How Do You Calculate Shear Stress in a Flowing Fluid?

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SUMMARY

The discussion focuses on calculating shear stress in a flowing fluid, specifically water at 20 degrees Celsius in a 0.8 cm diameter pipe with a velocity distribution described by the equation u(r) = 5[1 - r²/(16 x 10⁻⁶)] m/s. The correct shear stress at the pipe wall is established as 2.5 N/m², while the shear stress at a radius of 0.2 cm is 1.25 N/m². The participant initially miscalculated the shear stress at the wall due to an incorrect assumption about the velocity at that point, neglecting the no-slip condition which dictates that the velocity at the pipe wall is zero.

PREREQUISITES
  • Understanding of fluid mechanics principles, particularly shear stress calculations.
  • Familiarity with the no-slip condition in fluid flow.
  • Basic knowledge of calculus for differentiating velocity profiles.
  • Knowledge of the properties of water, including viscosity at room temperature.
NEXT STEPS
  • Study the concept of the no-slip condition in fluid dynamics.
  • Learn how to derive velocity profiles in cylindrical coordinates.
  • Explore the application of the shear stress formula in various fluid flow scenarios.
  • Investigate the effects of temperature on the viscosity of fluids, particularly water.
USEFUL FOR

Students in mechanical or civil engineering, fluid mechanics researchers, and professionals involved in hydraulic systems or pipe flow analysis.

math_04
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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

 
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math_04 said:
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?
 
The velocity should be 0 m/s at pipe wall right? Because of the no slip condition?
 
math_04 said:
The velocity should be 0 m/s at pipe wall right? Because of the no slip condition?

Yes. So where is r measured from?
 

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