Couette Flow Video Lecture | Fluid Mechanics for Mechanical Engineering

in this section we consider couette flow
which is a model flow in fluid mechanics
I will define some coordinates x and y
and consider the flow between two flat
plates a distance D apart the top plate
has velocity big V in the X direction
I'm going to consider a small element of
fluid of width Delta X and height Delta
Y and I shall draw that bigger on the
right of the picture now let's consider
the forces on this element of fluid
there is a shear stress at the bottom
but I will label tau a shear stress at
the top that I will label tau plus the
partial derivative d tau by dy times the
distance moved in the y-direction
between the bottom and top face which is
Delta Y in terms of the pressure if the
pressure on the left-hand surface is P
the pressure on the right-hand surface
is P plus partial DP by DX times the
distance moved in the x-direction which
is Delta X therefore the net force on
this control volume is equal to d tau by
dy times Delta Y times Delta X minus DP
by DX times Delta x times Delta Y and if
there is no acceleration which I shall
show later that must be equal to zero in
correct flow by definition of couette
flow the pressure gradient is zero which
implies that by the above equation d tau
by dy must also be equal to zero that is
the gradient of the shear stress with y
must be equal to zero in this particular
case tau only varies in the Y direction
so I can write this as an ordinary
derivative d tau by dy is equal to zero
and that's because the shear stress only
varies in the Y direction now we know
that we can model the shear stress in
terms of the velocity gradient d tau by
dy is equal to d by dy of our
expression for tau which is muga
viscosity times DV X where V is the
velocity in the X Direction dy and
that's equal to mu D - VX by dy squared
if and only if mu is uniform in space ie
does not change in space and that we
know must be equal to zero and you can
integrate that twice to show that the
solution is VX is B Y plus C the
constants B and C are evaluated from the
boundary conditions and if we set VX
equals 0 when y equals 0 then our
expression is that VX is equal to big V
divided by D times Y in other words we
have the expected result that the
velocity varies linearly with Y we need
to show that the flow is not
accelerating in order to do that we can
look at the acceleration of the fluid by
considering the material derivative of
the velocity field if we do that we see
that partial DV by DT is equal 0 because
the flow is steady ne D by DX term
except for pressure is zero because
nothing except pressure changes in the X
direction and V Y is zero throughout and
therefore we see that the fluid is not
accelerating anywhere and that means
that we were justified to consider the
forces around the control volume and set
the net force to zero