# Simple Laminar Flow Stress

1. Oct 1, 2006

### TFH

RE: non-compressible, laminar flow, newtonain fluid.

Viscosity is often defined by looking at a fluid between two plates with the upper plate moving at a small velocity relative to the lower plate. The plate moves in the x direction and a velocity gradiant is created. Viscosity is then defined as:

Fx/Axz= Sxy = mu dVx/dy.

Fx = force between adjaycent fluid layers in the x direction.
Axz = area between the fluid layers
Sxy = stress in the direction of the force (x) where y is the direction of the normal to the area.
mu = viscosity
dVx/dy = Velocity gradient in the y direction.

This is the standard way texts define viscosity (unless I've messed it up!).

Next this idea is generalized to multiple dimensions and the stress tensor is defined. This is where I need help. It is argued that for non rotating fluids that the stress tensor must be symmetric. In particular Sxy = Syx. I see this in one sense. Usually this idea is arrived at by noting that a small element would develop an infinite torque or moment if it was not symmetric....okay good. BUT:

Finally my problem: In the simple example, how does a force in the Y direction develop. How does a normal force to the direction of fluid flow, Fy, exist. By definition of a fluid, if Fy exists, the fluid must shear....Any help or reference would be most appreciated.

2. Oct 1, 2006

### Clausius2

That is not the definition of viscosity. And if it is, it is a naive one. Viscosity comes from the molecular level analysis in liquids, and from rotational and vibrational non equilibrium in gases.

The stress tensor is symmetric always, despites it is rotating or not. The symmetry of the stress tensor lies on the fact that each fluid particle is in rotational dynamic equilibrium .

I don't see your point. In your problem Fy=0, because there is no vertical velocity. You don't have any normal stress normal to the xz plane.

3. Oct 2, 2006