INSTRUCTOR: In this screen cast, we're going to consider
a steady state, one dimensional in the x direction.
So that means that our only velocity is u,
incompressible fluid that's between two horizontal
infinite, or modeled as infinite, parallel plates.
And what we're going to do with this,
is take our x component of the Navier-Stokes equations
and simplify it to terms that we can then integrate
and set boundary conditions for.
Okay, so lets go through these terms one by one.
Since it's steady state, there's no change in velocity
with respect to time.
Since it's one dimensional.
W is zero.
V is zero.
And by the continuity equation, u isn't zero,
but du dx is zero.
On the left hand side, we assume
that there's a pressure gradient
because nothing in the problem tells us that there isn't.
We know that gravity is in the y direction,
so that term is zero.
If du dx is zero, then d squared u dx squared is zero.
And finally, because we have two horizontal infinite plates,
we can say that u does not change with respect to z.
And so, what we're left with is that zero is equal
to minus dp dx plus mu d squared u dy squared.
So assuming that pressure is not a function of x
and that we can right it as a constant
and moving all the non-partial derivative terms
to one side.
What we end up with is
that one over mu dp dx
is equal to d squared u dy squared.
So, since this is a second order differential equation,
we're going to have to integrate it twice
to get our u of y.
So, lets do the first integration.
So du dy, and I've switched these two,
is equal to one over mu dp dx times y plus a constant.
And now, lets integrate this again and we find
that u of y is equal to one over two mu dp dx
times y squared, plus c1 times y, plus c2.
And so, this is going to be the equation
that we'll always start with when we have
these particular conditions.
So, let's take a look at one case.
So, let's say that we have stationary plate and fluid,
which is proven by the pressure gradient.
So these are parallel.
Flows through them.
And we're going to assume that they're infinite.
So basically, we have a plate here
and we have a plate here,
and we're gonna have fluid that's gonna flow through here
and have some kind of velocity profile.
So, we're gonna call this zero.
This is going to be h,
and this is going to be minus h.
So, because we have two constants, c1 and c2,
we're gonna need two boundary conditions.
So, we can say at y,
and that's what we mean about boundary conditions.
That means, let's find a certain y
and apply some form of the velocity to it.
So at y equals h, we know that u has to be equal to zero
because the plate is stationary.
In addition, at y equals minus h,
again, it's stationary.
So our u has to be equal to zero.
If now I made the origin, or the zero,
in the middle of a place, you don't necessarily have
to do that, but because the flow is going to be symmetrical
because both plates are stationary,
we often put the origin between the two plates.
Okay, so now we're going to look at a different situation.
So, we have a top plate that's moving
at a velocity u and a bottom plate that's stationary.
And again, we're gonna want to use boundary conditions
to find our u of y.
And again, we start with this equation
and our boundary conditions for this particular situation
are gonna be, and here it's not symmetric,
so we'll say that this is the zero,
and we'll say that the spacing in between is h
And sometimes you'll see this spacing as h
and sometimes as b.
So, we know, again, at y equals zero
that u equals zero, because the plate is stationary.
However, here, at y equals h, u is equal to the velocity
of the moving plate.
So if you notice, we still have the same
Navier-Stokes equation that we use.
However, what changes are the boundary conditions.
And finally, what we can look at
is if both plates are moving.
And so here, with both plates moving,
and we'll assume that they're moving at different speeds
from each other.
Now what we're gonna say is at y equals 0,
our u is equal to some constant u1.
And then at y equals h, our u is equal to some constant u2.
And so what we can do, is apply these boundary conditions
for any one of these situations to eventually solve
for our constants.
So we solve for c1 and c2.
And when we do that, if we go back
to our original equation here,
what we find is that we now have a velocity profile
that we can apply at any y.