Torque lost using Newton's law of viscosity

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SUMMARY

The discussion focuses on calculating the torque lost between two circular plates sliding over each other, utilizing Newton's law of viscosity. The key variables include the angular velocity (ω), the inner radius (R1), the outer radius (R2), the absolute viscosity (η), and the lubricant film thickness (e). The participant attempts to derive the torque using the formula τ = η ∂u / ∂z but encounters difficulties in determining the derivative ∂R / ∂z. The correct approach involves expressing the velocity profile as a function of both radius and height.

PREREQUISITES
  • Understanding of Newton's law of viscosity
  • Familiarity with angular velocity and its relationship to linear velocity
  • Basic knowledge of calculus, specifically partial derivatives
  • Concept of fluid dynamics, particularly in lubrication theory
NEXT STEPS
  • Study the derivation of velocity profiles in lubricated systems
  • Learn about the application of Newton's law of viscosity in engineering contexts
  • Explore the implications of torque loss in mechanical systems
  • Investigate the effects of varying lubricant film thickness on torque calculations
USEFUL FOR

Mechanical engineers, students studying fluid dynamics, and professionals involved in lubrication and tribology will benefit from this discussion.

felipe de carli
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Homework Statement



Two circular plates, as indicated in the figure, slide one over the other. Calculate the torque lost in this situation. The rotation w of the internal radius of the plates R1, the outer radius of the plates R2, the absolute viscosity of the lubricant between the plates (η) and the thickness of the lubricant film (e) are known. The speed profile is linear in the lubricant.

n0jtZJc.png


Homework Equations



Newton's law of viscosity:
τ = η ∂u / ∂z

The Attempt at a Solution



My first attempt was to get the angular velocity w from the formula v = ω R.
Substituting in the formula, I get τ = η w ∂R / ∂z.
I am stuck in here. How can I derive ∂R / ∂z ? I mean, this derivative should be equal 0, because they are not dependent. Is my approach correct?
Thank you in advance
 

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$$\frac{\partial u}{\partial z}=\frac{\omega R}{e}$$
 
felipe de carli said:
The rotation w of the internal radius
Some text missing in there?

I would start by finding u as a function of r and z.
 

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