# Shering stress

1. May 8, 2012

### Hybrid_Theory

1. The problem statement, all variables and given/known data
The velocity distribution for the floe of a newtonian fluid between 2 wide parallel plates is given by the equation u=3V/2[1-(y/h)^2] where V is the mean velocity. The fluid has a viscosity of 1.915Ns/m^2. When V=0.61m/s and h=5mm, determine:
A) the shearing stress acting on the bottom wall.
B) the shearing stress acting on a plane parallel to the wall.
C) the shearing stress at the centerline.

2. Relevant equations
ζ=μ du/dy

3. The attempt at a solution
For C) I worked out u=0.915[1-y^2/2.5X10^-3]
From that I worked out ζ=1.915 X 0.915[1-y^2/2.5X10^-3] X 1/dy.

I have no idea how to work out B).

I think A) =0 as there is no velocity there? I could be very wrong saying that though.

Last edited: May 8, 2012
2. May 8, 2012

### LawrenceC

Where is the origin of h? Bottom plate, top plate, center?

3. May 9, 2012

### Hybrid_Theory

Either bottom or top plate, doesn't make a diff in this question. It's like a radius if you will. It says the flow is symetrical.

4. May 10, 2012

### LawrenceC

I hit the h key instead of the y key. Typo. Where is the origin of y?

5. May 10, 2012

### LawrenceC

"For C) I worked out u=0.915[1-y^2/2.5X10^-3]
From that I worked out ζ=1.915 X 0.915[1-y^2/2.5X10^-3] X 1/dy."

How can this be when ζ=μ du/dy?

6. May 10, 2012

### Hybrid_Theory

Origin of y is the centerline.

I dunno, I'm really confused. :/

7. May 11, 2012

### LawrenceC

Here is some more help. The profile is parabolic with the velocity function given by u=V/2[1-(y/h)^2] where V is the average velocity.

The shear stress is given by:
S = mu*du/dy which is the viscosity multiplied by the velocity gradient in the direction perpendicular to the flow. If you plot the velocity profile you will note that when y is zero (centerline), the velocity is maximum. Furthermore the fluid velocity is zero at the walls where y= +h or -h. To determine shear, you take the derivative with respect to y. V, the average velocity, is constant.

So all you need to do for part C is take the derivative and plug in the value of y=h.