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Welcome to this lecture number nine, of our
course on fundamentals of transport processes.
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so far we have been discussing, initially
we discussed the vectors, and tensors, and
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their integral and differential theorems,
and then we looked at some results that are
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obtained, when these vectors operate on the
velocity field, a fluid velocity field. In
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the last class we had discussed the, rate
of deformation tensor. It is a rather important
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subject so we will briefly review it again,
before we proceed. So this is the rate of
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deformation tensor. Second order tensor, it
is the gradient of the velocity, it is also
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written as partially u i by partially x j.
And if I want to write it in expanded form,
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then I would write it as; sigma i is equal
to 1 to 3, sigma j is equal to 1 to 3, of
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partial u i by partial x j e i e j. So at
a given point in the fluid, if I mark a location
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x and I go a small distance delta x, to some
new location.
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You have a velocity at x, and this is the
velocity at x plus delta x. The difference
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in velocity delta u, is equal to
grad u, you can there have a delta x dotting
that. So basically the displacement dotting
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with grad u. The dot product is, between the
displacement vector, and the gradient, in
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the rate of deformation tensor. So this basically
gives you the rate at which, positions are
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moving related to each other, as the fluid
flows. It is a second order tensor, it has
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nine components; however as we discussed in
the last class, this can be decomposed into
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fundamental modes of deformation. So the gradient,
which is the second order tensor, can always
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be written as the sum of two parts; one is
the symmetric part, and the other is the anti
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symmetric part, where the symmetric part is
equal to half of the gradient of the velocity,
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plus x transpose.
The transpose is obtained just by interchanging
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rows and columns, or by interchanging the
two indices of the tensor, because one index
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represents the row, the other index represents
the column. The anti symmetric part, is equal
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to half of the matrix minus, its transpose,
half the matrix minus its transpose. This
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symmetric part can, once again be written
as the sum of two tensors; one is the symmetric
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traceless tensor, and the other is an isotropic
tensor. The isotropic part, is just equal
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is just propositional to the identity tensor.
It has equal diagonal elements, and zero half
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diagonal elements. The symmetric traceless
tensor, is symmetric, and the sum of its diagonal
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elements is equal to zero. So this can be
written as, this symmetric traceless part
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plus one-third delta i j times partial. In
the previous lecture, I have written this
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as the divergence of u.
Partial u k b partial x k is also the divergence,
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because there is a repeated index, so there
is a dot product. So this thing is a symmetric
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traceless tensor, what that means, is that
the sum of the diagonal part, diagonal elements
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of this tensor, is equal to zero. And I showed
you in the last lecture that the sum of the
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diagonal elements, of this tensor, is just
equal to, is e i i, because if I have e i
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i and I expand it out using my indicial notation,
there is one repeated index; that means there
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is a dot product. So there is one summation,
and no unit vector, so this is just equal
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to, summation i is equal to 1 to 3 of e i
i, which is basically the sum of the diagonal
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elements of separate matrix.
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So this rate of definition tensor can be separated
into three parts, is equal to the anti symmetric
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part plus the isotropic part which was
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plus e i j. The three parts, and I had shown
you, that there are specific forms of definition,
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associated with each of these three components.
This tensor I have separated into anti symmetric
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isotropic, and symmetric traceless, and if
you look at the local deformation has point
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to each individual part, we find for that
anti symmetric part. The deformation is basically
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a solid body rotation; that is if I am sitting
at some point, within the fluid at that center,
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and I look at the velocity with which nearby
points are moving relative to the point at
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which I am sitting. You find that the nearby
points are moving, in a rotational form; that
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is, they are there is a solid body rotation,
around this central point.
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So that is what, is captured by the anti symmetric
part, to the rate of deformation tensor. And
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we saw in the last class that, the rate of
deformation tensor, is related to the verticity.
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So the verticity, which is the curl of the
velocity, partial u k by partial x j; that
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is the verticity vector, can also be written
as epsilon i j k times a k j. So the curl
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of the velocity vector, is also equal to epsilon,
double dotted with the anti symmetric part
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of the rate of deformation tensor. So if I
take epsilon, and double dot it with a, anti
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symmetric part of the rate of deformation
tensor, i get a vector, whose direction is
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perpendicular to the plane of rotation, so
that is what you get from the anti symmetric
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part of the rate of deformation tensor. In
fact you can actually, in fact there is a
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half sitting in front, please note that.
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Now, that is because the anti symmetric part
has two partial u i by partial x j minus partial
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u j by partial x i, and once, let me just
go back and check. You can see that there
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is. I should make it correction here; half
of partial u k by partial x j minus partial
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u j by partial x k, this itself is the anti
symmetric part, so there should be no half
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in front here, because half of the difference
between these two, is just the anti symmetric
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part. So please correct that in the previous
lecture.
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And I had written a half here, so that half
should not be there. So that is the anti symmetric
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part of the rate of deformation tensor. The
second part was the symmetric traceless part,
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and that. I am sorry. The isotropic par, tand
that as I showed you corresponds to readily
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outward, or inward flow. If the divergence
of the velocity is positive; that means that
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the flow is readily outward. If the divergence
of velocity is negative, the flow is readily
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inward. And this divergence of velocity is
non zero, only if you have a source of fluid,
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as I said the volume has to increase for the
divergence to be non zero, because the divergence
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of velocity integrated over a volume, is equal
to integral of u dot n, over the surface.
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So that is non zero that means there is net
fluid, coming out of that surface, so there
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has to be a source, or the density has to
decrease. So unless, if the density is remaining
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a constant, and there is no source of fluid,
then this part will be identically equal to
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zero. And the third part, was pure extensional
string. We had seen in the last class that
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this, resulted in a deformation in which you
have expansion, along two axes, and compression
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along the perpendicular two axes. This is
expansion along two axes, compression along
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the perpendicular two axes.
So there is one extensional direction, and
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one compressional direction in two dimensions.
So that if I have some differential volume
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here, and if I have look at what this extensional
motion, how this that will deform, this differential
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volume. After some time I will have the volume
looking something like this. It is going to
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deform in such a way that it extends, along
one axis, compresses along the other axis,
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in such a way that the area, is preserved
in two dimensions or the volume, is persevered
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in three dimensions. And these two extensional
compression axes, also do not rotate, because
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rotational motion is associated with the anti
symmetric part. So the symmetric part has
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no rotational motion, so the axis remain the
same.
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There is no increase in volume, because the
trace was zero, because of that you cannot
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have any increase in volume. So deformation
is such a way that volume is persevered, as
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well as there is no rotation. So that is what
is represented by the symmetric traceless
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part of the rate of deformation tensor. This
is simplified picture, in two dimensions,
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if you go to three dimensions, you can have
you have a symmetric traceless tensor, which
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is a 3 by 3 matrix. Similarly, the anti symmetric
part is also a 3 by 3 matrix. However the
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anti symmetric part, this still holds. Note
that what I have written here is for three
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dimensions, this one third. In the last class
I showed you that for two dimensions, is equal
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to 1 by 2. So in general is equal to 1 by
d, where d is the dimensionality of the system.
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So in three dimensions these deformations
can be of different forms, so e i j is a symmetric
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traceless tensor, so it is symmetric e 1 1,
e 1 2, e 1 3, e 1 2. So it is a symmetric
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tensor, and the sum of the diagonals are all
zero; e 1 1 plus e 2 2 plus e 3 3 is equal
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to zero. Symmetric tensor, it is self ad joint,
these transpose is equal to itself, means
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it is a self ad joint matrix; that means that
it has eigen values, are real, and eigen vectors
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are orthogonal to each other, and the eigen
vectors form a complete basis at origin.
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So the eigen values of this tensor, basically
represents the rate at which, there is stretching
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or compression, along the principle directions.
The eigen vectors represent, the three principle
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directions, along which there is deformation.
So for example, in a flow field. You know
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that the eigen vectors of the symmetric matrix
at each and every point in space are all orthogonal
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to each other; that means that if I have one
particular point in space. I have three, in
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the most general case I will have three perpendicular
directions, represented by the eigen vectors
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at this point in space. So there are three
perpendicular directions, at one point in
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space, so there are two in the plane and one
that is coming out of the plane, and along
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each of these three perpendicular directions,
I have one eigen value, along each of these
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three perpendicular directions I have one
eigen value, which basically it gives you
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the rate at which there is extension or compression,
along that particular direction. Note that
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this is a traceless matrix; that means that
the sum of the diagonal elements is equal
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to zero. And we know from linear algebra,
that the sum of the diagonal elements, is
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also equal to, the sum of the eigen values.
Therefore, the sum of the three eigen values
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has to be equal to zero. There is the reflection
of the fact that volume is preserved, so either
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you have to have stretching in one direction
and compression in the other directions, in
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such a way that volumes do not change. Therefore,
you can have either two positive eigen values,
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and negative. You could have two negative
and one positive, or you could have, one zero,
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one positive, and one negative.
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If I have one positive eigen value, and two
negative eigen values; that means that at
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this point, along the three principle directions,
I am having extension in one direction, and
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compression in the other two directions, because
there are two negative eigen values, and two,
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and one positive eigen values. So, basically
flow was coming in, from two directions I
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was getting, is going out along the third
direction. So this is called uni-axial extension,
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there is extension along one axis, and there
is compression along two other axis. If two
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eigen values are negative, and one is positive.
The other option is for two eigen values to
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be positive, and one to be negative.
So in that case you have, stretching along
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two directions, and compression along the
third direction, has stretching along two
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directions, along the positive eigen values,
and compression along the third direction
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in such a way that the divergences the velocity
is equal to zero. So in that case, you will
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have, flow that looks like this, it will come
in along one direction, and it will flow out
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along two directions. The only other option
is to have, one zero, one positive, and one
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negative. If all three are zero, of course
there is no deformation, so you can have,
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so that is a trivial case.
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But you could have one positive, one negative,
and one zero. In that case, you have
deformation that is coming in, along one direction,
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it is going out, along one direction, and
there is no deformation in the third direction.
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Therefore, you just have fluid that is in
the plane, the same figure that I showed you
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in the previous lecture this one, deformation
in the plane; that is when the third eigen
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value is equal to zero, and the other two
have to be opposite in sign, and equal in
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magnitude. So you just have fluid that comes
in this way, deformation in the plane. So
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this is called uni-axial extension, extension
along one axis, this is by-axial extension,
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and this is what is called planar. So these
are different components of the rate of deformation
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tensor, and as we shown c a little later.
Rotation, solid body rotation, does not change
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the distance between nearby points in the
flow. So there cannot be a stress, due to
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the anti symmetric part of the rate of deformation
tensor. If the fluid is compressible, the
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isotropic part of the rate of deformation
tensor is identically equal to zero; so all
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of the stresses have to be only due to the
symmetric, traceless part of the rate of deformation
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tensor. One more topic in kinematics, before
we proceed to deriving equations of motion;
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and that is the substantial derivative.
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Now when you take the partial time derivative,
implicitly you are keeping the position the
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same, and finding out the change in property,
when between two subsequent instants of time.
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So the partial so for example, you have a
temperature field in this three dimensional
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system partial t with respect to temperature,
is equal to limit, as delta t equals to zero,
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of t at x 1 x 2 x 3 t plus delta t minus t
i divided by delta t. So that is the partial
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derivative, the difference in temperature
between two time instants, divided by the
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time interval in the limit as that interval
goes to zero. However if you have some kind
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of a fluid flow, if you have some kind of
a fluid flow, that is taking place. If I have
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some differential volume that is here, at
time t, after a small interval of time, it
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would have travelled to some other location.
It would travelled, and it would have deformed,
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so it would have travelled to some other location;
x plus delta x. That is because the mean fluid
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is carrying the properties, along with a,
the temperature is travelling along with the
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fluid itself, we have a hot parcel of fluid,
and there is a mean flow, that parcel is actually
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being convicted by that mean flow.
So in order to find out what is the temperature
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difference along the moving parcel of fluid.
This partial time derivative, is not the appropriate
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derivative, because this is in a fixed reference
frame. The fixed reference frame in is this
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what is called an eulerian, reference frame,
where as it is fixed in space. On the other
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hand, the reference that is moving, with the
mean fluid velocity, is what is called a lagrangian
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reference frame. So in the eulerian reference
frame, you would write the temperature as
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some function of position in time, absolute
position in the absolute time at each instant
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in time, is you just find out what is the
temperature, for the given position. In the
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lagrangean reference frame, you write it down,
as a function of the moving parcel of fluid,
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that is carrying its temperature along with
it.
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So in this case you write it as t as function
of, because the location of that parcel itself,
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is now a function of time. It has some one
value at this instant, one value at that instant,
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and it is moving in space, and therefore,
it is carrying the temperature or the concentration
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along with that, so that is the lagrangian
reference frame. Now in order to implement
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a lagrangian scheme, one has to know if a
parcel of fluid is at this location, at one
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instant of time ,where it was in the past,
what was the past location, of that parcel
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of fluid, which is at one particular instant
at this present location. That is usually
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a difficult task, because you have to trace
out the entire history of the fluid. However
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one can define a derivative, which is called
the substantial derivative, which contains
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in it the elements of the lagrangian moving
reference frame. And that is if a fluid element
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was at some location x at time t, after the
time t plus delta t it has moved to a new
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location.
So a difference in temperature, between the
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location at t plus delta t minus, at instant
t plus delta t minus the temperature at location
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x at time t, so that is the, how the substantial
derivative is defined. So, let us look at
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00:24:30,210 --> 00:24:37,210
that it is usually written as d t by d t is
equal to limit as delta t equals to 0 of t
191
00:24:43,309 --> 00:24:50,309
at x plus u delta t t plus delta t minus t
at x t. Note that I am taking the position
192
00:24:58,380 --> 00:25:05,380
x plus u delta t at time t plus delta t. Why
is that? Because if the fluid element was
193
00:25:06,230 --> 00:25:13,230
at location x at time t at time plus delta
t it has moved a small distance, time t plus
194
00:25:16,399 --> 00:25:23,399
delta t it has moved a small distance to some
new location x plus delta x, what is that
195
00:25:23,830 --> 00:25:29,909
new location x plus delta x? The new location,
the displacement, is going to be equal to,
196
00:25:29,909 --> 00:25:36,909
the velocity times, the time interval, the
distance moved is going to be, the displacement
197
00:25:37,200 --> 00:25:40,929
vector is going to be equal to the velocity
vector, times that time interval.
198
00:25:40,929 --> 00:25:47,309
So it is the difference in the temperature
between x plus u delta t and t plus delta
199
00:25:47,309 --> 00:25:53,799
t minus t at x t divided by delta t. So that
is the substantial derivative, a derivative
200
00:25:53,799 --> 00:26:00,799
in a reference frame that is moving with the
parcel of fluid. And you can easily evaluate
201
00:26:01,419 --> 00:26:07,409
this in terms of the partial derivatives by
just using the chain rule for differentiation,
202
00:26:07,409 --> 00:26:14,409
because x x 1 plus u 1 delta t at time t plus
delta t minus x 1 and t at time t. So I just
203
00:26:17,440 --> 00:26:22,049
evaluated by taking partial derivatives, and
what you will get is, this is partial t by
204
00:26:22,049 --> 00:26:29,049
partial t plus u 1 partial t by partial x
1 plus u 2, because the distance travelled
205
00:26:39,139 --> 00:26:44,720
was u 1 delta t. Therefore, the difference
in temperature is equal to u 1 delta t times
206
00:26:44,720 --> 00:26:49,000
partial t by partial x 1, and then i divide
by this delta t in the denominator, divide
207
00:26:49,000 --> 00:26:55,299
by this delta t in the denominator, and therefore,
i get u 1 times delta t partial t by partial
208
00:26:55,299 --> 00:27:00,669
x 1. So this is the substantial derivative,
which recognizes the fact that fluid elements
209
00:27:00,669 --> 00:27:05,679
are moving along with the mean velocity of
the fluid.
210
00:27:05,679 --> 00:27:12,679
I can also write this in vector notation,
as partial t by partial t plus u dot grad
211
00:27:21,820 --> 00:27:28,330
t partial t by partial t u dot grad t is u
1 partial t by partial x 1 plus u 2 partial
212
00:27:28,330 --> 00:27:34,090
t by partial x 2 plus u 3 partial t by partial
x 3. So this is the difference in temperature
213
00:27:34,090 --> 00:27:41,090
between the location x plus delta x at time
t plus delta t minus the temperature at location
214
00:27:42,909 --> 00:27:49,909
x at time t; that is this substantial derivative,
in a moving fluid, if for example, the temperature
215
00:27:51,580 --> 00:27:58,580
does not change, with time for a moving parcel
of fluid. In other words, if I have to neglect
216
00:28:01,000 --> 00:28:07,080
completely the diffusion, or the conduction
of temperature if the temperature is a constant,
217
00:28:07,080 --> 00:28:14,080
on a moving volume element of fluid, then
this derivative is 0. If the temperature is
218
00:28:14,919 --> 00:28:20,179
a constant in a moving element of fluid; this
is the substantial derivative at a zero, not
219
00:28:20,179 --> 00:28:26,529
the partial derivatives. That is because as
the fluid moves, this volume element that
220
00:28:26,529 --> 00:28:33,529
was at location x goes to x plus delta x,
some other fluid occupies the location x and
221
00:28:34,389 --> 00:28:41,389
there may be spatial variations between nearby
volume elements of fluid. If the temperature
222
00:28:41,779 --> 00:28:48,779
is not diffusing, then the temperature on
a moving element of fluid, is constant it
223
00:28:50,389 --> 00:28:54,389
may be specially varying.
So in that case it is a substantial derivative
224
00:28:54,389 --> 00:29:00,970
that will be zero, the partial derivative
in general will not be zero. So that is the
225
00:29:00,970 --> 00:29:06,279
substantial derivative, it is in a lagrangian
reference frame, and it recognizes the fact
226
00:29:06,279 --> 00:29:13,279
that, as the fluid moves, the fluid element
occupies a new location, in after an instant
227
00:29:13,590 --> 00:29:19,279
of time in comparison to the original location,
and you are taking the derivative on that
228
00:29:19,279 --> 00:29:26,279
moving volume element of fluid. So next we
go down to start deriving the conversion equations.
229
00:29:36,309 --> 00:29:41,830
In the previous lecture course on fundamentals
of transport processes one, we did this by
230
00:29:41,830 --> 00:29:46,899
actually writing out a differential volume
in different coordinate systems, looking at
231
00:29:46,899 --> 00:29:50,360
what comes in, and what goes out, and writing
the balance law
232
00:29:50,360 --> 00:29:57,360
The net change in the mass, the heat in a
differential volume, is equal to what comes
233
00:29:57,909 --> 00:30:04,909
in minus what goes out, plus source minus
sign, rather than do it that way, what we
234
00:30:06,070 --> 00:30:12,429
will do, is to look at the rate of change
of mass, on a moving differential volume,
235
00:30:12,429 --> 00:30:15,389
without reference to a specific coordinate
system.
236
00:30:15,389 --> 00:30:22,389
So for that, we have to do some back ground
work. If I have a volume d v with the surface
237
00:30:32,679 --> 00:30:39,679
t s, this volume d v with the surface d s,
on the surface. The net mass with in this
238
00:30:43,360 --> 00:30:50,360
volume, is going to be equal to integral over
the volume d v of the density within that
239
00:30:55,110 --> 00:31:02,110
volume, there is a net mass within the volume.
The mass conversion equation, basically states
240
00:31:02,659 --> 00:31:09,200
that, the mass within this volume has to be
conversed. The volume itself is moving as
241
00:31:09,200 --> 00:31:16,200
a function of time, so let us take this volume
d v, I will draw it bigger over here. This
242
00:31:23,690 --> 00:31:30,690
is the volume d v, with a surface d s. Now
this volume itself is a function of time,
243
00:31:36,659 --> 00:31:43,659
it is a fluid material volume, and has time
progresses, the fluid particles within this
244
00:31:48,460 --> 00:31:55,460
volume will move, and those on the surface
will also move. The fluid elements on within
245
00:32:00,919 --> 00:32:07,649
this volume, we will move as a function of
time, those on the surface will also move.
246
00:32:07,649 --> 00:32:14,120
So at some later time, at some time t plus
delta t, this volume would have gone to some
247
00:32:14,120 --> 00:32:20,889
other location. If the surface is that, the
volume itself is defined by the motion of
248
00:32:20,889 --> 00:32:27,889
the surface points; the points on the surface,
how they move? The points on the surface,
249
00:32:29,080 --> 00:32:34,289
move with the same velocity as the fluid velocity
on that surface, the points within move with
250
00:32:34,289 --> 00:32:39,269
the same velocity, as the fluid velocity of
within this volume. Note that the velocity
251
00:32:39,269 --> 00:32:43,679
is a vector that is defined, at each particular,
at each point in space, it is a continuous
252
00:32:43,679 --> 00:32:50,679
fluid. Now as this volume moves, the points
that are on the surface will continue to be
253
00:32:51,750 --> 00:32:56,370
on the surface, provided they are moving with
the same velocity, as the fluid velocity at
254
00:32:56,370 --> 00:33:02,450
that point. The points that are within will
continue to be within, once again provided
255
00:33:02,450 --> 00:33:07,080
the they are moving with the same velocity,
as the fluid velocity at that point. What
256
00:33:07,080 --> 00:33:12,159
that means is that points on the surface,
continue to be on the surface, the points
257
00:33:12,159 --> 00:33:19,159
within, continue to be within; that is because
the fluid velocity as I said, is a single
258
00:33:20,909 --> 00:33:27,909
valued function of position, at each point
the fluid velocity has only one value.
259
00:33:29,480 --> 00:33:36,480
If a point within has to cross and go outside
the surface, then obviously the velocity,
260
00:33:36,509 --> 00:33:42,559
it has to intersect with the trajectory of
a point on the surface. If a point within,
261
00:33:42,559 --> 00:33:49,399
has to travel and go outside the surface,
it has to intersect with the trajectory of
262
00:33:49,399 --> 00:33:56,090
some other surface point; that is travelling
on the surface, if it has to go from outside
263
00:33:56,090 --> 00:34:03,090
to inside. Similarly, for point that is outside
has to come inside, it has to cross trajectories.
264
00:34:03,259 --> 00:34:08,570
This crossing of trajectories is not permitted,
because at that particular location, it implies
265
00:34:08,570 --> 00:34:14,679
that the velocity has two values. Whereas,
we have defined the velocity vector a single
266
00:34:14,679 --> 00:34:21,679
valued function in space. Therefore, fluid
points cannot cross from inside to outside
267
00:34:22,030 --> 00:34:26,180
the surface, because the surface is moving
at the mean velocity at that location at the
268
00:34:26,180 --> 00:34:32,270
surface, inside is moving at the same mean
velocity as the inside points. And if it has
269
00:34:32,270 --> 00:34:36,660
to cross, that will a double valued function,
which it is not, it is a single valued function
270
00:34:36,660 --> 00:34:40,900
at each point in space.
Therefore, one cannot have the crossing of
271
00:34:40,900 --> 00:34:46,840
trajectories, what is outside is outside,
what is inside is inside, and what is on the
272
00:34:46,840 --> 00:34:52,490
surface, continuous to be on the surface,
provided it is a material point which is moving
273
00:34:52,490 --> 00:34:59,490
with the same velocity as the fluid velocity
on that surface. And within this moving surface,
274
00:34:59,890 --> 00:35:06,890
the mass is defined as the integral over this
volume, which is a function of time of integral
275
00:35:08,070 --> 00:35:15,070
d v times rho. Now the rate of change of mass,
is equal to the time derivative of the integral
276
00:35:30,100 --> 00:35:37,100
over this volume of d v times rho. The rate
of change of mass is the time derivative of
277
00:35:37,660 --> 00:35:43,890
the integral over this volume of d v times
rho. Now the volume itself is a function of
278
00:35:43,890 --> 00:35:50,840
time, the density also present general the
function of time and space. So you going to
279
00:35:50,840 --> 00:35:57,840
get contributions to this derivative, both
from the volume, the dependence of density
280
00:35:59,320 --> 00:36:06,170
on time, as well as from the fact that the
volume itself is changing, as time progresses.
281
00:36:06,170 --> 00:36:11,560
Now, how do we combined these two values,
so there are two contributions to this rate
282
00:36:11,560 --> 00:36:15,470
of change of mass.
We know that from mass conversion equation,
283
00:36:15,470 --> 00:36:19,410
the mass within that volume has to be conversed,
because points that are inside continue to
284
00:36:19,410 --> 00:36:23,940
be inside, points that are outside continue
to be outside. And therefore, I require that
285
00:36:23,940 --> 00:36:30,770
for mass conversion condition, this has to
be equal to zero, the rate of change of mass,
286
00:36:30,770 --> 00:36:35,150
because mass cannot be created or destroyed,
unless you take into account nuclear reactions
287
00:36:35,150 --> 00:36:42,150
for example, so we will work within the classical
regime, where this has to be equal to zero.
288
00:36:43,040 --> 00:36:48,790
And this relation that I have got, is for
a moving element of fluid, and I need to convert
289
00:36:48,790 --> 00:36:55,790
that into a relation for a fixed volume element.
So let us do that, so we go back to what we
290
00:37:02,830 --> 00:37:08,760
had earlier.
We have this volume, this is the volume at
291
00:37:08,760 --> 00:37:15,760
time t, and after a little bit of time t plus
delta t, this goes to some other volume. This
292
00:37:16,560 --> 00:37:23,560
is v at t plus delta t, what is the change
in this integral d v times rho between t and
293
00:37:26,820 --> 00:37:33,820
t plus delta t. There are two components to
this change in the volume, change in the mass;
294
00:37:34,680 --> 00:37:40,700
one is because the density within the volume
itself is changing at various points within
295
00:37:40,700 --> 00:37:47,040
the volume we have a density that is defined.
The density within the volume itself is changing,
296
00:37:47,040 --> 00:37:52,990
and the other is because the surface itself
has moved, so their certain parts of the surface,
297
00:37:52,990 --> 00:37:57,120
that have come in to the volume, which were
not previously in the volume.
298
00:37:57,120 --> 00:38:02,190
That has to be incorporated into the mass
within this volume, the certain parts of the
299
00:38:02,190 --> 00:38:06,250
surface that were previously within the volume,
but have now left, so that is to be subtracted
300
00:38:06,250 --> 00:38:13,250
from the mass. So there are two contributions;
one is t to the change in density within this
301
00:38:13,460 --> 00:38:18,840
volume itself, the other is because certain
regions are come in, and some other regions
302
00:38:18,840 --> 00:38:25,410
are left. So the rate, the change in mass
due to the volume, change in density itself,
303
00:38:25,410 --> 00:38:32,150
is just integral d v of partial rho by partial
t, so that is the change in mass, because
304
00:38:32,150 --> 00:38:37,910
the density is changing, within this differential
volume. In addition there is certain parts
305
00:38:37,910 --> 00:38:44,600
that have come in, and certain parts that
have left; so for example, if I expand out
306
00:38:44,600 --> 00:38:51,600
a small region over here. Surface, which was
initially at this location, has now moved
307
00:38:58,810 --> 00:39:04,500
to a new location, because of the fluid velocity,
the fluid velocity could in general be in
308
00:39:04,500 --> 00:39:10,190
some direction at this location. This is the
fluid velocity u, in some general direction,
309
00:39:10,190 --> 00:39:17,190
with respect to the surface, because the fluid
is moving, the surface is also moved. So you
310
00:39:18,810 --> 00:39:25,810
should take this little patches surface, I
will call it as d s, because of this little
311
00:39:26,050 --> 00:39:32,210
patches surface, how much of surface, how
much of volume has come in, to this differential
312
00:39:32,210 --> 00:39:36,940
volume, because the surface is moved in this
way.
313
00:39:36,940 --> 00:39:43,940
The volume that has come in, is basically
equal to this distance, which is the perpendicular
314
00:39:46,590 --> 00:39:51,770
distance of the surface, the perpendicular
distance that the surface has travelled; that
315
00:39:51,770 --> 00:39:58,770
is this perpendicular distance, that the surface
is travelled. What is the perpendicular distance
316
00:39:58,780 --> 00:40:05,290
that the surface is travelled in a time delta
t, perpendicular distance travelled is equal
317
00:40:05,290 --> 00:40:12,290
to the velocity, along the perpendicular times
time. So perpendicular distance travelled,
318
00:40:19,920 --> 00:40:26,830
is equal to the velocity times time; the velocity
along the normal direction. so the velocity
319
00:40:26,830 --> 00:40:32,800
along the normal direction is u dotted with
n times delta t, so that is the distance that
320
00:40:32,800 --> 00:40:36,050
is travelled. Note that u dot n is a velocity,
is normal velocity.
321
00:40:36,050 --> 00:40:43,050
The component of the velocity u, along the
unit normal to the surface n. Note that this
322
00:40:46,370 --> 00:40:53,370
is the outward unit normal to the surface,
so that is the distance travelled. The total
323
00:40:53,420 --> 00:40:59,370
volume that has come in is equal to the distance
travelled, times that surface patch. So the
324
00:40:59,370 --> 00:41:06,370
total volume that has come in, into area,
so volume that is come in, is this times d
325
00:41:06,770 --> 00:41:13,520
s. So there is a total volume that has come
in, because the surface is travelled. Now
326
00:41:13,520 --> 00:41:20,520
the mass that has come in is just equal to
these times, the density itself, density at
327
00:41:21,460 --> 00:41:25,020
that particular location; that is the mass
that is come in.
328
00:41:25,020 --> 00:41:32,020
So therefore, the mass in, let me just get
rid of this, so that we can write it all the
329
00:41:40,170 --> 00:41:47,170
same, is equal to rho d s times u dot m delta
t; that is the mass that has come in for this
330
00:41:53,050 --> 00:41:59,890
little patch of surface, it has travelled,
the distance u dot n delta t. So that is the
331
00:41:59,890 --> 00:42:06,890
mass m unit time delta t. Therefore, the rate
of mass m, is going to be equal to this mass
332
00:42:11,590 --> 00:42:18,230
m divided by the time, this going to be equal
to rho t s u dot n, this is the rate at which
333
00:42:18,230 --> 00:42:25,230
mass is coming in, for this little patch of
surface, because the surface is moving. Now
334
00:42:28,620 --> 00:42:33,890
if you take a patch of surface on the other
side. If you take a patch of surface on the
335
00:42:33,890 --> 00:42:40,570
other side, this surface initially it was
something like this, so if I take this little
336
00:42:40,570 --> 00:42:47,570
patch, this surface initially it was here,
and after some time it goes to some other
337
00:42:47,990 --> 00:42:53,500
location, after some time it goes to some
other location.
338
00:42:53,500 --> 00:42:59,100
So obviously some volume elements have left
this volume, it has left behind some volume,
339
00:42:59,100 --> 00:43:04,900
as it is travelling in this direction, and
therefore, the mass has left this volume,
340
00:43:04,900 --> 00:43:11,900
mass has been reduced from this volume, what
is the mass that is reduced, the argument
341
00:43:13,140 --> 00:43:20,140
is exactly the same, I take this little patch
of surface. Now the outward unit normal, is
342
00:43:21,410 --> 00:43:26,740
in this direction; that is the outward unit
normal, at that point on the surface, it is
343
00:43:26,740 --> 00:43:33,740
in this direction, may be plotted on the original
surface for simplicity. The outward unit normal
344
00:43:34,410 --> 00:43:40,810
in this direction, and the velocity vector
is in some direction like this u vector is
345
00:43:40,810 --> 00:43:47,810
in this direction. The amount that has left
is once again u dot n d s, there is a rate
346
00:43:47,910 --> 00:43:54,910
at which mass is left. Except that now u dot
n is negative, because u and n the angle between
347
00:43:56,040 --> 00:44:01,340
them is greater than 90 degrees, if the volume
l, if the velocity is going to be opposite,
348
00:44:01,340 --> 00:44:07,420
to the unit normal in such a way that, there
is some volume that is left behind as the
349
00:44:07,420 --> 00:44:13,120
surface moves.
Therefore, for volume elements that are leaving
350
00:44:13,120 --> 00:44:18,000
the surface, u dot n is automatically negative,
because the velocity is in opposite, the component
351
00:44:18,000 --> 00:44:23,160
of the velocity along the unit normal, is
along the inward unit normal, it’s opposite
352
00:44:23,160 --> 00:44:28,170
to the outward unit normal, so u dot n is
automatically negative. And therefore, rho
353
00:44:28,170 --> 00:44:34,960
s rho d s times u dot n, will automatically
be negative, and therefore, this expression
354
00:44:34,960 --> 00:44:41,960
gives you the rate at which mass comes in,
and mass goes out. If mass is coming in, and
355
00:44:42,690 --> 00:44:47,860
if you define n is the outward unit normal,
u dot n will be positive that will add mass
356
00:44:47,860 --> 00:44:54,860
to the volume. If it is leaving, then u dot
n will automatically be negative. So this
357
00:44:55,950 --> 00:45:01,550
is the second component, for each patch of
surface d s, we get this, therefore, the total,
358
00:45:01,550 --> 00:45:08,550
is just going to be equal to, integral over
the surface, of rho u dot n. So there are
359
00:45:11,840 --> 00:45:17,960
two components; one because the, there is
a change in density within this volume itself.
360
00:45:17,960 --> 00:45:24,390
The other is, because as the volume is moving
in space, as a function of time. There are
361
00:45:24,390 --> 00:45:27,840
elements, volume elements that are coming
in, to this differential volume, there is
362
00:45:27,840 --> 00:45:34,060
volume elements, which are leaving this differential
volume. This thing is what is called the Leibnitz
363
00:45:34,060 --> 00:45:41,060
rule t, this identity that integral d v rho
over something that is function of time. This
364
00:45:54,450 --> 00:46:01,450
is called the Leibnitz rule, and it works
not just for density, but for all other things.
365
00:46:03,780 --> 00:46:09,400
It works for momentum, for energy, and so
on. In all such cases, you have a change in
366
00:46:09,400 --> 00:46:13,890
that quantity, because one that quantity is
changing within the volume itself.
367
00:46:13,890 --> 00:46:19,300
Second is, because as the volume is moving,
there are several regions that are being struck
368
00:46:19,300 --> 00:46:24,210
in, and there are certain regions that are
being left behind, inside the account for
369
00:46:24,210 --> 00:46:27,320
both those regions that has being swept in,
as well as the regions that have being left
370
00:46:27,320 --> 00:46:32,350
behind. And that those regions that have being
swept in, and the regions that are being left
371
00:46:32,350 --> 00:46:37,910
behind, are incorporated in this second surface
integral term, there are both incorporated
372
00:46:37,910 --> 00:46:44,910
and this second surface integral term. So
that is the Leibnitz rule, and we know that
373
00:46:45,330 --> 00:46:52,330
the rate of change of mass has to be equal
to zero, for a mass conversion.
374
00:46:53,500 --> 00:47:00,500
Integral v f t d v times rho, the rate of
change of that, which I had derived as integral
375
00:47:10,240 --> 00:47:17,240
d v times partial rho by partial t plus integral
over the surface d s of rho u dot n, this
376
00:47:19,800 --> 00:47:26,800
has to be equal to zero. For the second term
here I can use the divergence theorem, vector
377
00:47:28,940 --> 00:47:34,360
dot n integrated over a surface, is equal
to the divergence of that vector, integrated
378
00:47:34,360 --> 00:47:41,360
over the volume. So this integral d v partial
rho by partial t plus integral d v of del
379
00:47:45,280 --> 00:47:52,280
dot rho u, this is equal to 0. And this has
to be true for every volume within the fluid,
380
00:47:56,010 --> 00:47:59,340
because it is true at each and every point
within the fluid. And therefore, it has to
381
00:47:59,340 --> 00:48:04,340
be at every, if it is true for each and every
volume that I can consider; this is true at
382
00:48:04,340 --> 00:48:11,340
each point within the fluid. And that means
that the mass conservation equation; partial
383
00:48:18,390 --> 00:48:25,390
rho by partial t plus del dot rho u is equal
to zero, that is the equation for the conservation
384
00:48:27,220 --> 00:48:32,040
of mass.
In indicial notation, I will write this as,
385
00:48:32,040 --> 00:48:39,040
partial rho by partial t plus partial by partial
x i of rho u i is equal to zero, dot product
386
00:48:42,870 --> 00:48:49,520
repeated index it is a scalar equation for
the density. I can also write it as use differentiation
387
00:48:49,520 --> 00:48:56,520
by chain rule, to write this as. This differentiation
by chain rule, and you can identify this first
388
00:49:10,010 --> 00:49:15,830
two terms, as the substantial derivative.
These first two terms are identical to the
389
00:49:15,830 --> 00:49:22,830
substantial derivative, so I can also write
this as d rho by d t plus rho partial u i
390
00:49:24,310 --> 00:49:31,310
by partial x i is equal to 0. If the density
is a constant, then d rho by d t in a moving
391
00:49:34,800 --> 00:49:40,690
reference frame has to be 0, and therefore,
I have the mass conservation equation for
392
00:49:40,690 --> 00:49:47,690
constant density, incompressible. Density
is a constant, I have partial u i by partial
393
00:49:51,170 --> 00:49:58,080
x i is equal to zero, over the divergence
of velocity is equal to zero. Divergence of
394
00:49:58,080 --> 00:50:03,020
velocity is equal to 0, as I just discuss
the isotropic part of rate of deformation,
395
00:50:03,020 --> 00:50:08,950
is tensor is zero. There is no radially outward
or inward flow.
396
00:50:08,950 --> 00:50:15,950
I can derive the other conservation equations,
just as easily. If I wanted to derive an equation
397
00:50:23,940 --> 00:50:30,540
for concentration field, what I would say
is that for this moving volume element for
398
00:50:30,540 --> 00:50:37,540
this moving volume element d by d t of the
integral d v times the concentration, of the
399
00:50:39,870 --> 00:50:46,870
solute. The concentration of the solute, in
this moving reference frame is equal to an
400
00:50:47,780 --> 00:50:54,640
integral over the surface, is equal to an
integral over the surface, of the flux, times
401
00:50:54,640 --> 00:51:01,640
a unit normal. So if I have some volume element,
which is moving, and I have some flux on the
402
00:51:03,170 --> 00:51:10,090
surface. Then d by d t of the integral of
the volume times, the concentration is equal
403
00:51:10,090 --> 00:51:17,090
to the net flow into this volume, due to the
flux is acting on the surface. The flux will
404
00:51:17,800 --> 00:51:23,190
increase the concentration within this volume,
if it is directed inwards. So the concentration,
405
00:51:23,190 --> 00:51:27,330
so the mass within this volume is going to
increase, if the flux is directed in the direction
406
00:51:27,330 --> 00:51:33,480
of opposite, to the outward unit normal.
The mass within this volume is going to increase,
407
00:51:33,480 --> 00:51:39,040
if the flux is directed outward, along the
inward unit normal, or opposite to the outward
408
00:51:39,040 --> 00:51:44,700
unit normal. I have defined my unit normal
n as the outward unit normal in this case,
409
00:51:44,700 --> 00:51:51,700
and therefore, this is equal to minus integral
over the surface of q dot n. When q dot is
410
00:51:52,490 --> 00:51:57,940
negative, the mass increases, so when q and
n are in opposite directions, the mass increases.
411
00:51:57,940 --> 00:52:03,000
So this equal to minus integral over the surface,
of the mass flux; that is the amount of material
412
00:52:03,000 --> 00:52:10,000
coming in, per unit surface area per unit
time. Simplify this using the Leibnitz rule
413
00:52:10,220 --> 00:52:17,220
once again, I get a integral d v of partial
c by partial t plus integral d s of c u dot
414
00:52:22,880 --> 00:52:29,880
n minus integral over the surface of q dot
n. And now, we use the divergence theorem,
415
00:52:36,270 --> 00:52:40,600
for both the convective part, as well as for
the flux.
416
00:52:40,600 --> 00:52:47,600
So then I get integral d v partial c by partial
t plus integral over the volume of the divergence
417
00:52:49,230 --> 00:52:56,230
of concentration times the velocity is equal
to minus integral over the volume of the divergence
418
00:53:01,580 --> 00:53:08,580
of the flux itself, of the flux, the divergence
of the flux. And this has to be true for each
419
00:53:14,050 --> 00:53:19,930
and every differential volume; that means
it has to be true at each point in space.
420
00:53:19,930 --> 00:53:26,930
And therefore the concentration equation becomes
partial c by partial t plus
minus partial q i by partial x i or d c by
421
00:53:35,280 --> 00:53:42,280
d t plus del dot u c is equal to minus divergence
of q. Recall we got exactly the same expression
422
00:53:45,440 --> 00:53:51,120
by doing our differential balances of a volumes
in the fundamentals of transport processes
423
00:53:51,120 --> 00:53:56,280
one. There will be the differentially balances
over cubic volumes, spherical volumes, cylindrical
424
00:53:56,280 --> 00:54:01,520
volumes. In this case we get the exact same
expression without reference to any underline
425
00:54:01,520 --> 00:54:08,240
coordinate system.
And then, we have a constitutive relation
426
00:54:08,240 --> 00:54:15,240
for the, I should use j for the mass flux.
Let me just use j for the mass flux to avoid
427
00:54:20,450 --> 00:54:27,450
any confusion, q is better use for the heat
flux, so this as j, this is j i this is j,
428
00:54:31,760 --> 00:54:38,060
and this is delta j. And then we have the
constitutive relation j is equal to minus
429
00:54:38,060 --> 00:54:44,240
d times that gradient of c, where d is the
diffusion coefficient, and the flux is in
430
00:54:44,240 --> 00:54:50,200
the direction of decreasing concentration,
so it is opposite to the gradient vector,
431
00:54:50,200 --> 00:54:54,340
gradient vector gives you the direction in
which there is a maximum increase in concentration.
432
00:54:54,340 --> 00:54:58,390
In the constant of proportionalities, the
diffusion coefficient, with dimensions of
433
00:54:58,390 --> 00:55:04,090
n square by time. And so from this, you will
just get the equation, partial c by partial
434
00:55:04,090 --> 00:55:11,090
t plus del dot u c is equal to the divergence
of d grad c. And is equal to d times the Laplacean,
435
00:55:16,840 --> 00:55:23,400
provided if the diffusion coefficient is in
a point of position, is equal to d times the
436
00:55:23,400 --> 00:55:29,810
Laplacean, if the diffusion coefficient is
in a point of position, and this is the exact
437
00:55:29,810 --> 00:55:35,980
expression that we got for the concentration
equation, in the previous course.
438
00:55:35,980 --> 00:55:42,980
Similarly, one can get an expression for the
temperature equation, and if I write it out
439
00:55:43,880 --> 00:55:50,880
in in this form we will get rho c v into partial
t by partial t plus d by d x i of u times
440
00:55:55,060 --> 00:56:02,060
t is equal to of k partial t by partial x
i or attentively rho c v into
plus del u t is equal to del dot k grad t.
441
00:56:25,150 --> 00:56:32,150
And in case, the thermal conductivity is in
a point of position, this can also be written
442
00:56:32,730 --> 00:56:39,730
as del square, sorry del square is the Laplacian,
and I had shown you in the previous lecture,
443
00:56:42,170 --> 00:56:47,400
how we derive the Laplacian in the different
coordinate systems, spherical, cylindrical,
444
00:56:47,400 --> 00:56:53,250
as well as Cartesian. In spherical and cylindrical,
you have to account for the fact that, the
445
00:56:53,250 --> 00:56:57,790
unit vectors dependent upon position, and
therefore, the dependence of unit vectors
446
00:56:57,790 --> 00:57:02,460
in position, has to be taken into account,
and we derived explicit expressions, for each
447
00:57:02,460 --> 00:57:03,800
of these.
448
00:57:03,800 --> 00:57:10,310
So, these are the mass heat conservation equations.
Note that in the previous course, we had actually
449
00:57:10,310 --> 00:57:16,060
taken a differential volume, taken it surfaces,
six of them, found out what came in, what
450
00:57:16,060 --> 00:57:23,060
went out, and on that basis, determined what
the conservation equation was; that is in
451
00:57:25,010 --> 00:57:29,810
this case we do not have to do any of those.
We just use the divergence theorem, and use
452
00:57:29,810 --> 00:57:36,810
the fact that, the Leibnitz rule can be applied
for a differential volume. In order to relate
453
00:57:38,200 --> 00:57:42,330
the change in a moving reference frame, to
that in a fixed reference frame, and that
454
00:57:42,330 --> 00:57:47,860
is all we require in order to find out the
conservation equations. So the next step is
455
00:57:47,860 --> 00:57:53,610
to proceed in determine conservation equations
for the fluid momentum. This procedure as
456
00:57:53,610 --> 00:57:58,450
you can see is much simpler, had I done it
the way that I done it previously.
457
00:57:58,450 --> 00:58:04,380
I would have to have three components of momentum,
and for each of those, I had to have three
458
00:58:04,380 --> 00:58:08,960
components of fluxes. Whereas, here I am just
going to treat both momentum and flux, as
459
00:58:08,960 --> 00:58:13,770
objects in themselves, independent of the
underline coordinate system. So we will proceed
460
00:58:13,770 --> 00:58:19,810
with the conservation equations for, momentum
for the fluid in the next lecture. So kindly
461
00:58:19,810 --> 00:58:25,270
review what we will done here; that is how
do you relate the changes in a moving reference
462
00:58:25,270 --> 00:58:28,740
frame, to that in a fixed reference frame,
and we will proceed in the next lecture, to
463
00:58:28,740 --> 00:58:33,850
deal with fluid momentum conservation equation.
So we see you then.