# Kinematic viscosity confusion

1. Aug 25, 2011

### DerrickStorm

Fluids with a relatively high viscosity have a high resistance to flow. Fluids with a relatively low viscosity have very little resistance to flow.

Kinematic viscosity is measured in mm2/s. (This is area per second.)

If one "thick" oil and one "thin" oil are poured down onto a flat plane. The thinner oil covers a relatively larger area in than the thick oil does in the same amount of time.

The "thinner" oil clearly therefore covers a larger area per second. So does the thinner oil have a higher viscosity? I know this is incorrect.

How would one make sense of this given the existing unit of kinematic viscosity? Or is the example misleading? Some feedback would be most appreciated. :)

DJ Storm

2. Aug 25, 2011

The important thing to note is that viscosity is effectively just a proportionality constant relating shear stresses to velocity gradients. As such, the units are derived out of necessity for the proportionality to work, not from some physically measurable quantity like how much of a surface it wets at a given rate.

3. Sep 4, 2011

### DerrickStorm

Thanks for the reply however its not very helpful. I've read that viscosity can be described as an ability to dissipate momentum. Is there a way to tie this description to the unit in a way that makes sense practically? An example or analogy would be helpful.

4. Sep 5, 2011

You pour syrup down a chute and it moves slowly and seems to resist flowing down the chute. It is more viscous than water, which readily flows down the same chute.

5. Sep 5, 2011

### Andy Resnick

Perhaps you are confused by the unit area/time in viscosity (e.g., Stokes). You are correct that viscosity is a measure of momentum dissipation. Recall also that viscosity, in one dimensional flow, is used to relate the momentum of neighboring layers of moving fluid (e.g., du/dy). The momentum diffuses normal to the direction of flow and so conceptually, the flux of momentum out of one layer into another layer is expressed as area/time.

Does that help?

6. Apr 4, 2013

### DerrickStorm

Thanks Andy, once I read "normal to the direction of flow" it made complete sense.