Free and hindered settling of particles

Terminal velocity from the force balance, Stokes/intermediate/Newton regimes with the K criterion, and hindered settling by Richardson–Zaki.

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Why it matters

The terminal settling velocity is the single number behind thickeners, clarifiers, classifiers, elutriators, cyclones and centrifuges. Get the flow regime wrong — for example, apply Stokes' law to a millimetre-sized grain — and the predicted velocity can be off by orders of magnitude, so a settler sized on it will fail.

Key ideas

Forces on a settling particle. A particle released in a still fluid accelerates under gravity minus buoyancy until the drag force balances it. From then on it moves at the constant terminal velocity u_t. For a sphere of diameter D_p the force balance is gravity − buoyancy = drag: (π/6)D_p³(ρ_p − ρ)g = C_D·(π/4)D_p²·ρ·u_t²/2. The drag coefficient C_D depends on the particle Reynolds number Re_p = ρ·u_t·D_p/μ, built with the fluid density and viscosity.

Regimes for spheres.

  • Stokes' law region, Re_p below about 1 (strictly < 0.3 for 3% accuracy, often quoted < 1 or < 2 in exams): C_D = 24/Re_p, and u_t = g·D_p²(ρ_p − ρ)/(18μ). Velocity grows with D_p², falls with μ, and is independent of fluid density except through (ρ_p − ρ).
  • Intermediate region, roughly 1 < Re_p < 1000: C_D falls gradually (a common fit is C_D ≈ 18.5/Re_p^0.6); solve by trial or from the C_D–Re chart.
  • Newton's law region, about 1000 < Re_p < 2 × 10⁵: C_D ≈ 0.44 is constant and u_t = 1.75·√(g·D_p(ρ_p − ρ)/ρ). Velocity grows only with √D_p and is independent of viscosity.

Choosing the regime without trial and error. The criterion K = D_p·[g·ρ(ρ_p − ρ)/μ²]^(1/3) contains no velocity. K < 2.6 → Stokes; 68.9 < K < 2360 → Newton; in between, intermediate. Always confirm by computing Re_p afterwards.

Non-spherical particles have higher drag and lower u_t than spheres of equal volume; sphericity-dependent C_D curves are in your data book. Wall effects slow a particle when D_p is not small compared with the vessel diameter.

Free versus hindered settling. Free settling means a particle is far from walls and from other particles (dilute suspensions, solids below about 1% by volume). In a concentrated suspension, settling particles displace fluid that must flow upwards past their neighbours; the effective density and viscosity of the suspension rise and particles interfere. This is hindered settling, and the velocity is lower than u_t. The Richardson–Zaki equation correlates it with the void fraction ε (= 1 − volume fraction of solids): the exponent n is about 4.65 in the Stokes region and about 2.4 in the Newton region, varying in between.

Equal-settling particles. Two particles of different density but the same terminal velocity are "equal settling". Their size ratio is the basis of gravity classification and jigging, covered in a later topic.

Formulas

F_g − F_b = (π/6)·D_p³·(ρ_p − ρ)·g, F_D = C_D·A_p·ρ·u_t²/2 with A_p = π·D_p²/4 General: u_t = √[4·g·D_p·(ρ_p − ρ) / (3·C_D·ρ)] Stokes: u_t = g·D_p²·(ρ_p − ρ) / (18·μ), valid for Re_p < about 1 Newton: u_t = 1.75·√[g·D_p·(ρ_p − ρ) / ρ], valid for 1000 < Re_p < 2 × 10⁵ Re_p = ρ·u_t·D_p / μ K = D_p·[g·ρ·(ρ_p − ρ) / μ²]^(1/3): K < 2.6 Stokes, 68.9 < K < 2360 Newton Hindered (Richardson–Zaki): u_s = u_t·εⁿ, n ≈ 4.65 (Stokes region) to 2.4 (Newton region)

  • u_t: terminal velocity (m/s); D_p: particle diameter (m); ρ_p, ρ: particle and fluid density (kg/m³); μ: fluid viscosity (Pa·s); g = 9.81 m/s²; C_D: drag coefficient (−); ε: void fraction (−); u_s: hindered settling velocity (m/s).

Worked examples

Example 1 (standard): Stokes regime plus hindered settling. Silica particles, D_p = 50 µm, ρ_p = 2650 kg/m³, settle in water (ρ = 1000 kg/m³, μ = 1 × 10⁻³ Pa·s). Find u_t, then the hindered velocity at 20% solids by volume.

  1. K = 50 × 10⁻⁶ × [9.81 × 1000 × 1650/(10⁻³)²]^(1/3) = 50 × 10⁻⁶ × 25 296 = 1.26 < 2.6 → Stokes.
  2. u_t = 9.81 × (50 × 10⁻⁶)² × 1650/(18 × 10⁻³) = 2.25 × 10⁻³ m/s (2.25 mm/s).
  3. Check: Re_p = 1000 × 2.25 × 10⁻³ × 50 × 10⁻⁶/10⁻³ = 0.11. Stokes confirmed.
  4. ε = 1 − 0.20 = 0.80; u_s = u_t·ε^4.65 = 2.25 × 10⁻³ × 0.354 = 7.97 × 10⁻⁴ m/s. The particles settle at only about a third of their free velocity.

Example 2 (GATE level): Newton regime. A glass sphere of 5 mm diameter (ρ_p = 2500 kg/m³) falls in water (ρ = 1000 kg/m³, μ = 10⁻³ Pa·s). Find u_t.

  1. K = 5 × 10⁻³ × [9.81 × 1000 × 1500/10⁻⁶]^(1/3) = 5 × 10⁻³ × 24 505 = 122.5 → Newton region.
  2. u_t = 1.75 × √(9.81 × 5 × 10⁻³ × 1500/1000) = 1.75 × √0.07358 = 1.75 × 0.2713 = 0.475 m/s.
  3. Check: Re_p = 1000 × 0.475 × 5 × 10⁻³/10⁻³ = 2370, within 1000–2 × 10⁵. Correct regime.
  4. Had Stokes' law been applied blindly, u_t = 9.81 × (5 × 10⁻³)² × 1500/(18 × 10⁻³) = 20.4 m/s — about 40 times too high.

Common mistakes

  • Applying Stokes' law to large or dense particles without checking Re_p.
  • Using the particle density in Re_p; it is the fluid density.
  • Writing (2/9)·r²… and then substituting the diameter for r. The diameter form has 18 in the denominator.
  • Using the solids volume fraction instead of the void fraction ε in Richardson–Zaki.
  • Assuming hindered settling is faster because particles "drag each other down" — it is slower (except for flocculated clusters, which behave as larger particles).

For GATE CH

Expect: terminal velocity in the Stokes or Newton region, often with a regime check; the size of a particle with a given settling velocity; ratio of settling velocities of two particles; hindered settling with Richardson–Zaki; equal-settling size ratio. Practise the K criterion and always state the Re_p check in your working.

Quick check

  1. In the Stokes region, how does u_t change if the particle diameter doubles?
  2. In the Newton region, how does u_t depend on viscosity?
  3. What is the value of n in u_s = u_t·εⁿ at low Reynolds number?
  4. Which density is used in the particle Reynolds number? Answers: 1. It increases four times. 2. It is independent of viscosity. 3. About 4.65. 4. The fluid density.

Interview questions

All Mechanical Operations interview questions

Try answering each one aloud before you open it.

  1. 1.What is free settling of particles?Concept

    Free settling of particles occurs when particles fall through a fluid without any interference from other particles. The motion is influenced only by gravity and the fluid's resistance, and the particles are sufficiently spaced apart so that they do not affect each other's movement.

  2. 2.What is hindered settling of particles?Concept

    Hindered settling occurs when particles settle through a fluid but are close enough to each other that their movements are affected by the presence of other particles. This results in a slower settling velocity compared to free settling due to increased resistance and interactions among particles.

  3. 3.Explain the difference between free and hindered settling.Concept

    The main difference between free and hindered settling is the interaction between particles. In free settling, particles are far apart and settle independently, while in hindered settling, particles are close together, causing interactions that slow down their settling velocity. Hindered settling typically occurs at higher particle concentrations.

  4. 4.Why is the concept of terminal velocity important in the settling of particles?Application

    Terminal velocity is important because it represents the constant speed that a particle reaches when the gravitational force pulling it down is balanced by the drag force of the fluid. Understanding terminal velocity helps in designing equipment for separation processes, such as sedimentation tanks, where knowing the settling speed is crucial for efficiency.

  5. 5.What factors affect the settling velocity of particles in a fluid?Application

    The settling velocity of particles is affected by several factors, including particle size, shape, and density, as well as the viscosity and density of the fluid. Larger, denser particles tend to settle faster, while higher fluid viscosity slows down the settling process.

  6. 6.How does particle concentration affect the settling process?Application

    As particle concentration increases, the likelihood of particle interactions also increases, leading to hindered settling. This results in a reduced settling velocity compared to free settling, as the particles create a network that increases resistance to movement.

  7. 7.What happens if the fluid viscosity increases during the settling process?Application

    If the fluid viscosity increases, the drag force on the particles also increases, which slows down their settling velocity. This can lead to longer settling times and may require adjustments in process design to achieve the desired separation efficiency.

  8. 8.Calculate the terminal velocity of a 100 µm spherical particle of density 2500 kg/m³ settling in water (ρ = 1000 kg/m³, μ = 0.001 Pa·s), and check the regime.Numerical

    Assume Stokes' law: u_t = g·D²(ρ_p − ρ)/(18μ) = 9.81 × (10⁻⁴)² × 1500/(18 × 10⁻³) = 8.2 × 10⁻³ m/s. Check: Re_p = ρ·u_t·D/μ = 1000 × 8.2 × 10⁻³ × 10⁻⁴/10⁻³ = 0.82, below about 1, so Stokes' law is acceptable (the criterion K = 2.45 < 2.6 says the same). For a much larger particle, say 10 mm, Stokes' law would be badly wrong and Newton's law u_t = 1.75√(gD(ρ_p − ρ)/ρ) should be used.

  9. 9.A particle settles in a fluid at a terminal velocity of 2 mm/s. The fluid density is 1200 kg/m³, the particle density 2600 kg/m³ and the fluid viscosity 0.002 Pa·s. Assuming Stokes' law, find the particle diameter.Numerical

    Rearrange Stokes' law: D = √[18μ·u_t/(g(ρ_p − ρ))] = √[18 × 0.002 × 0.002/(9.81 × 1400)] = √(5.24 × 10⁻⁹) = 7.24 × 10⁻⁵ m, about 72 µm. Check Re_p = 1200 × 0.002 × 7.24 × 10⁻⁵/0.002 = 0.087, well inside the Stokes region, so the assumption holds.

  10. 10.Explain why sedimentation tanks are designed with specific dimensions and flow rates.Application

    Sedimentation tanks are designed with specific dimensions and flow rates to optimize the settling process. The dimensions ensure that particles have enough time to settle before the fluid exits the tank, while the flow rate is controlled to prevent turbulence that could resuspend settled particles. Proper design ensures efficient separation and reduces the need for additional treatment processes.

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