Fluid mechanics: calc the acceleration of a particle at a point.

[Zarathuztra]

Homework Statement

A velocity field is given by [itex]\vec{V}[/itex]= [Ax[itex]^{3}[/itex] + Bxy[itex]^{2}[/itex]][itex]\hat{i}[/itex] + [Ay[itex]^{3}[/itex] + Bx[itex]^{2}[/itex]y][itex]\hat{j}[/itex]; A=0.2 m[itex]^{-2}[/itex]s[itex]^{-1}[/itex], B is a constant, and the
coordinates are measured in meters. Determine the value and
units for B if this velocity field is to represent an incompressible
flow. Calculate the acceleration of a fluid particle at point
(x, y)=(2, 1). Evaluate the component of particle acceleration
normal to the velocity vector at this point.

Homework Equations

u=[itex]\frac{\partial\Psi}{\partial y}[/itex] v=-[itex]\frac{\partial\Psi}{\partial x}[/itex]

The Attempt at a Solution

I used the above equation to get value equations u and v, there is something I'm missing, a bit of reasoning that has to be made using the given information to determine a constraint that exists since the fluid is incompressible. This should link the rates of change to each other. But that's where I get stuck, having trouble hanging on to all the concepts..

u = [2Bxy][itex]\hat{i}[/itex] + [.6y[itex]^{2}[/itex] + Bx[itex]^{2}[/itex]][itex]\hat{j}[/itex]
v = -[.6x[itex]^{2}[/itex] + By[itex]^{2}[/itex]][itex]\hat{i}[/itex] - [2Bxy][itex]\hat{j}[/itex]

Assumptions:
1. Incompressible flow
2. B is constant

 

[mfb]

"Incompressible" links the derivatives of the velocities in both directions.
This gives you one constraint and allows to calculate B.

 

[Zarathuztra]

so du = -dv? increases of u lead to decreases of v.

 

[mfb]

Not du = -dv, but du/dx = -dv/dy (or the other way round?) should work.

 

[Zarathuztra]

when I try [itex]\frac{du}{dx}[/itex] = -[itex]\frac{dv}{dy}[/itex] I get

[2By][itex]\hat{i}[/itex] + [2Bx][itex]\hat{j}[/itex] = [2By][itex]\hat{i}[/itex] + [2Bx][itex]\hat{j}[/itex]

which doesn't help solve for B, unless I'm messing up somewhere or forgetting something.
The same happens when I take [itex]\frac{du}{dy}[/itex] = -[itex]\frac{dv}{dy}[/itex]

[2Bx][itex]\hat{i}[/itex] + [1.2y][itex]\hat{j}[/itex] = [1.2x][itex]\hat{i}[/itex] + [2By][itex]\hat{j}[/itex]

 

[mfb]

The second equation can be solved for B.

 

[Zarathuztra]

I'm getting B = 1.2/2
does that look right?

 

[mfb]

I think so. It solves the equations for i and j at the same time, which looks good.