:18,460 --> 00:33:25,460
if the fluid itself has large molecules like
polymeric molecules, then the molecular dimension,
279
00:33:32,190 --> 00:33:39,190
then the molecular dimension becomes comparable
to the channel dimensions. Channel width.
280
00:33:43,399 --> 00:33:50,399
So again, continuum approximation becomes
questionable; the use of continuum approximation.
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00:33:52,389 --> 00:33:59,389
So, there are cases where the length scales;
the molecular length scales becomes comparable
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00:34:00,649 --> 00:34:04,909
to the macroscopic scales as in rarified flows
or micro fluidic flows.
283
00:34:04,909 --> 00:34:10,450
In such cases, it is not obvious that whether
one can successfully use continuum approximation,
284
00:34:10,450 --> 00:34:15,810
and there are always other more fundamental
approaches where one takes a full molecular
285
00:34:15,810 --> 00:34:22,100
approach, full molecular detail, but those
are; obviously, much more involved. And they
286
00:34:22,100 --> 00:34:26,790
are specific to the kind of problems that
one addresses.
287
00:34:26,790 --> 00:34:33,790
But in general, continuum approximation can
be safely used in many engineering applications
288
00:34:34,140 --> 00:34:40,910
and almost all engineering applications, conventional
engineering applications where the macroscopic
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00:34:40,910 --> 00:34:47,520
scale is very large compared to the pipe diameter
or the tank conveys of fluid is stored or
290
00:34:47,520 --> 00:34:52,650
if you have flow passed sphere, the diameter
of the sphere or which the fluid is flowing.
291
00:34:52,650 --> 00:34:58,170
If those macroscopic dimensions are very very
large compared to molecular dimensions, which
292
00:34:58,170 --> 00:35:05,170
is normally a valid in engineering circumstances,
then one can safely apply the continuum approximation.
293
00:35:05,570 --> 00:35:12,570
So, this completes my discussion of what the
frame work that we are going to use. So, once
294
00:35:13,960 --> 00:35:18,500
you use that frame work; the continuum frame
work, we will define velocity as a function
295
00:35:18,500 --> 00:35:25,500
of all the three co-ordinates in the fluid
in the co-ordinate system and time, pressure
296
00:35:27,060 --> 00:35:33,690
as a function of all the three co-ordinates
and time, and density as a function of smooth
297
00:35:33,690 --> 00:35:39,600
functions of all the three co-ordinates and
time. These are called fields; continuum fields.
298
00:35:39,600 --> 00:35:46,600
So, this goes by name of fields. So, and we
assume that they all can be differentiated
299
00:35:50,710 --> 00:35:57,710
easily. They are differentiable, smooth and
continuous.
300
00:36:03,450 --> 00:36:09,790
Now that we have understood the limitations
and validity of continuum approximation, it
301
00:36:09,790 --> 00:36:16,790
is a right time to define what a fluid is.
So, in order to understand what a fluid is,
302
00:36:21,770 --> 00:36:28,620
it is useful to contrast the mechanical behavior
of fluid with that of solid. So, let us begin
303
00:36:28,620 --> 00:36:35,620
with so that we can compare and contrast the
true behavior how a fluid responds under applied
304
00:36:36,290 --> 00:36:39,480
forces, and how a solid responds in the applied
forces.
305
00:36:39,480 --> 00:36:46,480
So, let me say, fluid verses solid. So, let
us first consider the case of solid which
306
00:36:50,130 --> 00:36:57,130
is elastic in nature. For example, a piece
of rubber; a piece of rubber that one uses
307
00:36:58,580 --> 00:37:05,580
to erase. So, example is eraser.
So, imagine let us take a block of solid.
308
00:37:07,540 --> 00:37:13,370
So, this is a block of solid. This is a piece
of rubber that we want to use in our thought
309
00:37:13,370 --> 00:37:20,370
experiment. So, this dimension, let us say
w is very large compared to this dimension
310
00:37:22,600 --> 00:37:29,600
which is h, w is very large, it is about let
us say twenty or fifty times h. So, we need
311
00:37:29,930 --> 00:37:35,950
not worry about variations in the third direction.
So, let us just consider the solid to be in
312
00:37:35,950 --> 00:37:42,640
the plane. So, let just consider the piece
of solid here.
313
00:37:42,640 --> 00:37:47,830
And of course, I am putting a co-ordinate
system as I surely not analyze any problem;
314
00:37:47,830 --> 00:37:54,830
we have to first set up a co-ordinate system.
So, this means, the z is the co-ordinate that
315
00:37:55,920 --> 00:38:02,920
is coming out of the board. So, variations
in z co-ordinate are not important. So, we
316
00:38:12,010 --> 00:38:19,010
are considering a solid. Therefore, and we
are taking a cross section at any plane perpendicular
317
00:38:19,680 --> 00:38:23,670
to the z direction. So, it will be in the
x y plane.
318
00:38:23,670 --> 00:38:30,670
Now imagine that this piece of rubber is placed
between two plates. The top plate, this is
319
00:38:32,920 --> 00:38:39,920
the bottom plate, and let us says the bottom
plate is stationary. This is a solid and the
320
00:38:44,320 --> 00:38:51,320
top plate, you are applying an external force
and this force is being applied in the x direction.
321
00:38:55,610 --> 00:39:01,180
This is the x co-ordinate. The force is being
applied in the x direction. Now in mechanics,
322
00:39:01,180 --> 00:39:06,520
in both fluid and solid mechanics, if you
have a surface, remember I am just drawing
323
00:39:06,520 --> 00:39:13,520
here the cross section at any plane that is
perpendicular to z. This force is actually
324
00:39:13,530 --> 00:39:20,530
applied on a surface, on the top surface of
this piece of solid, and let if the area of
325
00:39:26,430 --> 00:39:33,430
the top surface is a, then this f x by a is
called the stress.
326
00:39:34,710 --> 00:39:41,710
Now, one normally distinguishes between two
types of stresses. This is called the shear
327
00:39:42,410 --> 00:39:49,410
or tangential stress because this is tangential
to the surface on which the force is acting.
328
00:39:53,590 --> 00:40:00,590
So, if you have a surface, this is the surface.
The surface has unit normal let us say j,
329
00:40:02,620 --> 00:40:09,620
j is the normal in the… So, if you have
a co-ordinate system, i j and k are the three
330
00:40:09,930 --> 00:40:15,520
unit normals in the x y and z directions.
So, unit normal of this surface is j.
331
00:40:15,520 --> 00:40:22,520
But the direction of the force is parallel
to the surface and not in the direction of
332
00:40:23,210 --> 00:40:27,300
the normal. It is perpendicular to the normal,
its parallel to the surface. So, these are
333
00:40:27,300 --> 00:40:34,300
called shear stresses. So, we are applying
a shear stress; a force that is applied parallel
334
00:40:35,740 --> 00:40:41,320
to the surface per unit area.
Now if this is in contrast to, suppose if
335
00:40:41,320 --> 00:40:48,320
I have a surface again in of course, unit
normal is in the j direction, and if I apply
336
00:40:48,460 --> 00:40:55,460
a force like here, this is a compressive force.
This is called a normal stress because the
337
00:40:59,470 --> 00:41:03,160
force is applied along the normal. So, here
it is applied in the direction opposite to
338
00:41:03,160 --> 00:41:08,350
the normal. So, whether the magnitude; the
direction of the force can be either in the
339
00:41:08,350 --> 00:41:13,570
plus j or minus j direction; that that is
immaterial, but essentially it is a force
340
00:41:13,570 --> 00:41:19,430
in the direction of the normal. So, it is
a normal stress whereas, a tangential if the
341
00:41:19,430 --> 00:41:20,930
force applied tangential. It is called shear
stress.
342
00:41:20,930 --> 00:41:25,600
So, this is the experiment we are doing. We
are taking a piece of rubber and then we are
343
00:41:25,600 --> 00:41:29,420
placing it between two plates. The bottom
plate is stationary, top plate you are applying
344
00:41:29,420 --> 00:41:33,860
a tangential force and tangential force per
unit area is a stress.
345
00:41:33,860 --> 00:41:39,180
Now, if you have to do this experiment, what
is going to happen? That is the thing, that
346
00:41:39,180 --> 00:41:40,700
is the question we are going to ask.
347
00:41:40,700 --> 00:41:47,700
So, you have these two plates, and you have
a material. The thickness of this material
348
00:41:52,840 --> 00:41:58,430
is h and as you know, this is the x y plane.
We are not worrying about the variation in
349
00:41:58,430 --> 00:42:05,340
the z direction.
So, you have let say before applying stress
350
00:42:05,340 --> 00:42:12,340
any stress, the solid material is in the stress
free state. What we do is that we take this
351
00:42:23,000 --> 00:42:29,330
piece of material which is not being acted
upon the any stress; it is in a stress free
352
00:42:29,330 --> 00:42:35,880
state. So, let us apply, let us mark various
points with the colored with the color dye
353
00:42:35,880 --> 00:42:39,790
or whatever.
Let us imagine that we can mark points on
354
00:42:39,790 --> 00:42:46,790
a cross section of the solid with some colored
material and imagine what will happen to this,
355
00:42:48,120 --> 00:42:55,120
the same system, when you now start applying
stress. So, this is the location of the points
356
00:43:06,280 --> 00:43:13,280
in the undeformed state or the stress free
state; undeformed location of points
357
00:43:14,710 --> 00:43:21,710
Now, you are going to apply a stress in this
direction. Now there is no stress that is
358
00:43:24,960 --> 00:43:29,800
being acted in the bottom direction, bottom
plates sorry, and some stress is being applied
359
00:43:29,800 --> 00:43:34,560
in the x direction in the top plate f x. So,
what will happen to these points?
360
00:43:34,560 --> 00:43:41,560
In a solid, a solid responds to apply forces
by undergoing some deformations. So, all these
361
00:43:43,390 --> 00:43:50,390
points will move. By deformation we mean that,
these points will start moving from the undeformed
362
00:43:52,090 --> 00:43:59,090
locations.
So, the green points refer to deformed location.
363
00:43:59,710 --> 00:44:06,710
We say that the solid undergo, solid responds
to applied stresses by undergoing deformation.
364
00:44:09,430 --> 00:44:14,180
By deformation we mean, this various points
that were there in the solid before the application
365
00:44:14,180 --> 00:44:19,420
of stresses will start moving to some other
new point, will move to some other new point.
366
00:44:19,420 --> 00:44:25,410
But they will stop moving because, after you
start applying stress, once the solid starts
367
00:44:25,410 --> 00:44:29,980
deforming, internal stress is developed in
the solid which will resist further deformation.
368
00:44:29,980 --> 00:44:35,570
Solid will not deform continuously. The points
will deform to some extent and this stop,
369
00:44:35,570 --> 00:44:40,600
and this is a very quick process in a solid.
So, it deforms and then it stops deforming,
370
00:44:40,600 --> 00:44:47,600
and the deformed location of points is given
by the green points. Now, so let me do this
371
00:44:52,230 --> 00:44:59,230
in a more quantitative way. So, you have the
undeformed thing and the deformed thing. So,
372
00:44:59,530 --> 00:45:05,270
this angle, since there is no force in the
bottom plate, no force, so, this point will
373
00:45:05,270 --> 00:45:12,270
remain here itself whereas, this point would
have moved because you are applying a force
374
00:45:12,330 --> 00:45:16,720
here, a tangential force here.
Now, this deformation can be characterized
375
00:45:16,720 --> 00:45:23,720
by an angle delta alpha because if you apply
a very small force, then that means that the
376
00:45:25,520 --> 00:45:32,520
amount of deflection of this blue line; of
this let us draw it with a green line.
377
00:45:33,840 --> 00:45:39,160
The amount of deflection that you will experience
is very small. So, this delta alpha is a measure
378
00:45:39,160 --> 00:45:46,160
of what is called strain in the solid because
a solid undergoes some deformation upon in
379
00:45:48,700 --> 00:45:53,370
response to the applied stresses, but it does
not continue to deform, it merely stops deforming
380
00:45:53,370 --> 00:45:58,320
because of the internal stresses.
So, in a solid, if you do experiments, suppose
381
00:45:58,320 --> 00:46:04,650
you have to do an experiment. Let us say you
are applying f x 1; a force f x 1 and you
382
00:46:04,650 --> 00:46:10,060
divide by the area or is process force is
applied, and then you will find that if you
383
00:46:10,060 --> 00:46:15,690
have to do this experiments…, so, let us
let us write this in the left side, in the
384
00:46:15,690 --> 00:46:18,700
right side.
So, suppose you have to apply a force f x
385
00:46:18,700 --> 00:46:25,610
1, and you are applying it over an area, and
if you measure that deflection delta alpha,
386
00:46:25,610 --> 00:46:32,180
if you apply a value of f x 1, let us say
you find a deflection delta alpha 1, and f
387
00:46:32,180 --> 00:46:38,480
x 2 and you will find some other deflection
delta alpha 2, and if you were to plot all
388
00:46:38,480 --> 00:46:45,480
these, you will find that f x by a is proportional
to delta alpha. You will get a straight line.
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00:46:49,310 --> 00:46:56,310
When there is no force, there is no delta
alpha. So, it starts with 0 and it is a linear.
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00:46:56,720 --> 00:47:03,060
It is a straight line, it is directly proportional.
So, you will find for elastic solids, if you
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00:47:03,060 --> 00:47:10,060
have to do this experiments for purely elastic
solids, this f x by a is called the stress,
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00:47:10,830 --> 00:47:17,830
and stress is denoted by the greek letter
sigma and you will have x being the direction
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00:47:19,790 --> 00:47:26,790
of the force which is one subscript and y
is the direction of the normal over which
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00:47:29,930 --> 00:47:36,930
the force is acting.
Remember that we are having the surface whose
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00:47:38,630 --> 00:47:43,860
unit normal is in the j direction, and the
force is acting in the i direction which is
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00:47:43,860 --> 00:47:48,580
x. That is why there are two indices; one
is a direction of the force, this is called
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00:47:48,580 --> 00:47:52,740
a shear stress.
This is a shear stress because the force is
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00:47:52,740 --> 00:47:57,330
acting tangentially to the surface. It is
not acting in the normal direction of the
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00:47:57,330 --> 00:48:04,330
surface. So, shear stress force per unit area.
You will find that this is proportional to
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00:48:04,820 --> 00:48:10,300
since I am using alpha, so, I do not want
to use… So, let me use… This is the proportionality
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00:48:10,300 --> 00:48:17,300
sign. It is proportional to delta alpha.
So, this is what one will find, if you are
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00:48:17,850 --> 00:48:23,810
to do this experiment for a solid. The stress
is directly proportional to strain in a solid.
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00:48:23,810 --> 00:48:30,810
So, for an elastic solid, stress is proportional
to strain. So, let us draw this picture again.
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00:48:38,780 --> 00:48:45,780
So, you have these two plates and you have
a piece of solid; rubber let us say, and then
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00:48:47,190 --> 00:48:54,190
you found that before deformation, things
were in the red line. After deformation, things
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00:48:54,900 --> 00:49:01,900
were in the green line. And this angle is
delta alpha. Now this distance is h. Let this
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00:49:04,310 --> 00:49:10,720
displacement be delta l.
So, this is the force f x that is acting on
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00:49:10,720 --> 00:49:17,720
a surface whose unit normal is in the y direction.
Now by geometry, tan delta alpha is delta
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00:49:22,330 --> 00:49:29,330
l by h, but when delta alpha is very very
small, for small forces, you will expect that
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00:49:31,310 --> 00:49:38,310
the deflection delta alpha and displacement
l is small is very small, then tan delta alpha
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00:49:42,140 --> 00:49:49,140
is roughly equal to delta alpha.
Remember that tan delta alpha is sin delta
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00:49:50,420 --> 00:49:57,420
alpha by cos delta alpha, and as delta alpha
becomes small or tends to 0, cos delta alpha
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00:49:59,400 --> 00:50:06,400
tends to 1 and sin delta alpha is proportional
to delta alpha; it goes as delta alpha.
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00:50:07,940 --> 00:50:14,940
Therefore, tan delta alpha will go as delta
alpha. So, this equation means that delta
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00:50:16,790 --> 00:50:23,790
alpha is approximately delta l by h. So, sigma
x y is proportional to delta l by h. This
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00:50:26,470 --> 00:50:33,470
is called the strain and the constant of proportionality
is called the modules of elasticity. This
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00:50:40,860 --> 00:50:46,810
is for an elastic solid.
This is an equivalent of Hooks law of elasticity.
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00:50:46,810 --> 00:50:53,610
It is essentially some statement of Hooks
law of elasticity. For elastic solid, the
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00:50:53,610 --> 00:50:58,440
stress is directly proportional to strain
and the strain is measured by this deflection
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00:50:58,440 --> 00:51:05,440
delta alpha and the constant of proportionality
is called the elastic modules or modules of
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00:51:05,590 --> 00:51:12,590
elasticity. Now stress has units of pascals,
because its force per unit area. This is dimensionless.
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00:51:14,930 --> 00:51:19,080
This is unit less or dimensionless because
this is ratio of two lengths. This has no
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00:51:19,080 --> 00:51:26,080
unit, it is a pure number. So, g will have
units of pascals.
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00:51:28,100 --> 00:51:34,630
Now this is what a solid is. A solid responds
by undergoing some deformation or strain impressed
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00:51:34,630 --> 00:51:39,310
upon application of some shear stress. And
the stress is directly proportional to the
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00:51:39,310 --> 00:51:44,010
strain in solid.
In the next lecture, what we will do is we
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00:51:44,010 --> 00:51:49,500
will contrast the mechanical behavior of fluid
upon application of external stresses, and
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00:51:49,500 --> 00:51:56,500
see how it is different from a solid. So,
we will see you soon in the next lecture.