Particle characterisation: size, shape and sphericity

Why it matters

Almost every mechanical operation — screening, crushing, settling, filtration, fluidisation, conveying — depends on how big the particles are and what shape they have. The drag on a particle, the pressure drop through a packed bed or filter cake, and the energy needed to grind a solid all scale with particle size and with surface area per unit volume. A one-number shape factor, sphericity, lets you carry irregular real particles into equations that were derived for spheres.

Key ideas

Size of an irregular particle. A sphere has one obvious size, its diameter. A real particle does not, so we use an equivalent diameter — the diameter of a sphere that matches the particle in one chosen property:

• Volume-equivalent diameter D_v: sphere of the same volume.
• Surface-equivalent diameter D_s: sphere of the same surface area.
• Sieve diameter: the side of the smallest square aperture the particle passes through; in practice the mean of two adjacent screen apertures.
• Stokes (sedimentation) diameter: sphere of the same density that settles at the same terminal velocity in the laminar region.
• Projected-area diameter: circle with the same area as the particle's image (microscopy). Different methods give different "sizes" for the same particle, so always state which diameter you are using. For granular solids coarser than about 75 µm, sieving is standard; finer powders are measured by sedimentation, laser diffraction or microscopy/image analysis.

Shape. Shape controls packing (bed voidage), flowability, drag and how a cake forms in a filter. Shape can be described by aspect ratio, roundness or angularity, but in unit-operation calculations the most used single descriptor is sphericity.

Sphericity ψ (Wadell). The surface area of a sphere having the same volume as the particle, divided by the actual surface area of the particle. Because a sphere has the least surface for a given volume, 0 < ψ ≤ 1, with ψ = 1 only for a sphere. Any calculation giving ψ > 1 contains an error. Typical values: cube 0.806, cylinder with L = D 0.874, crushed solids about 0.6–0.8, flakes and mica much lower (often below 0.3). Values for real materials come from tables in your data book.

Specific surface. Surface area per unit particle volume, S_v = A_p / V_p. For a sphere S_v = 6/D. For a non-spherical particle with volume-equivalent diameter D_v, S_v = 6/(ψ·D_v). This 6/(ψ·D) term is exactly what appears in the Kozeny–Carman and Ergun equations for packed beds and filter cakes, which is why sphericity matters downstream.

Connection to neighbouring topics. Mean diameters for a whole sample (next topic) are built from the single-particle sizes defined here; settling, packed-bed flow and filtration all use 6/(ψ·D).

Formulas

D_v = (6·V_p / π)^(1/3)

• D_v: volume-equivalent diameter (m); V_p: particle volume (m³).

ψ = (surface area of sphere of same volume) / A_p = π·D_v² / A_p ψ = π^(1/3)·(6·V_p)^(2/3) / A_p

• ψ: sphericity (dimensionless, 0 < ψ ≤ 1); A_p: actual particle surface area (m²).

S_v = A_p / V_p = 6 / (ψ·D_v)

• S_v: specific surface per unit particle volume (m⁻¹). For a sphere ψ = 1 and S_v = 6/D.

a_w = 6 / (ψ·ρ_p·D_v)

• a_w: specific surface per unit mass (m²/kg); ρ_p: particle density (kg/m³).

N = m / (ρ_p·a·D_v³) with a = V_p / D_v³ (volume shape factor; a = π/6 for a sphere)

• N: number of particles in mass m (kg).

Worked examples

Example 1 (standard): sphericity of a cube. Given: a cube of side a = 2 mm = 2 × 10⁻³ m.

1. V_p = a³ = 8 × 10⁻⁹ m³; A_p = 6a² = 2.4 × 10⁻⁵ m².
2. D_v = (6·V_p/π)^(1/3) = (6 × 8 × 10⁻⁹ / π)^(1/3) = 2.481 × 10⁻³ m.
3. Surface of equal-volume sphere = π·D_v² = π × (2.481 × 10⁻³)² = 1.934 × 10⁻⁵ m².
4. ψ = 1.934 × 10⁻⁵ / 2.4 × 10⁻⁵ = 0.806. Note that the answer does not depend on a: sphericity depends only on shape.

Example 2 (GATE level): cylinder pellet and its specific surface. Given: catalyst pellets are cylinders with D = 1 mm and L = 3 mm. Find ψ and the specific surface S_v.

1. V_p = (π/4)·D²·L = (π/4) × (10⁻³)² × 3 × 10⁻³ = 2.356 × 10⁻⁹ m³.
2. A_p = π·D·L + 2·(π/4)·D² = π × 10⁻³ × 3 × 10⁻³ + (π/2) × 10⁻⁶ = 9.425 × 10⁻⁶ + 1.571 × 10⁻⁶ = 1.0996 × 10⁻⁵ m².
3. D_v = (6 × 2.356 × 10⁻⁹ / π)^(1/3) = 1.651 × 10⁻³ m.
4. ψ = π·D_v² / A_p = π × (1.651 × 10⁻³)² / 1.0996 × 10⁻⁵ = 8.563 × 10⁻⁶ / 1.0996 × 10⁻⁵ = 0.779.
5. S_v = A_p / V_p = 1.0996 × 10⁻⁵ / 2.356 × 10⁻⁹ = 4667 m⁻¹. Check: 6/(ψ·D_v) = 6 / (0.779 × 1.651 × 10⁻³) ≈ 4667 m⁻¹, the same, as it must be.

Common mistakes

• Getting ψ > 1. That is physically impossible; it means V and A are inconsistent or the formula was inverted.
• Mixing units: if V is in mm³ then A must be in mm². ψ is a ratio, so any consistent unit works.
• Using the sieve aperture as if it were D_v. Sieve size, volume diameter and Stokes diameter differ for non-spherical particles.
• Forgetting ψ in 6/(ψ·D) when computing bed surface area or Kozeny–Carman pressure drop; using 6/D overestimates permeability for crushed solids.
• Writing S_v = 4πr²/(4/3 πr³) = 3/r and then substituting the diameter for r. In diameter form it is 6/D.

For GATE CH

Expect short numericals: sphericity of a cube, cylinder or other simple shape; equivalent diameter from a given volume; specific surface of a particle or of a bed using 6/(ψ·D); number of particles in a given mass. Conceptual questions test the bounds of ψ, which shape has the highest sphericity, and which equivalent diameter each measuring method gives. Practise computing ψ for cubes and cylinders quickly and carrying ψ into packed-bed and filtration equations.

Quick check

1. What is the sphericity of a cylinder whose length equals its diameter?
2. Can sphericity be greater than 1? Why?
3. Write the specific surface of a particle of sphericity ψ and volume-equivalent diameter D_v.
4. Which equivalent diameter does sedimentation analysis report? Answers: 1. 0.874. 2. No — a sphere has the minimum surface for a given volume, so ψ ≤ 1. 3. S_v = 6/(ψ·D_v). 4. The Stokes diameter.