Particle size distribution and mean diameters

Why it matters

A real powder is never one size. To size a filter, predict a bed pressure drop, estimate the surface available for reaction or check whether a grinding step met its target, you must reduce a whole distribution to the right "average" — and the right average depends on the property you care about. Using a number mean where a surface mean is needed can be wrong by a large margin.

Key ideas

Describing the distribution. Data come either as counts (microscopy, particle counters: nᵢ particles of size Dᵢ) or as mass fractions (screen analysis: xᵢ = mass fraction retained in increment i, with Dᵢ = mean of the two apertures bounding that increment). Mass and volume fractions are the same when all particles have the same density.

• Differential (frequency) distribution: fraction in each size interval, plotted against size.
• Cumulative distribution: fraction finer (undersize) or coarser (oversize) than a size D. The two cumulative curves are mirror images and sum to 1 at every size.
• Median, D₅₀: the size at which the cumulative curve crosses 50%. Mass-median and number-median are different numbers for the same sample.
• Spread: a narrow distribution behaves nearly like uniform particles; a wide one packs densely (fines fill voids) and gives low permeability.

Mean diameters as ratios of moments. Define the k-th moment Mₖ = Σ nᵢ·Dᵢᵏ. Every mean diameter in common use is a ratio of moments, written D_pq = (M_p / M_q)^(1/(p−q)):

• D₁₀ arithmetic (number) mean — what you get by averaging sizes particle by particle.
• D₂₀ surface mean — the sphere whose surface equals the average surface.
• D₃₀ volume mean — the sphere whose volume equals the average particle volume.
• D₃₂ Sauter or volume-surface mean — the sphere with the same volume-to-surface ratio as the whole sample. It is the correct mean for anything controlled by specific surface: packed-bed and filter-cake flow, mass transfer, sprays and droplets, catalyst activity.
• D₄₃ mass (volume-weighted) mean — dominated by the coarse particles; the natural mean of a screen analysis. For any non-uniform sample D₁₀ < D₂₀ < D₃₀ < D₃₂ < D₄₃; they are all equal only when every particle is the same size. The higher the moments, the more weight on large particles.

From screen-analysis (mass fraction) data the same means are computed without counts, because nᵢ is proportional to xᵢ/Dᵢ³. This gives the compact forms in the Formulas section. The specific surface of the mixture uses D₃₂ and the sphericity from the previous topic.

Formulas

Mₖ = Σ nᵢ·Dᵢᵏ, D_pq = (M_p / M_q)^(1/(p−q))

• nᵢ: number of particles of size Dᵢ; Dᵢ: diameter (m).

From counts: D₁₀ = Σ nᵢDᵢ / Σ nᵢ D₂₀ = (Σ nᵢDᵢ² / Σ nᵢ)^(1/2) D₃₀ = (Σ nᵢDᵢ³ / Σ nᵢ)^(1/3) D₃₂ = Σ nᵢDᵢ³ / Σ nᵢDᵢ² (Sauter) D₄₃ = Σ nᵢDᵢ⁴ / Σ nᵢDᵢ³ (mass mean)

From mass fractions xᵢ (Σxᵢ = 1): D₃₂ = 1 / Σ(xᵢ/Dᵢ) D₄₃ = Σ xᵢ·Dᵢ D₃₀ = [1 / Σ(xᵢ/Dᵢ³)]^(1/3) D₁₀ = Σ(xᵢ/Dᵢ²) / Σ(xᵢ/Dᵢ³)

Specific surface and particle population of a mixture: A_w = (6 / (ψ·ρ_p))·Σ(xᵢ/Dᵢ) = 6 / (ψ·ρ_p·D₃₂) N_w = (1 / (a·ρ_p))·Σ(xᵢ/Dᵢ³)

• A_w: surface per unit mass (m²/kg); N_w: number of particles per kg (kg⁻¹); ψ: sphericity; ρ_p: particle density (kg/m³); a: volume shape factor V_p/D³ (π/6 for spheres; for real solids from your data book).

Worked examples

Example 1 (standard): means from count data. Given: 10 particles of 2 mm, 20 of 3 mm, 30 of 5 mm.

1. Moments: M₀ = 60, M₁ = 10·2 + 20·3 + 30·5 = 230 mm, M₂ = 40 + 180 + 750 = 970 mm², M₃ = 80 + 540 + 3750 = 4370 mm³, M₄ = 160 + 1620 + 18750 = 20530 mm⁴.
2. D₁₀ = 230/60 = 3.83 mm.
3. D₂₀ = (970/60)^(1/2) = 4.02 mm.
4. D₃₀ = (4370/60)^(1/3) = 4.18 mm.
5. D₃₂ = 4370/970 = 4.51 mm (Sauter).
6. D₄₃ = 20530/4370 = 4.70 mm. The ordering D₁₀ < D₂₀ < D₃₀ < D₃₂ < D₄₃ holds, as it must.

Example 2 (GATE level): screen analysis to specific surface. Given: a crushed sand (ρ_p = 2650 kg/m³, ψ = 0.8) has mass fractions 0.2, 0.5 and 0.3 in increments of mean size 1.0, 0.5 and 0.25 mm. Find D₃₂, D₄₃ and the specific surface A_w.

1. Σ(xᵢ/Dᵢ) = 0.2/1.0×10⁻³ + 0.5/0.5×10⁻³ + 0.3/0.25×10⁻³ = 200 + 1000 + 1200 = 2400 m⁻¹.
2. D₃₂ = 1/2400 = 4.17 × 10⁻⁴ m = 0.417 mm.
3. D₄₃ = Σ xᵢDᵢ = 0.2 × 1.0 + 0.5 × 0.5 + 0.3 × 0.25 = 0.525 mm.
4. A_w = 6/(ψ·ρ_p) × Σ(xᵢ/Dᵢ) = 6/(0.8 × 2650) × 2400 = 6.79 m²/kg.
5. For comparison, D₃₀ = [1/Σ(xᵢ/Dᵢ³)]^(1/3) with Σ(xᵢ/Dᵢ³) = 2.34 × 10¹⁰ m⁻³ gives D₃₀ = 0.350 mm. Taking the particles as spheres (a = π/6), N_w = 2.34 × 10¹⁰/(0.5236 × 2650) ≈ 1.69 × 10⁷ particles per kg. Notice how strongly the fine 0.25 mm fraction controls the surface even though it is only 30% of the mass.

Common mistakes

• Calling Σ nD³/Σ nD² the "volume mean". It is the Sauter (volume-surface) mean D₃₂; the volume mean is (Σ nD³/Σ n)^(1/3).
• Applying count formulas to mass fractions. With screen data, use the xᵢ forms or convert with nᵢ ∝ xᵢ/Dᵢ³.
• Using the aperture of the retaining screen instead of the mean of the two bounding apertures for Dᵢ.
• Picking the arithmetic mean for a surface-controlled process. Pressure drop, reaction and mass transfer need D₃₂.
• Reading a cumulative oversize curve as if it were undersize; check which way the curve rises.

For GATE CH

Typical questions give a count table or a screen analysis and ask for one named mean (most often the Sauter mean or mass mean), the specific surface A_w, or the number of particles per unit mass. Conceptual questions test the ordering of means, which mean suits which process, and reading a median from a cumulative plot. Practise the mass-fraction forms until they are automatic, and always check the ordering of your answers.

Quick check

1. Which mean diameter equals 1/Σ(xᵢ/Dᵢ)?
2. For a non-uniform sample, which is larger: D₁₀ or D₄₃?
3. What does the 50% point on a cumulative mass-undersize curve give?
4. Which mean should you use in the Kozeny–Carman equation? Answers: 1. The Sauter mean D₃₂. 2. D₄₃. 3. The mass-median diameter. 4. The Sauter mean D₃₂ (with sphericity).