This example shows you how to use boundary conditions and solve for constants using those
boundary conditions.
This is more complicated problem.
Then perhaps you have seen before, but once you realize what the boundary conditions are
and also you keep track of your math you should be able to do a problem like this any time.
So we have 2 immiscible fluids.
So they each have there own viscosity, and they come through a pipe with a circular cross
section, and there is a pressure difference, and as I said there are 2 different viscositys.
One viscosity mu1 that fluid forms an inner core, and the one mu2 forms an outer shell
around the core and we are assuming that the interface between them is at D/4.
So a general solution for a pipe flow with a radial velocity distribution has been obtained
by using the Naiver-Stokes equations, and we come out with u equals r^2/(4mu) times
dp/dx plus (c1/mu) ln of r plus c2.
However we have two different velocity profiles so we will call that u1, nd u2 is going to
be exactly the same.
Except our viscositys are different, and our constants will be different as well.
So we are going to need 4 boundary conditions, because we have 4 constants, and lets draw
a picture.
There is the inside.
There is the outside, and the first boundary conditions that we are going to look at is
that r=0.
So the very center of the pipe.
The velocity has to be finite.
If that is the case the only way that can happen is if c1=0 because ln(0) is infinite.
So the only way to make that, keep the velocity finite is to get rid of that whole term all
together.
We also have at r=D/4.
So that is the interface between the 2.
We know that u1 is going to equal u2, and we also know.
This is our third boundary condition that the shear stresses are going to be equal.
Our final boundary condition is going to no slip condition when r=D/2, which is at the
wall, the pipe wall.
Our velocity equals 0.
So we have already solved for C1, and there are number of ways to go about this using
any one of your boundary conditions.
You can solve for your constants.
In this particular case using tau1=tau2 is probably the easiest boundary condition to
work with, and the reason for that is.
Since tau equals mu*du/dr, when we take a derivative of either one of our velocities
we are going to end up dropping out some of our constants.
So lets do that here.
Lets look at tau1.
So tau1 is mu1*du1/dr is (r/(2mu1)) (dp/dx) and notice that the mu1s drop out.
So our tau1 equals (r/2)dp/dx.
Now lets look at our tau2.
So that is mu*du2/sr, which is (r/2mu2)(dp/dx), but now we have to keep in that C3.
We have gotten rid of C1 already.
Mu2/r, again we can rewrite this as tau2 equals (r/2)(dp/dx) plus C3/r, and if you take a
look at them.
Tau1 has to equal tau2.
That means r/2 regardless of what the r is (dp/dx) has to equal our r/2(dp/dx) plus c3/r,
and the only way for that to happen is for that to be 0.
So C3=0.
So now we have taken care of 2 of our constants.
They are both 0, C1, and C3.
So the next thing to do is to look at our boundary conditions.
At r=D/2, u2=0.
So we can write our u2.
(D/2)^2 that is our r^2, over 4*mu2 times dp/dx plus C4 equals 0/ That solves for C4,
which is minus (D^2/16mu2)(dp/dx).
So now we have 3 out of the four.
So we are left with of course with the hardest for last, and this is at our r=D/4, u1=u2,
and we should always leave this one for last.
Just because when we have all the constants in it, we haven's solved for it we are going
to have equations with to many unknowns, and we are going to have to have more information
in order to solve it.
So here we are with our final boundary condition and lets set them equal to each other.
So this is at r = D/4.
So that is (D/4)^2 over 4mu1*(dp/dx), plus C2.
So that is the only constant that we have not solved for.
Equals (D/4)^2/(4mu2)*(dp/dx).
Now our constant minus D^2 over 16mu2*(dp/dx).
So that lets us solve for C2, which is D^2/64(dp/dx).
So I am pulling that out of each on of the expressions.
1/mu2, minus 4/mu2, minus 1/mu1, and when we put this together we get our( D^2/64)(dp/dx),
and a took the minus sign out there, and we are left with 3mu1 plus mu2, all over mu1
times mu2.
So when we put all of our constants together that we have solved for, and put them back
into our u1 and u2.
We end up with the following equations.
So our u1 is (1/4mu1)(dp/dx)(r^2-(D/2)^2, and here is our (mu2+3mu1)/(4mu2).
Our u2 is (1/4mu2)(dp/dx)(r^2-(D/2)^2), and the difference in complexity between these
two is because we had a boundary condition were u2=0 at the wall.