A slurry contains 30% solids by volume (spherical sand) settling in a viscous oil. The measured hindered settling velocity is 4.44 μm/s. If the Richardson–Zaki index n = 4.5, what is the single-particle terminal velocity in the same fluid?

Introduction / Context:Hindered settling describes the reduced settling velocity of particles in concentrated suspensions. The Richardson–Zaki correlation relates hindered velocity to the single-particle terminal velocity and solids fraction. This is a staple calculation in sedimentation and classification design.

Given Data / Assumptions:

• Solids volume fraction, φ = 0.30 (therefore fluid fraction = 1 − φ = 0.70).
• Hindered settling velocity, V_h = 4.44 μm/s.
• Richardson–Zaki index, n = 4.5 (appropriate to the regime and Re).

Concept / Approach:The correlation is V_h = V_t * (1 − φ)^n, where V_t is the single-particle terminal velocity. Rearranging gives V_t = V_h / (1 − φ)^n. We substitute the given numbers and compute.

Step-by-Step Solution:

Verification / Alternative check:The terminal velocity must exceed the hindered velocity because crowding impedes motion; 22.1 μm/s is indeed greater than 4.44 μm/s, consistent with physics.

Why Other Options Are Wrong:

Common Pitfalls:Using φ (solids) instead of (1 − φ); misapplying n for the wrong Reynolds regime; unit mistakes converting μm/s.

Final Answer:22.1 μm/s