# Pressure in the z-axis

1. Nov 2, 2004

### Fritz

If you have a rectangular fluid element verticle z-axis and a horizontal x-axis, pressure is constant along the x=axis, but p=f(z) along the z-axis.

If $$\frac{dP}{dz}=-density.g$$

is the integral of $$\frac{dP}{dz}$$ equal to p(z) - p(z1) = -density.g(z-z1)?

Last edited: Nov 2, 2004
2. Nov 2, 2004

### Tide

That would be true only if the denisity is constant. Also, be careful with your signs on the right hand side.

3. Nov 2, 2004

### Fritz

Assume it is a liquid and therefore (practically) incompressible.

4. Nov 2, 2004

### arildno

Incompressibility does not entail that the density field is constant.

5. Nov 2, 2004

### Fritz

Density = mass / volume.

If the volume doesn't change (incompressible), surely the mass wouldn't change?

6. Nov 2, 2004

### arildno

In continuum mechanics, we use density FIELDS, rather than particle densities.

Incompressibility entails that the density of A FLUID PARTICLE remains constant; but because the individual fluid particles may jump around in space, it does not follow that the density FIELD is constant.

The field measures the density of whatever particle happens to be AT A FIXED POINT IN SPACE; since it may be different fluid particles which occupy that point at DIFFERENT TIMES, the density field, evaluated at that point may change in time.

If you suppose in addition that all particles has the SAME density, then the density field will be constant.
Get it?