# On fluid mechanics

1. May 20, 2005

### Raparicio

Dear Friends,

I'm studying the 2D flows, and I have any questions.

In a square, there's a grid of flow in horizontal an vertical linear flow.

The flow that entry the square must be the same that exits form it.

$$\frac {\partial \Psi_e}{\partial x} -\frac {\partial \Psi_O}{\partial x}= \frac {\partial \Psi} {\partial x}= 0$$

Becouse we have presure, we must apply navier-stokes'

$$\rho [ \frac {\partial {v_x}}{\partial t} + ( \vec{v_x} \frac {\partial}{ \partial x} ) \vec{v_x} ] = \mu \frac {\partial^2} {\partial x^2} \vec {v_x} - \frac {\partial p} {\partial x}$$

How can I vinculate the presure with the flow in x and y directions?

.

2. May 20, 2005

### learningphysics

I think Navier Stokes in 2D would be:

$$\rho [ \frac {\partial {v_x}}{\partial t} + ( \vec{v_x} \frac {\partial}{ \partial x} ) \vec{v_x} + ( \vec{v_y} \frac {\partial}{ \partial y} ) \vec{v_x} ] = \mu \frac {\partial^2} {\partial x^2} \vec {v_x} + \mu \frac {\partial^2} {\partial y^2} \vec {v_x} - \frac {\partial p} {\partial x}$$

and
$$\frac {\partial}{ \partial x} \vec{v_x} + \frac {\partial}{ \partial y} \vec{v_y} = 0$$

3. May 21, 2005

### Raparicio

Aha!

Ah! Thanks learningphysics. Now I'm trying to extract for this formula the flow. Is it possible?