So lets consider an oil that is flowing in a pipe.
The pipe has a diameter of 0.03 m, and the oil has a viscosity of 0.40 N*s/m^2, with
a density of 900 kg/m^3.
What we are looking for is what is the volumetric flow rate if you have a pressure drop of 30
kPa in a 10 meter section of pipe.
Now normally the first thing you would do is calculate the Re number so that you can
determine whether the flow was laminar or turbulent, but here because we don't have
flow rate therefore we don't have the velocity.
We are going to have to assume laminar flow, and use the equation, and then go back and
check whether it is actually laminar.
So for laminar flow our Q or our flow rate equals delta p times pi times the diameter
to the 4th divided by 128 times mu times l.
If we put in in our pressure drop remember that is 30 kPa.
So that 30 times 10 to the 3 Pa.
Our diameter is 0.03, our mu is 0.40 Ns/m^2, and our length is 10 m.
So if we calculate this we find that the flow rate
is 1.5x10^-4 m^3/s.
Now we have to go back and check the Re number to see that this is an equation that we can
use.
So we find our average velocity, which we would use in the Re number, which is our Q
over A. Which is 1.5x10^4 m^3/s, divided by pi D^2, divided
by 4, and when we calculate this we find that our velocity is 0.21 m/s.
This is our average velocity.
So our Re number which is rho, v, D, over mu, and by the way it is important that we
use D as the length scale in the Re number for pipe flow, not l or the length of the
pipe, and so this equals our 900 kg/m^3, times 0.21 m/s, times 0.03 m, all divided by 0.40
Ns/m^2, and when we calculate that we find that the Re number is 14.2, which is certainly
less then 2100, which is the critical Re number and we can use the equation that we used.