Tuesday, September 13, 2011

MrsDrPoe: Potential Flow, Part I

It's Thesis Tuesday on the blog, and today we'll be going down a bit of a rabbit trail involving Potential Flow.  

If we make the assumption that a flow field is irrotational, rotation (and vorticity) in the field is equal to zero and d/dx(v) = d/dy(u), d/dy(w) = d/dz(v), and d/dz(u) = d/dx(w).  Irrotational flow occurrs in regions far from surfaces (inviscid flow); the Bernoulli Equation holds true between any two points in the flow field, not just in two points along a streamline.

We will define some scalar function PHI as the velocity potential function: V = Grad(PHI) or u = d/dx(PHI), v = d/dy(PHI), w = d/dz(PHI).  Applying these definitions for velocity to our relations for irrotational velocity fields (i.e. d/dx(d/dy(PHI)) = d/dx(d/dy(PHI))), we can see that this function does describe irrotational flow.  If we apply the velocity potential function to the continuity equation, we end up with: 

d/dx(d/dx(PHI)) + d/dy(d/dy(PHI)) + d/dz(d/dz(PHI)) = 0


This expression is known as the Laplace Equation.  Inviscid, incompressible, irrotational flow governed by the Laplace Equation is often called potential flow.  Lines of constant PHI are called equipotential lines.


The streamfunction, PSI, is another special two-dimensional function defined as: u = d/dy(PSI), v = d/dx(PSI).  The value of PSI is constant along streamlines.  Applying this to the continuity equation gives:


d/dx(d/dx(PSI)) + d/dy(d/dy(PSI)) = 0


which is the 2D Laplace Equation.  It should be apparent that the streamfunction and the velocity potential function are related.  Streamlines and equipotential lines are actually perpendicular to each other; they form a flow net that can be used to help visualize the flowfield.


Let's look at a two basic potential flows:


Uniform Flow


For uniform flow with constant velocity (U) in the x-direction:


To determine the potential function for this flow, we must examine our relationships between PHI, u, and v.  Integration will be necessary, and it is crucial that the results from integration both parts of the definition must match.

d/dx(PHI) = U

int(1)dPHI = int(U)dx

PHI = Ux + c 
 (If we take the derivative of this function with respect to y, we obtain zero; thus this function holds true for this flow.)

d/dy(PHI) = 0 

To determine the streamfunction we examine the relationships between PSI, u, and v.  Again integration and obtaining matching results are necessary.


d/dy(PSI) = U

int(1)dPSI = int(U)dy

PSI = Uy+c 
(Again this function matches the additional constraint below.)

d/dx(PSI) = 0


For uniform flow with constant velocity alpha degrees from the horizon:


Potential function:


d/dx(PHI) = U*cos(alpha)

int(1)dPHI = int(U*cos(alpha))dx

PHI = U*cos(alpha)*x + c 

d/dy(PHI) = U*sin(alpha)

int(1)dPHI = int(U*sin(alpha))dy

PHI = U*sin(alpha)*y + c

PHI = U*cos(alpha)*x + U*sin(alpha)*y 
(This function matches both constraints placed on the flow.)


Stream function:


d/dx(PSI) = U*sin(alpha)

int(1)dPSI = int(U*sin(alpha))dx

PSI = U*sin(alpha)*x + c 

d/dy(PSI) = U*cos(alpha)

int(1)dPSI = int(U*cos(alpha))dy

PSI = U*cos(alpha)*y + c

PSI = U*sin(alpha)*x + U*cos(alpha)*y
(This function also matches both constraints placed on the flow.)




Now that we've determined the velocity potential function and the streamfunction for these flows, we can use this knowledge to examine any potential flow.  Next week, we'll look at a couple of flow examples in cylindrical coordinates, but until then, happy studying!

0 comments:

Post a Comment