Particle settling by Stokes’ law: If a 10 μm sphere settles with velocity v in a Newtonian oil (SG 0.9, μ = 10 poise), what is the settling velocity of a 20 μm sphere in the same oil (all else equal)?

Introduction / Context:In the creeping-flow regime (Reynolds number much less than 1), Stokes’ law gives the terminal settling velocity of small spheres in a viscous fluid. Understanding how velocity scales with particle diameter is essential in sedimentation and centrifugation design.

Given Data / Assumptions:

• Spherical particles with diameters d1 = 10 μm and d2 = 20 μm.
• Same oil, so identical density and viscosity.
• Stokes regime applies (laminar drag, Re << 1).

Concept / Approach:Under Stokes’ law, terminal velocity v_t is proportional to d^2 for a given density difference and viscosity. Thus, doubling the diameter multiplies v_t by 2^2 = 4.

Step-by-Step Solution:Let v1 ∝ d1^2 and v2 ∝ d2^2.Given d2 = 2 * d1, then v2 / v1 = (d2/d1)^2 = 2^2 = 4.Therefore, v2 = 4 * v1.

Verification / Alternative check:Dimensionally and from the Stokes expression v_t = (g * (ρ_p - ρ_f) * d^2) / (18 * μ), the quadratic dependence on diameter is explicit, confirming the 4x scaling.

Why Other Options Are Wrong:

• Same or one fourth: contradict the d^2 proportionality.
• Twice: would correspond to linear scaling, not Stokes regime.
• Eight times: would imply cubic scaling, incorrect for Stokes settling.

Common Pitfalls:Applying turbulent or intermediate drag correlations instead of Stokes law; forgetting that shape and agglomeration can alter effective drag.

Final Answer:Four times that of the 10 μm particle