Flow past immersed bodies and terminal settling velocity

Why it matters

Thickeners, clarifiers, classifiers, cyclones, spray dryers, slurry pipelines, fluidised beds and pneumatic conveyors all depend on how fast a particle or drop moves relative to the fluid. The terminal settling velocity, obtained from a force balance and the drag curve of a sphere, is the starting point for sizing all of them.

Key ideas

Drag on an immersed body. A body moving relative to a fluid feels a drag force F_D = C_D·A_p·ρu²/2, with A_p the projected area normal to the flow (πd²/4 for a sphere). Drag is the sum of skin friction and form drag; for spheres both depend on the particle Reynolds number Re_p = ρ·u·d/μ, where u is the relative velocity and ρ, μ are fluid properties.

The standard drag curve for a sphere (C_D vs Re_p).

• Stokes' law regime (Re_p < about 0.3; often quoted as < 1): creeping flow, no separation, C_D = 24/Re_p. Drag F_D = 3πμud: one-third form drag, two-thirds skin friction.
• Intermediate regime (about 1 < Re_p < 1000): C_D falls gradually; correlations such as C_D = 18.5/Re_p^0.6 or Schiller–Naumann C_D = (24/Re_p)(1 + 0.15·Re_p^0.687).
• Newton's law regime (about 1000 < Re_p < 2 × 10⁵): separated wake, C_D ≈ 0.44, nearly constant.
• Drag crisis (Re_p ≈ 3 × 10⁵): boundary layer becomes turbulent, C_D drops to about 0.1. Rarely reached by settling particles. Regime limits differ slightly between textbooks; use the ones given in your course.

Terminal settling velocity. A particle released in a still fluid accelerates until gravity is balanced by buoyancy plus drag. Force balance (per particle): (π/6)d³(ρ_p − ρ)g = C_D(πd²/4)(ρu_t²/2), giving u_t = √[4gd(ρ_p − ρ)/(3C_D·ρ)]. Because C_D depends on u_t through Re_p, the solution is in general iterative:

1. Assume a regime (Stokes is a good first guess for fine particles).
2. Compute u_t, then Re_p.
3. Check that Re_p lies in the assumed regime; otherwise switch and repeat.

Criterion K (avoids trial and error). K = d·[gρ(ρ_p − ρ)/μ²]^(1/3). Stokes' law applies for K < 2.6; Newton's law for 68.9 < K < 2360; between is the intermediate range. (This is the McCabe–Smith form; based on Re = 0.3 and 1000.)

Real systems.

• Non-spherical particles: use sphericity ψ and equivalent diameter; C_D is higher than for a sphere.
• Hindered settling: in concentrated suspensions particles settle more slowly because of upflow of displaced fluid and higher effective viscosity. Richardson–Zaki: u = u_t·εⁿ, with ε voidage and n ≈ 4.65 in the Stokes regime, ≈ 2.4 in the Newton regime.
• Wall effect: significant when particle diameter is a noticeable fraction of the container diameter.
• Centrifugal settling: replace g by rω² (cyclones, centrifuges).
• Light particles or bubbles (ρ_p < ρ) rise with the same equations using |ρ_p − ρ|.

Formulas

• F_D = C_D·A_p·ρ·u²/2 — drag (N); A_p projected area (m²), u relative velocity (m/s).
• Re_p = ρ·u·d/μ — particle Reynolds number; ρ, μ of the fluid.
• C_D = 24/Re_p, F_D = 3·π·μ·u·d — Stokes regime.
• u_t = √[4·g·d·(ρ_p − ρ)/(3·C_D·ρ)] — general terminal velocity (m/s); d (m), ρ_p particle density (kg/m³).
• u_t = g·d²·(ρ_p − ρ)/(18·μ) — Stokes' law, Re_p < ≈0.3–1.
• u_t = 1.74·√[g·d·(ρ_p − ρ)/ρ] — Newton's law (C_D = 0.44), 1000 < Re_p < 2 × 10⁵.
• K = d·[g·ρ·(ρ_p − ρ)/μ²]^(1/3) — regime criterion: K < 2.6 Stokes; 68.9 < K < 2360 Newton.
• u = u_t·εⁿ — Richardson–Zaki hindered settling.

Worked examples

Example 1 (standard). Find the terminal velocity of a 50 µm sand particle (ρ_p = 2650 kg/m³) in water (ρ = 1000 kg/m³, μ = 1 × 10⁻³ Pa·s).

1. Assume Stokes regime: u_t = g·d²·(ρ_p − ρ)/(18·μ).
2. u_t = 9.81 × (50 × 10⁻⁶)² × 1650 / (18 × 10⁻³) = 9.81 × 2.5 × 10⁻⁹ × 1650 / 0.018 = 2.25 × 10⁻³ m/s.
3. Check: Re_p = 1000 × 2.25 × 10⁻³ × 50 × 10⁻⁶ / 10⁻³ = 0.11 < 0.3 → Stokes valid. u_t ≈ 2.25 mm/s.

Example 2 (GATE level). A 4 mm glass bead (ρ_p = 2500 kg/m³) falls in water (ρ = 1000 kg/m³, μ = 1 × 10⁻³ Pa·s). Find its terminal velocity, and the largest bead of this glass for which Stokes' law still applies.

1. A careless Stokes calculation gives u_t = 9.81 × (0.004)² × 1500 / 0.018 = 13.1 m/s, with Re_p ≈ 52 000 — clearly invalid.
2. Criterion: [gρ(ρ_p − ρ)/μ²]^(1/3) = (9.81 × 1000 × 1500 / 10⁻⁶)^(1/3) = 24 505 m⁻¹. K = 0.004 × 24 505 = 98.0 → Newton regime (68.9 < K < 2360).
3. u_t = 1.74·√[g·d·(ρ_p − ρ)/ρ] = 1.74 × √(9.81 × 0.004 × 1.5) = 1.74 × 0.2426 = 0.422 m/s.
4. Check: Re_p = 1000 × 0.422 × 0.004 / 10⁻³ = 1690 → within the Newton range ✓.
5. Largest Stokes bead: K = 2.6 → d = 2.6/24 505 = 1.06 × 10⁻⁴ m. u_t ≈ 0.42 m/s; Stokes' law holds only for beads smaller than about 0.11 mm.

Common mistakes

• Using particle density in Re_p; it uses fluid density and viscosity.
• Applying Stokes' law to millimetre-size particles in water, giving absurd velocities.
• Forgetting buoyancy: the driving force uses (ρ_p − ρ), not ρ_p.
• Using surface area instead of projected area in the drag equation.
• Not checking the assumed regime at the end of the calculation.

For GATE CH

Expect terminal-velocity numericals in Stokes and Newton regimes, ratio questions (how u_t changes with d, μ or Δρ), the regime criterion K, the diameter at which a regime limit is reached, hindered settling with Richardson–Zaki, and classification problems (which particles are carried over by an upflowing fluid). Centrifugal settling extends the same equations.

Quick check

1. In Stokes' regime, how does u_t change if d doubles?
2. In Newton's regime, how does u_t change if d doubles?
3. What C_D applies at Re_p = 0.1?
4. Which density appears in the particle Reynolds number?

Answers: 1. Four times. 2. √2 = 1.41 times. 3. 24/0.1 = 240. 4. The fluid density.