00:00:25
So far, we have seen the analysis of ER and
MR fluid flow in a rectangular passage.
00:00:34
Now, today we will see how the formulation
looks if we consider an annular passage.
00:00:41
So, an annular passage looks like this.
00:00:46
So, there is two concentric cylindrical domains.
The inner cylinder is a solid cylinder and the
00:00:55
outer cylinder is a hollow cylinder. So, the inner
cylinder is between inside the outer cylinder.
00:01:03
So, when the inner cylinder
moves or remains fixed,
00:01:07
when the inner cylinder moves it gives rise to a
shear mode, when the inner shear cylinder and the
00:01:12
outer cylinder is fixed that is a flow mode. Now,
the fluid is flowing in this region, so between
00:01:18
the inner cylinder and outer cylinder that annular
region is filled with the fluid. So, when both
00:01:25
the cylinders are fixed, and the fluid is driven
by a pressure, it is a valve mode or flow mode.
00:01:30
When the inner cylinder is moving and that is
causing the fluid flow that is a shear mode.
00:01:36
So, we will analyze a shear mode here.
And we are doing a shear mode analysis.
00:02:02
Now to do this,
00:02:06
again, let us take a small fluid element.
So we are taking a small fluid element here
00:02:16
which looks like this.
00:02:30
And we have the pressure P and
pressure is increased to P plus
00:02:38
del P by del x into dx.
Now, we are taking a fluid, an element in
00:02:46
this region of length delta x, and we are taking
a cross-sectional view of the delta x region.
00:02:55
So, from the outside, it looks like this curved,
but if I take a cross-section throughout the
00:03:00
center, through the center, it looks like
this, and we are analyzing the flow there.
00:03:07
Now this is our shear at the top delta, delta
r by dr, now here our consideration is that
00:03:24
consideration is - this is at the center and the
distance from the center to the top or bottom is r
00:03:34
and this dimension is our x and the velocity along
x direction is called u. So, here we have tau.
00:03:49
So, this is our solid cylinder inside.
This is the outer boundary of the outer cylinder,
00:03:58
the top boundary of the outer cylinder, the
bottom boundary of the outer cylinder and
00:04:03
this is our annular region. We are studying only
one annular region and that is sufficient for
00:04:09
understanding the flow behavior. Now, here we
need to find out the mass of fluid element.
00:04:22
So, mass of the fluid
00:04:28
here is 2 pi r dr dx rho, where rho is
the fluid density. Now, 2 pi r dr dx.
00:04:51
So, 2 pi, so, this is our as we know
this is our dr. So, this dimension is dr,
00:05:02
and if I multiply this dr by 2 pi r, that gives
me the total cross-sectional area of the annular
00:05:11
region considering the 360-degree revolution. And
then if I do 2 pi r dr dx that gives me the volume
00:05:19
of the fluid element that means, the volume of the
fluid element throughout this considering a length
00:05:27
delta x. So, that is the volume of the fluid
element and then we have after having the
00:05:36
expression of the mass of the fluid we
can now write the equation of motion.
00:05:43
So, the force balance - so, the total force
is equal to the mass into acceleration.
00:05:50
If we do that, then finally,
the equation that we get is
00:05:56
minus, and this quantity we may write as dm.
00:06:06
So minus of
00:06:14
dm multiplied by del u by del t, plus 2 pi
r dr P, plus tau plus del tau by del r dr, 2
00:06:35
pi multiplied by r plus dr dx multiplied by tau
2 pi r dx, then we have minus P plus del P by
00:06:54
del x into dx into 2 pi r is equal
to 2 pi r dr is equal to 0.
00:07:03
So, we are finding out the total force
due to this pressure here minus the
00:07:08
total force due to this pressure here.
The total force due to this pressure is
00:07:11
this pressure multiplied by this thickness,
and then we multiply that by 2 pi, 2 pi r.
00:07:21
Similarly, we find out the total pressure here
due to the total force due to this pressure,
00:07:25
we find out the total shear stress, the total
force due to this shear stress, and that is
00:07:32
this shear stress multiplied by this length, and
then we multiplied by the periphery, 2 pi r.
00:07:39
And similarly, we find out the total force
due to this stress and that net unbalance is
00:07:45
equated with mass into acceleration. And after
doing that, finally, the equation simplifies to
00:07:55
minus rho del u by del t, plus tau into tau by
r plus del tau by del r minus del P by del x.
00:08:12
And then we assume that the flow is
quasi-steady. So, del u by del t is 0.
00:08:24
And that gives me del T by del r plus
tau by r is equal to del P by del x.
00:08:35
So, this was the governing differential equation,
00:08:37
and that is valid for both the
shear mode and the valve mode,
00:08:45
valid for both shear and mode or another mode
now the only thing is that when we are in the
00:08:58
shear mode, del P by del x is 0 because it
is not pressure-driven. So, in shear mode,
00:09:08
we have del tau by del r
plus tau by r equal to 0.
00:09:14
Now, first, we will do the
solution for 0 applied field.
00:09:25
Now, in the zero applied field, we know that the
fluid constitutive relation is tau is equal to
00:09:31
mu multiplied by del u by del r,
and if that is so, if I put this
00:09:39
in this equation, then the governing
differential equation becomes mu into
00:09:45
del 2 u by del r 2 plus mu by r
multiplied by del u by del r equal to 0
00:09:57
and this can be written as del by del
r of r multiplied by del u by del r.
00:10:07
Because if I take mu out then del 2 u by del
r 2 plus 1 by r into del u by del r is our
00:10:18
derivative of r into del u
by del r with respect to r.
00:10:27
Because if I take my mu out and then if I put this
r here, it becomes r into del u by del r 2 plus
00:10:41
del u by del r, and that is
equal to the derivative of
00:10:44
r into del u by del r with respect to r.
00:10:49
Now, I can integrate this equation,
and after integrating, we get
00:10:54
u as a function of r as A natural log of r
plus B. So, where A and B are the constants.
00:11:05
And then we can apply the boundary
conditions. So, the boundary conditions are
00:11:19
at r is equal to ri, we have velocity
00:11:32
is equal to u0, so that is the velocity at
which the inner cylinder is moving, and r
00:11:39
is equal to r0, that means outer radius, where the
velocity is 0, because if I go back to the diagram
00:11:49
if I say that the inner radius is 0. ri and
the outer radius is ro, and we say that the
00:11:58
inner cylinder is moving with a velocity u0 and
the outer cylinder is fixed. So, at the inner
00:12:04
cylinder, because of the no-slip condition, the
boundary condition is u is equal to u0 here,
00:12:10
and here, because of the no-slip condition,
the boundary condition is u equal to 0.
00:12:19
Now, after putting the boundary conditions, I can
evaluate the constants. So, these are now a, b.
00:12:31
So, a is equal to u0 divided by the natural log
of ri by r0 and this gives me that b is equal to
00:12:47
minus of u0 natural log of r0 divided by ri by r0.
So, finally, my u as a function of r becomes u0
00:13:05
divided by the natural log of ri by r0
multiplied by the natural log of r by r0.
00:13:16
Now, if this is so, then the shear stress tau
is equal to mu into del u by del r, and that
00:13:31
becomes minus mu u into u0 r
into a natural log of r0 by ri.
00:13:49
Now, we can find out the total force F0. So, the
total force F0 is the force at the surface of the
00:13:58
inner cylinder, and that force is - to
evaluate the force, I have to evaluate the
00:14:04
shear stress at r, which is equal to ri, and
then multiply by just 2 pi ri into length.
00:14:11
So, here we are considering that
the length of the cylinder is L.
00:14:24
If this is my fluid element dx, and if I say that
00:14:31
this dx is a part of the cylinder.
So,
00:14:39
the length of the cylinder is supposed as
L. If I do a cross-sectional view here,
00:14:55
that becomes our total L. So, this
total L, this is dx is just this part.
00:15:28
So that gives me that the force F0 is
00:15:38
2 pi ri multiplied by R into tau evaluated at
ri, and that gives me 2 pi into L into mu into u0
00:15:55
divided by the natural log of r0 by ri.
And then, with this, we can find out the
00:16:05
expression for the active damping,
the damping coefficient. So, the
00:16:19
damping coefficient becomes F0 is
equal to c-equivalent into u0.
00:16:31
That is the force and the damping coefficient is
F0 divided by u0. And finally, the expression is
00:16:41
capital gamma into u where capital gamma is 2
pi L divided by the natural log of r0 by ri.
00:16:54
So, that was for the inactive case when I have
zero applied field, so yield stress is 0.
00:17:02
Now, we will consider a non-zero applied field.
00:17:18
Now, in the non-zero applied field, we have
the fluid constitutive relation is - tau i
00:17:25
into, we know is the sin of gamma
dot plus mu into gamma dot.
00:17:37
So, when the gamma dot is negative, it is
minus tau. When the gamma dot is positive,
00:17:42
it is plus tau i. So, it is tau equal to tau y
plus mu into gamma dot, when gamma dot is positive
00:17:54
and minus tau i plus mu into gamma
dot, when gamma dot is negative.
00:18:11
Now, if I put this constitutive relation, then my
governing differential equation becomes this. This
00:18:26
is equal to tau y by r, and
again, this can be written as
00:18:33
partial of r into del u by del r with
respect to r is equal to tau y by r.
00:18:44
And then finally, on being integrated,
this gives me u as a function of r as
00:18:52
tau y divided by mu into r plus
C natural log of r plus D.
00:19:06
And then, we can evaluate the constants applying
the same boundary conditions that at u is equal to
00:19:15
ri, I mean, at r is equal to ri u is u0 and
at r is equal to ro outer boundary u is 0.
00:19:30
So, these boundary conditions give me
that - applying the boundary conditions,
00:19:42
applying above
00:19:52
boundary conditions, we get C equal to minus 1
by the natural log of ro by ri multiplied by u0
00:20:10
plus tau I divided by mu multiplied by r outer
minus r inner. And then, we have D is equal to
00:20:25
u0 divided by the natural log of r0 by ri plus tau
y by mu, and this is multiplied with minus r0 plus
00:20:42
r minus ri divided by the natural log of r0 by ri.
00:20:52
And then after putting that in the expression
of u, we get u as a function of r as minus tau
00:21:02
y by mu multiplied by ro minus r plus mu
0 plus tau i by mu multiplied by ro minus
00:21:17
ri divided by natural log of r outer by r inner
multiplied by the natural log of r0 by r.
00:21:31
Now, from here, we can find
out tau as a function of r as
00:21:36
minus of mu divided by r multiplied by
the natural log of r zero by ri into
00:21:54
u zero plus tau y by mu multiplied by ro
minus ri, and that finally can be written as
00:22:11
minus mu into u0 divided by r into
natural log of r0 by ri into 1 plus Bi
00:22:26
and where we know that Bi is the Bingham number.
So, Bi is the Bingham number,
00:22:42
and that Bingham number is the ratio of the
yield stress by the viscous stress. So, this is
00:22:48
tau y multiplied by d by mu u0. So, tau y is the
yield stress, and mu u0 by d also has a dimension
00:22:59
of stress, and that is the viscous stress.
So, yield stress by viscous stress is our Bi,
00:23:03
a Bingham number, and that is how
we express our shear stress tau.
00:23:10
And then, from here again, we can
find out the force F0 as 2 pi r0
00:23:19
multiplied by length into tau at ri,
and that gives me this as mu into 1 plus
00:23:29
Bi into capital gamma u0, and we
have already defined capital gamma.
00:23:37
So, this quantity helps us finding
out the active damping coefficient,
00:23:49
and we can write that as F0 by u0 and that we
get as mu into 1 plus Bi into capital gamma,
00:24:03
and then if we do c equivalent of the active case
by the inactive case, we get this as 1 plus Bi
00:24:15
because mu capital gamma was c equivalent 0,
I mean, c equivalent for the inactive case.
00:24:21
So, that is how we analyze it for the
00:24:26
annular passage. So, we can see that
the method is same only due to the
00:24:31
geometry we have to be careful. So, instead of
working on x and y, we are working on x and r,
00:24:38
and accordingly, everything falls into place.
Now, this same procedure can be extended to the
00:24:43
analysis of the flow mode or the valve mode.
Now, in the next week, we will again look into
00:24:51
flow mode analysis and valve mode analysis,
considering again rectangular case,
00:24:55
but using different constitutive relations.
So, this brings us to the end of this lecture
00:25:00
and this week.
Thank you.
![[ Sad Story ] - I didn't like my father](https://ruhnrawpgxrzdrgaohnf.supabase.co/storage/v1/object/public/images/video/ynSdkDZw2FQ/thumbnail.jpg)
