In a cyclone separator, consider a small spherical particle of diameter Dp located at radius r from the rotation axis. For Stokes regime slip, what is the radial slip velocity of the particle (relative to gas) when the flow rotates at angular speed ω?\nUse standard symbols: ρ_p (particle density), ρ_f (fluid density), μ (fluid viscosity).

Introduction / Context:
The motion of fine particles inside a cyclone involves a balance between outward centrifugal forcing and inward drag. In the laminar (Stokes) slip regime, an analytical expression relates the radial slip velocity to particle size, fluid properties, and rotational kinematics. This question tests recognition of that canonical relation.

Given Data / Assumptions:

• Spherical particle of diameter Dp at radial location r.
• Angular speed of rotating flow field is ω (forced vortex idealisation locally).
• Low Reynolds number slip of the particle relative to gas; Stokes drag applies.
• Fluid viscosity μ; densities ρ_p and ρ_f.

Concept / Approach:
Radial slip arises from the net outward body force due to centrifugal acceleration acting on the excess mass of the particle over displaced fluid. For Stokes drag, drag force = 3 * π * μ * Dp * v_r. Effective outward body force per particle volume equals (ρ_p - ρ_f) * a_c, where a_c = ω^2 * r is centrifugal acceleration. Equating net body force to Stokes drag on a sphere yields the well known closed form for v_r.

Step-by-Step Solution:

Verification / Alternative check:
Dimensional check: numerator has kg * m^-3 * m^2 * s^-2 * m = kg * m^0 * s^-2; denominator has Pa * s = kg * m^-1 * s^-1, and 18 is dimensionless. Result reduces to m * s^-1, consistent with velocity.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing tangential gas velocity with radial slip; forgetting buoyancy correction; using turbulent drag instead of Stokes where Re_p is small.

Final Answer:
v_r = ((ρ_p - ρ_f) * Dp^2 * ω^2 * r) / (18 * μ)