Dimensional Analysis:Relating Particle Diam, Viscosity, & Weight

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SUMMARY

The discussion focuses on the relationship between terminal velocity (ut) of a spherical particle in a fluid and its diameter (d), dynamic viscosity (μ), and buoyancy weight (W), defined as the product of the density difference (∆ρ) and gravitational acceleration (g). The participant initially struggled with the complexity of the variables but received guidance to simplify the weight term to W = ∆ρ⋅g. This clarification led to the correct application of Stokes' law in determining the relationship among these variables.

PREREQUISITES
  • Understanding of Stokes' law in fluid dynamics
  • Knowledge of dynamic viscosity (μ) and its significance
  • Familiarity with buoyancy concepts and density differences (∆ρ)
  • Basic principles of terminal velocity in fluid mechanics
NEXT STEPS
  • Study the derivation and applications of Stokes' law in various fluid scenarios
  • Explore the impact of particle diameter (d) on terminal velocity in different fluids
  • Investigate the role of dynamic viscosity (μ) in fluid resistance and flow behavior
  • Examine buoyancy effects in fluid mechanics and their mathematical representations
USEFUL FOR

Students in physics or engineering, particularly those studying fluid dynamics, as well as researchers and professionals analyzing particle behavior in fluids.

evoke1l1
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Homework Statement


It is found the terminal velocity ut of a spherical particle in a fluid depends upon the diameter d of particle, the dynamic viscosity μ of fluid and the buoyancy weight W of the particle [given by the difference in density between the particle and the fluid (∆ρ) × gravitational acceleration (g)]. Determine the nature of the relationship between these variables.

Homework Equations



The Attempt at a Solution

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I'm not quite sure if I'm on the right track and where to go next from here? Any guidance or a nudge in the right direction would be appreciated. I can't help but think I have gone slightly wrong due to the number of indices to solve for.
 
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Perhaps you should make the weight a single term? That is, treat ∆ρ⋅g as one term. That, after all, is how it is described in the problem statement.
 
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Thank you. That was a big help. I obviously missed taking ∆ρ⋅g as W and got stuck in a rut. The end equation I've now determined to be as per Stokes law.

Many thanks.
 

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