There is no single 'size'
Ask a colleague the size of a powder and the honest answer is a question back: which size? A real batch contains particles spanning a wide range, and most of them are not neat spheres anyway — their shape, or crystal habit, may be needles, plates, or rough blocks. To talk about an irregular particle's size at all, we borrow the idea of an equivalent diameter: the diameter of a sphere that matches the particle in some chosen property — its volume, its surface, or how fast it settles.
Because the choice of property differs, two instruments can report different numbers for the same powder — and both are correct. This is why a result is meaningless without naming the method. A mean diameter from settling will not equal one from a microscope counting projected outlines.
Reading a size distribution
Since a powder spans a range, we describe it with a particle size distribution — a curve showing how much material falls in each size band. The most quoted landmarks are percentiles read off the cumulative curve: d10, d50, and d90. The d50 (also called the median) is the size below which half the material lies; d10 and d90 mark the fine and coarse ends. Together they tell you both the typical size and how spread out the batch is.
The same particles can also be weighted differently. A number distribution treats every particle equally, so it is dominated by the many tiny fines. A volume (or mass) distribution weights by how much material each particle contains, so a single big particle counts as much as thousands of small ones. The volume-weighted mean therefore sits far above the number-weighted mean for the same powder — neither is wrong, but you must know which you are reading.
Why number-mean and volume-mean differ — a 2-size mix
Suppose 1000 particles of 1 µm and 1 particle of 100 µm.
Number-mean diameter:
(1000×1 + 1×100) / 1001 = 1100 / 1001 ≈ 1.10 µm
→ dominated by the many fines
Volume-mean (weight each particle by its volume ∝ d³):
Vol of 1 µm particle ∝ 1³ = 1 (×1000 = 1000)
Vol of 100 µm particle ∝ 100³ = 1,000,000
Volume-weighted mean d = Σ(d·d³)/Σ(d³)
= (1000×1 + 1,000,000×100) / (1000 + 1,000,000)
≈ 100,001,000 / 1,001,000 ≈ 99.9 µm
→ dominated by the single large particle
Same powder, ~1 µm vs ~100 µm depending on weighting.Two everyday instruments
The oldest workhorse is sieve analysis: a stack of mesh screens, coarsest on top, shaken so particles fall until each is trapped by a mesh it cannot pass. Weighing the powder on each sieve gives a mass distribution directly. It is cheap, robust, and intuitive, but limited to roughly above 40–50 µm and biased by elongated particles that slip through end-on.
For finer powders the modern standard is laser diffraction: particles passing a laser beam scatter light at angles that depend on their size — small particles scatter wide, large ones scatter narrow. Detectors capture the pattern and software reconstructs a volume distribution in seconds, over a huge range from sub-micron to millimetre. It is fast and reproducible, though it assumes particles behave optically like spheres.