1. Group 16 Dustin Welch Ross Martin Brandon Neary
Determining Rotation and Compressibility in a Flow Given the Velocity Profile

2. Determine whether the following velocity profile is irrotational or not. Is this an incompressible flow? V = (A/r - B) ê + (-B + C)êr A, B, and C are constants. ê and êr are the unit vectors in the  and r directions respectively.

3. Sketch this velocity profile.ê êr r 
As you can see, the velocity changes in the radial
direction by (A/r - B), and in the -direction ( is
in radians) by (-B + C).

4. First, recall the vorticity equation:  = ½(  v)
For an irrotational case,  = 0 = (  v)
(  v) for polar coordinates is expressed as: 1/r [(rv)/r - (vr)/]êz
We must now substitute our velocity profile components into (  v) : rv = A - Br and vr = -B + C => (rv)/r = -B and (vr)/ = -B

5. Now, substituting into (  v) : (  v) = 1/r[(-B) - (-B)] = 0
Thus, our flow is indeed irrotational.
Now we must determine if the flow is incompressible.
For incompressible flow, ( · v) = (from the continuity equation)
Express ( · v) in polar coordinates: /r [(rvr)/r + (v)/]

6. rvr = -Br + Cr and rv = A/r - B => (rvr)/r = -B + C and (v)/ = -A/r²
Our continuity equation can now after substitution be expressed as: ( · v) = -B/r + C/r - A/r³
This does not equal ‘0’, and thus the flow is not incompressible.