In the last lecture, we studied the kinematics
of fluid motion so as to understand the deformation
that a fluid element goes through as flow
takes place. And, we have considered three
different possible deformations, components
of deformation: one of which is a pure solid
body rotation.
For example, if this is a solid body; you
can have pure translation in the x direction
or in the y direction. This is one form of
a motion that a fluid element goes through.
And, in the case of pure translation in any
direction, the fluid element does not get
any deformation, it remains the same. You
can have pure rotation along the diagonal
of this. And, here again we can see that,
the shape does not change there is no noticeable
difference, deformation in this. But, we can
also have a pure stretching, so that the volume,
it becomes elongated in this direction or
it becomes elongated in this direction. And,
there can be a change in the cross-sectional
area. And, there can be a change in the shape
from an initial rectangular thing into more
rhombus like thing or a general quadrilateral
like thing. So, these are different kinds
of deformations that are possible.
And, we have seen that, the rates of these
deformations can be expressed in terms of
velocity gradients – in terms of linear
combinations of velocity gradients. And, this
is an important aspect because it means that,
in the general three dimensional case, all
the nine velocity gradients; so, that is,
dou u by dou x, dou u by dou y, dou u by dou
z; and, similarly for the three v components
and the three w components; all the nine velocity
gradients, matrix together can define the
general rate of deformation that is possible
for a fluid element. And, we would like to
relate this rate of deformation in a linear
way to the stress that is induced by relative
motion; and, so that the idea would be that,
the stresses that are present in the conservation
of linear momentum can be replaced by these
velocity gradients. And, since the velocity
components are already part of the variables,
we can then replace the stresses by velocities
and thereby create a system of equations,
which is closed in the sense that, we have
as many equations as there are, the number
of variables.
So, here to this extent, in the last class,
we decomposed the stress tensor into a hydrostatic
component and a viscous stress component,
which arises only because of relative motion.
The hydrostatic component is present; this
pressure is compressible is present, even
when there is no relative motion; whereas,
this is supposed to be present only when there
is relative motion. And, given that, relative
motion produces strains – strain rate, which
is given by dou u by dou u k by dou x m in
the general case. So, this can be written
as dou u by dou x dou u by dou y dou u by
dou z. And similarly, dou v by dou x dou v
by dou y dou v by dou z and dou w by dou x.
These are the nine components that are occurring
here. And together, this is the strain rate
components. And, the idea of, of putting it
here is to make proportional relation between
the stresses that are induced by relative
motion and the strain rates that are a result
of relative motion. We note that the distortion
of the rectangular element is possible only
when the four corners of this – the ABCD's
have different velocities. If they have the
same velocities; then, they are just translating
in a certain direction without any distortion.
When there is relative motion within the fluid
element, within the fluid continuum; when
ABCD have different velocities; then, there
is relative motion and that relative motion
is present in the form of non-zero velocity
gradient components here. And, these give
rise to different deformations like we have
mentioned that, delta alpha plus delta div,
the rate of deformation is given by delta
beta and the average of these two. And, this
is expressed as half of dou u by dou y minus
dou v by dou x and so on like this. So, this
is the shear rate and this is expressed in,
in this particular form here, perhaps with
a negative sign here. And so, this is a linear
combination of two elements that are coming
here dou u by dou y and dou v by dou x.
So, we are now saying that, we are seeking
a relation – a linear relation between tau
i j and epsilon k m. This is a hypothesis.
And, why are we seeking a linear relation?
Because a linear relation is a simplest possible
that we can have, other than having no relation
at all. If it is a non-linear variation; then,
it means that, there will be this plus some
multiplication of this and that involves addition
constraints and so on. So, this is the simplest
possible. And, the simplest possible appears
to work for very common fluids like air and
water. So, that is a advantage of this. The
simplest common fluids that we can come across
obey empirically, based on empirical observation,
we can say that, they obey this hypothesis,
and, which is therefore, this hypothesis is
very useful practically. But, there are number
of other common fluids, more complicated fluids
like blood, which contains lots of white corpuscles
and red corpuscles and, those kind of additional
things, which do not obey this, this linear
hypothesis.
Similarly, many polymeric liquids, which have
elongated oriented molecular structure with
long chain molecules, do not obey this – this
particular assumption. But, this is a very
useful assumption and it is an empirical hypothesis
– it is a hypothesis, which is back by empirical
observations of the goodness of this hypothesis.
So, with this thing, we would like to have
a mathematical formulation for this, so that
we can rewrite this tau i j in terms of this
deformation strain rate tensor, involving
the velocity gradients, so that if you substitute
this into this. Then, the stresses would disappear
and you will have only velocity gradients;
and, the velocity gradients are not new variables.
So, in that sense, that is where we are heading.
And, we mentioned that, this is a tensor and
this is a tensor. A general relation would
have A i j k m epsilon k m. And, this is a
nine by nine matrix. And so that, and we also
mentioned that, this is – these are material
properties like viscosity that we are familiar
with. But, this many material properties are
very difficult to get empirically. And, we
also would like to – may study this particular
thing further and take advantage of special
features of the matrices that we are trying
to relate. And, what are the special features.
We know from angular momentum that, tau i
j is symmetric. So, that means that, there
are only six independent components. And,
in the special case, where we do some matrix
operations and rotate this and then write
these in the principal coordinates system;
then, we will only have tau a tau b b and
tau c c. And, all these things are 0. So,
that is, we can – this is in the general
case of x and y directions. And, when we do
principal component analysis and we put this
in a diagonal form then, we will have nonzero
components in the shear stresses and only
normal stresses.
So, this is a special coordinate system aligned
not in the i j, but in the – in principal
coordinate direction here. And, here we have
only three normal stresses compo – three
normal stress components – only three nonzero
normal stresses. And so, that means that,
here in this special thing, which we are doing
by matrix multiplication or matrix operations
on this, so that we are not introducing any
new features into this; we are looking at
three independent components here. So, instead
of six independent components, we have three
independent components.
And similarly, the epsilon here k m is dou
u k by dou x m. And, in general, this is not
symmetric because it is not necessary that,
dou u by dou y is the same as dou v by dou
x and – in that. But, when we look at our
objective, is not to take relation between
this and this, but to seek a relation, linear
relation between stress and the deformation
rate tensor. Now, we have said earlier that,
if there is a pure solid body rotation here,
the shape remains always rectangular and there
is no deformation in this. So, you could say
that, pure solid body rotation does not give
rise to any change of shape; it does not deform
the rectangular element. Similarly, pure translation
will not deform this. So, the only thing that
is actually changing the shape and thereby
creating strain on this is the shear strain
rate and the extensional strain rate.
So, the shear strain rate is expressed as
half dou u i by dou x j plus dou u j by dou
x i okay in the general case. And, we can
see that, this is symmetric. And, extensional
strain rate 
is dou u k by dou x k. So, that is dou u by
dou x, dou v by dou y, dou v by dou z. And,
those are these components here. And, shear
strain rate has this plus this going together
and this plus this going together and this
plus this going together. So, the same component
– it is a summation of – it is a summation
of these two and summation of these two and
summation of these two. So, that means that,
a combination of these two – the deformation
rate tensor D i j is symmetric. So, if D i
j is symmetric. Then, this is also has six
independent components.
Now, this also can be decomposed into the
principal components into – this can also
be rotated and – so that we have only D
p p, D q q and D r r; all others are 0. So,
we are looking at essentially a matrix rotation
– matrix rotation operation in such a way
that, a general tensor with all the nine components
are converted into a special orientation in
which only the diagonal elements are nonzero
and all the others are zero. So, this means
that, now we have a set of coordinate axis
in which we have only three stress components.
And another set of coordinate axis in which
only three nonzero components of the strain
rate – deformation rate tensor. And, we
would like to have a relation between this
tau i j versus D prime k m; where, these are
these – special rotate – specially rotated
operations here.
Now, when you are seeking a relation between
these two, which is linear; then, it is necessary
that the principal coordinates of these and
the principal coordinates of these are the
same. So, principal axes of both must be aligned.
If these are not aligned, then you cannot
have a linear relation. So, that means that,
we can now say that, we are seeking a relation
between not 9 by 9 components, not even 6
by 6 components; we are – we can only have
– if we say that we are going to have a
linear relation between this and this; mathematically,
you can only have three independent components
here and three independent components here.
So, that means that, you can only have nine
independent values, which will describe this
general relation, which is, which is applicable
in any coordinate system, but, the same constituents
must also be applicable in the principal axis
coordinate system here. So, in that sense,
we have only nine independent components.
And what are these nine independent components?
We can put these as sigma, tau 1 1, tau 2
2, tau 3 3, because only diagonal terms are
there; and similarly, D 1 1, D 2 2 and D 3
3. Now, what kind of linear – generic linear
relation that is possible?
So, we can write for example, tau 1 1 is equal
to a 1 times D 1 1 plus a 2 times D 2 2 plus
a 3 times D 3 3. So, this is for different
values of a 1, a 2, a 3, which are constants.
We have a linear relation in which tau 1 1
is expressed as a linear combination of all
the three possible variables on the right-hand
side. So, when you have all the three possible
– all the three variables on the right-hand
side here; then, that is a most generic relation
that is possible. If you put D 1 times D 2;
then, that becomes – maybe that is not a
linear relation there. Similarly, we can say
that, tau 2 2 is also related only to the
same D 1 D 2 D 3, but using a different set
of constants, for example, a 4 D 1 1 plus
a 5 D 2 2 plus a 6 D 3 3. And, tau 3 3 is
also written in terms of the same things.
So, you have a 7 D 1 1 plus a 8 D 2 2 plus
a 9 D 3 3. So, here we have nine constants
a 1 to a 9, which define a general linear
relation between the three rotative stress
components and the three deformation strain
rate components.
So, now, these, you cannot reduce it further
without assumptions. So, here we make one
more assumption, which is that, the fluid
is isotropic. So, when we say the fluid is
isotropic; then, what we are saying is that,
application of a force on this produces a
deformation in a solid body. And, in a case
of a fluid, it produces a rate of deformation.
And, what we are saying is that, if we apply
a tensile stress in this direction; then,
in a solid body, it produces a deformation
in this direction, in this direction and also
in the z direction. So, in an isotropic medium,
it does not matter whether you apply the tensile
stress here and here or in this direction
or in this direction; the response of this
surface is the same in all directions.
So, that means that, if you apply a tensile
force here; then, the body elongates here
and then it also contracts on this side and
contracts on this side. The contraction along
this side, the z direction must be the same
as the contraction in the y direction. And
similarly, if you now apply the stress in
this direction along the direction which you
applied the stress, you have a certain extension
stretching. And, in the other two directions,
there is a small contraction. And so, the
contraction that is produced by the application
of a stress in this direction in the z direction;
so, that is, you apply a stress in the y direction,
that is, a tensile stress; it produces a contraction
in the z direction.
Now, similarly, you apply the same stress
– same tensile stress along the x direction.
It again produces a contraction in z direction.
If the body is isotropic, it does not matter
whether the contraction produced, is produced
by application of a tensile stress in the
y direction or x direction. If the stress
is the same, then it produces the same amount
of contraction in the z direction, okay. Similarly,
if you apply in the z direction certain tensile
stress or a compressive stress, and that produces
a corresponding deformation in the transverse
direction; then, the amount of deformation
that takes place, that is produced, must be
the same if the same tensile or compressive
stress is applied in the other direction;
so, in the other orthogonal direction.
So, what we are saying is that, in an isotropic
medium, we can – we are allowed to distinguish
between a deformation, which is produced in
the direction of application of the stress
and a deformation which is produced in the
transverse direction. In the transverse direction,
you cannot distinguish between y and z. So,
there can be one deformation, which is in
the direction of the stress and there is the
same deformation, which is produced by the
same amount of stress in the other two directions.
The two transverse directions are similar.
So, that is the idea that we can take it here.
And, that kind of deformation, that kind of
meaning, can be rewritten in this way – that
is, tau 1 1 is b 1 times D 1 1 plus D 2 2
plus D 3 3 plus b 2 times D 1 1; tau 2 2 is
b 1 times D 1 1 plus D 2 2 plus D 3 3 plus
b 2 times D 2 2; tau 3 3 is b 1 times D 1
1 plus D 2 2 plus D 3 3 plus b 2 times D 3
3.
Now, you look at this relation here. In this
relation here, it is possible for tau 1 1,
tau 2 2 and tau 3 3 to be different, because
although they are related to the same set
of variables on the right-hand side, D 1 1,
D 2 2, D 3 3 here; even though for a given
set of values of D 1 1, D 2 2 and D 3 3, this
contribution is the same. There is also a
contribution coming from this; so, tau 1 1,
these three stresses will be equal only if
these three are equal. If these three are
different; then, you can have different stresses
that are possible. So, this is a generic description
of a linear relation between the principal
stress components and the principal deformation
rate tensor components involving only two
constants: b 1 and b 2, oaky, and which is
isotropic, because you have a deformation
D 1 1 associated in the direction of stress
and you have something like a transverse component
which is coming here. So, in that sense, it
– this is a relation between the same three
components on the left-hand side and the same
three components on the right-hand side, which
is written with only two independent constants.
And, this kind of relation is applicable for
an isotropic medium, okay.
So, now, what we are saying is that, this
linear relation between a symmetric stress
tensor and a symmetric deformation rate tensor
here, for an isotropic medium, would have
only two independent constants. And normally,
we put this as lambda here and this as mu.
This is our dynamic viscosity. And, this is
known as the second coefficient of viscosity.
And, this relation expressed in – in the
principal coordinates will be like this. But,
when it is transformed from the principal
coordinates into general i j coordinates,
can be written as tau i j equal to mu dou
u i by dou x j plus dou u j by dou x i plus
lambda times dou u k by dou x k. This is a
relation between the stress induced by the
relative motion and the strain rates – deformation
strain rates that are induced by the stresses.
And, that relation is linear and it is for
an isotropic medium, involving the two coefficients
mu and lambda here. And so, delta i j here
– the kronecker delta function. So, this
is the expression that we have.
Here usually this particular thing dou u k
by dou x k is very small for most fluids.
So, this becomes negligible. And so, we do
not need to really know the value of lambda
in most cases. So, this – this can be obtained
for some simple gas species as having certain
value. But, otherwise, this is not possible
to; it is not possible to get an accurate
estimate for this; whereas, viscosity is something
that can be measured in the special cases
of where we impose a velocity gradient by
rotation between two cylinders and so on.
We can get estimation for the velocity. So,
this is easily measured. So, we measure the
viscosity and then we make use of this relation
here. So, this relation now can be substituted
into this. And, that gives us minus p delta
i j plus mu times this and lambda times this.
And, what we then have is that, the stresses
on this side are expressed in terms of pressure
and velocity gradients. So, velocity gradients
are not new variables; and, material properties,
which are the two viscosity coefficients:
the first coefficient of viscosity and the
second coefficient of viscosity.
So, in the next lecture, we will see how when
these are substituted into the governing equation.
We will end up with the conservation of linear
momentum, which has only extra variable as
a pressure, okay, so that the three momentum
equations for the three velocity components
plus the continuity equation together constitute
a set of four equations.
Thank you.
