Settling regimes and size dependence: the terminal velocity of a particle moving in a fluid varies as d_p^n. What is n for Newton’s law regime (constant drag coefficient, high Re)?

Difficulty: Medium

Correct Answer: 0.5

Explanation:


Introduction:
Terminal settling velocity depends on particle size differently in different drag regimes. Recognizing the exponent n in v_t ∝ d_p^n helps choose the correct correlation for classifier and cyclone design when Reynolds numbers are high and drag coefficient is nearly constant (Newton regime).


Given Data / Assumptions:

  • Single rigid particle settling in a fluid.
  • Newton’s law regime (high particle Reynolds number).
  • Drag coefficient C_d ≈ constant for the relevant Re range.


Concept / Approach:
At terminal velocity, drag balances apparent weight: (1/2) C_d ρ_f A v_t^2 = (π/6) d_p^3 (ρ_p − ρ_f) g. Solving for v_t gives v_t ∝ sqrt(d_p). Hence n = 0.5 in the Newton regime, contrasting with Stokes regime (n = 2) at low Re.


Step-by-Step Solution:
Write force balance: (1/2) C_d ρ_f (π d_p^2/4) v_t^2 = (π d_p^3/6) (ρ_p − ρ_f) g.Cancel common factors and rearrange: v_t^2 ∝ d_p.Therefore v_t ∝ d_p^0.5, i.e., n = 0.5.


Verification / Alternative check:
Dimensional analysis at high Re with constant C_d leads to v_t ∝ [ (ρ_p − ρ_f) g d_p / ρ_f ]^0.5, again yielding the square-root dependence.


Why Other Options Are Wrong:

  • n = 1, 1.5, 2, 3: correspond to other, non-Newtonian or low-Re situations; Stokes gives n = 2, not applicable here.


Common Pitfalls:
Applying Stokes’ n = 2 beyond its creeping-flow limit; always check Reynolds number regime before choosing n.


Final Answer:
0.5

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