How Do You Calculate Shear Stress in a Flowing Fluid?

[math_04]

Homework Statement

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

Homework Equations

Shear stress = viscosity x du/dr

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...

Homework Statement

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[haruspex]

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.

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

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?

 

[math_04]

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

 

[haruspex]

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

Yes. So where is r measured from?

 

[HalfLostAndSearching]

First, let's make sure we have the correct equation for shear stress. The equation you used, shear stress = viscosity x du/dr, is actually the equation for shear rate. The correct equation for shear stress is:

Shear stress = viscosity x (du/dr)

So, for part (a), when r = 0, the shear stress will also be 0. This is because at the pipe wall, there is no velocity gradient (du/dr = 0) and therefore no shear stress.

For part (b), you have the correct equation, but you have made a mistake in your units. The units for viscosity are Pa·s, so you need to convert your velocity from m/s to cm/s in order to get the correct units for shear stress, which is Pa. So the correct calculation would be:

Shear stress = (1.1x10^-3 Pa·s) x (-10/16x10^-6 s^-1) x (5 cm/s) = -3.44 Pa = -3.44 N/m^2

For part (c), at the centerline of the pipe, r = 0.4 cm. So the calculation would be:

Shear stress = (1.1x10^-3 Pa·s) x (-10/16x10^-6 s^-1) x (5 cm/s) = -6.88 Pa = -6.88 N/m^2

So the correct answers are:

(a) Shear stress at pipe wall = 0 N/m^2
(b) Shear stress at r = 0.2 cm = -3.44 N/m^2
(c) Shear stress at centerline = -6.88 N/m^2