This is a shear stress problem where we have crude oil.
It has a specific gravity of 0.85, a viscosity of 2.15*10^-3 pound force second per feet
squared and it flows steadily down a surface inclined with theta equal to 30 degrees below
the horizontal.
It has a film of thickness h which is 0.125 in.
The velocity profile is given below.
Determine the magnitude of the shear stress that acts on the surface.
Our velocity profile mu equals rho times g over mu times "h" "y" minus "y" squared over
2 times the sin of theta.
Our governing equation is the definition of shear stress or tau which is mu du/dy.
Basically in order to solve this problem first we have to find du/dy.
Once we determine that we have to look at the magnitude of the shear stress that acts
on the surface itself.
The first thing we are going to do is find du/dy.
This is a relatively simple integration.
rho times g divided by mu times sin of theta.
Those are all constant so not a function of "y".
All this times "h" minus "y".
To find the shear stress we multiply this by mu and we get that tau equals rho time
g times sin of theta times (h-y).
What is interesting about this is because of the velocity profile the shear stress is
completely independent of the viscosity.
Now we have our equation for shear stress.
Now we have to find it at the surface.
That means that "y" equals 0.
If we have this film of thickness where "h" is 0.125 in.
At y=0 that is where the surface is and the film equals 0.
Now we put our numbers in.
Tau equals 0.85 times 1.94 slugs per feet cubed times 32.2 ft per second squared
times 0.125 in.
To make sure we are in the right units we have that 1 foot equals 12 in.
We multiply this by the sin of theta.
In our case that is 30 degrees.
The sine of 30 degrees is 0.5.
When we plug all of our calculations together we find that the shear stress at the surface
is 0.277 pound force per foot squared.