When P
L
=0 and P o
=1500 Pa the velocity profile assuming the z axis run parallel to the plates:
Carlos Sarria
MANE 6960
HW 03
After obtaining an expression for the velocity, we integrate to get the volumetric flow:
Q=1.25E-6 m 3 /sec/unit width
When we reverse the pressure gradient, we get the opposite:
After obtaining an expression for the velocity, we integrate to get the volumetric flow:
Q=1.25E-6 m 3 /sec/unit width

For P
0
=1500 Pa and P
L
=0, the velocity profile looks like:
It is clear that the layers of fluid nearby the moving plate are moving at the velocity of the plate. As you move away from the bottom plate, the velocity decreases until it is zero at the opposite plate in a somewhat
After integrating the velocity profile expression we get the volumetric flow rate:
Q=5.12E-4 m 3 /sec/unit width
For P
0
=15000 Pa and P
L
=0, the velocity profile looks like:
The fluid layers near the moving bottom plate will move at the plate’s velocity
As we increase the pressure gradient, it becomes evident that the velocity of the fluid increases significantly, with its peak value halfway between the two plates.
Q=6.25E-4 m 3 /sec/unit width
For P o
=0 and P
L
=1500Pa the velocity profile looks like:

The profile is very similar to the situation where P
0
=1500 Pa and P
L
=0, and this is due to the fact that the velocity of the bottom plate is a much bigger driver of the velocity of the fluid compare to the pressure gradient.
Q=4.9E-4 m 3 /sec/unit width
When P
0
=0 Pa and P
L
=15000, the velocity profile looks like this:
Now we see that the velocity profile is not linear. Near the bottom the plate the fluid layers move with a velocity close to the plate’s velocity. Then as expected, the fluid’s velocity decreases until it reaches zero on the opposite side.
Q=-3.7E-4 m 3 /sec/unit width

For the first part, when P
0
=1500 Pa and P
L
=0, the numerical solution for the velocity field is:
It does not match the exact solution. One could argue that maybe the mesh is not fine enough. To estimate the volumetric flow rate, we can also integrate in COMSOL to obtain:
This value also does not match the analytical solutions. A more thorough investigation is needed to find out the root cause of the discrepancy.
When we reverse the pressures, we obtain the following velocity profile:
As expected, it’s the opposite, but once again, the magnitudes don’t match the analytical solutions.
For part b, when P
0
=1500 Pa and P
L
=0 we obtain a velocity profile that closely matches the analytical solutions:

The calculated volumetric flow rate is:
Again, this is similar to the value obtained by the analytical solutions.
When P
0
=15000 Pa and P
L
=0 we obtain the following velocity profile:
It doesn’t quite match the analytical solution. The volumetric flow rate is:
When P
0
=0 Pa and P
L
=1500 we obtain the following velocity profile:

Again, this looks similar to the analytical solutions. The calculated volumetric flow is:
Volumetric flow is also similar to the analytical solutions.
When P
0
=0 Pa and P
L
=15000 we obtain the following velocity profile:
This plot suggests that there is a change in velocity direction, which can also be seen in the streamlines plot below. This does not match the analytical solutions.

The calculated volumetric flow for this case is:
Which makes sense, since due to the high pressure gradient in the negative X direction, very little fluid is flowing from the moving plate inertia.
For Part C, the velocity profile looks like: