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

[Zarathuztra]

Homework Statement

A velocity field is given by \vec{V}= [Ax^{3} + Bxy^{2}]\hat{i} + [Ay^{3} + Bx^{2}y]\hat{j}; A=0.2 m^{-2}s^{-1}, 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=\frac{\partial\Psi}{\partial y} v=-\frac{\partial\Psi}{\partial x}

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

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 \frac{du}{dx} = -\frac{dv}{dy} I get

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

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

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

 

[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.