# Fluid mechanics question

1. Feb 7, 2014

### math_04

1. The problem statement, all variables and given/known data

Water at 20 degrees C flows in a 0.8cm diameter pipe with a velocity distribution of u(r) = 5[1-r2/(16x10-6)]m/s. Calculate the shear stress on

(a) pipe wall
(b) at a radius where r = 0.2 cm
(c) at centerline of pipe

2. Relevant equations

Shear stress = viscosity x du/dr

3. The attempt at a solution

So what I did first was get du/dr, which is -10/(16x10-6)r = du/dr

I used the value of viscosity for water at room temperature and pressure which is 1.1x10-3

And then I just plugged everything in so shear stress at wall = 1.1x10-3x -10/16x10-6x0 = 0N/m2 at pipe wall. It is x0 because it is the pipe wall so r = 0m

for part (b) I got 1.1x10-3x (-10/16x10-6) x 0.2x10-2 = -1.38N/m2

Looking at the answers, it seems I got it wrong. part (a) answer is 2.5N/m2 and part(b) is 1.25N/m2

Can anyone please tell me where I got it wrong? Somehow I feel like I made a careless mistake somewhere....

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 7, 2014

### haruspex

What makes you think r is 0 at the pipe wall? What would you expect the velocity to be at the pipe wall? For what r is u(r) equal to that?

3. Feb 8, 2014

### math_04

The velocity should be 0 m/s at pipe wall right? Because of the no slip condition?

4. Feb 8, 2014

### haruspex

Yes. So where is r measured from?