Consider steady-state, fully developed flow
in a pipe where the flow is in the x-direction.
So because it is steady state du/dt=0, because
it is fully developed du/dx=0, and what that
means is we have no local or convector acceleration.
Therefore newtons second law simplifies to
the sum of the forces equal 0.
Instead of mass times acceleration.
So lets look at a fluid element, and do a
force balance on it, and again we are assuming
steady state, and fully developed flow.
So we have our pressure force p1 times pi*r^2
on this surface.
On this surface we have p1 minus delta p or
the change in pressure times pi*r^2.
So there is our normal forces.
We also have our tangential force, which is
our shear stress, and this is tau times 2pirL,
and one thing you need to be careful of are
these areas, because this is a tangential
force the area that it is acting on is 2*pi*r*L,
as were the normal forces are acting on the
cross-section area of the pipe.
So when we sum these forces, p1*pi*r^2 minus
and this is minus because it is in the negative
x-direction.
(p1-delta p)pi*r^2 minus, and again it is
minus because of the coordinate system, tau
times 2pi*r*L. This equals 0.
So if we divided through by pi*r^2 and we
simplify it we end up with delta p minus 2*tau*L
over r equals 0.
Or delta p per length equals 2*tau divided
by r.
Now lets look at the pipe wall.
At the pipe wall r equals D divided by 2,
and our tau is our shear stress at the wall.
So we can then say that delta p equals 4*L
times the shear stress at the wall divided
by D. So here is a relationship between the
shear stress at the wall, and our pressure
drop.
Now we can go further.
Remember that tau equals minus mu du/dr.
The reason it is negative is because tau is
greater then 0 when du/dr is less then 0,
which is the case as the velocity decreases
from the center.
So putting this together we can write du/dr
equals negative delta p divided by 2 times
mu, times L all times r, and you will end
up with our u of r equals minus delta p over
4*mu*L) all times r^2 plus C1.
So in order to solve for that C1 we need a
boundary condition.
In other words we need to relate u to some
r.
So one of the things we could say is what
happens when r=0.
When r=0 it is a maximum.
So du/dr equals 0, but if we try and put that
in to our equation it is not going to help
us solve for C1.
So instead what we say is that r=D/2 u=0,
and that is because of the no slip condition.
So lets put those boundary conditions in.
So u equals 0, equals negative delta p over
4*mu*L, and now D^2 over 4 plus C1.
This then allows us to solve for C1.
C1 equals delta p*D^2 over 16*mu times the
L. Next we put that in our equation for u
of r, and that equals minus delta p over 4*mu*L
times r^2 plus delta p*D^2 over 16*mu*L. So
we are going to divided by delta p*D^2 over
16*mu*L. So if we do that.
If we divided this through and we get u of
r equals delta p*D^2 divided by 16*mu*L*(1-4*r^2/D^2)
or we can write this as delta p*D^2 over 16*mu*L*(1-(2r/D)^2),
and when r equals 0 we get the center line
velocity Vc, which is delta p*D^2 over 16*mu*L
