Reference ID: MET-9196 | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The number-average diameter is a fundamental particle-size statistic that weights every particle equally, regardless of its mass or volume. In process engineering it is indispensable for:
Fluidisation studies—pressure-drop correlations rely on the actual count of particles.
Reaction kinetics—surface-area limited reactions scale with the number of particles present.
Quality control—comparing batch-to-batch consistency when the population, not the mass, is the critical attribute.
Methodology & Formulas
Convert each mass fraction \(m_i\) (g) to kilograms:
\[ m_{i,\text{kg}} = \frac{m_i}{1000} \]
Convert each sieve diameter \(d_i\) (µm) to metres:
\[ d_{i,\text{m}} = \frac{d_i}{10^{6}} \]
Compute the spherical volume of one particle:
\[ V_i = \frac{\pi\,d_{i,\text{m}}^{3}}{6} \]
Determine the number of particles in the fraction:
\[ n_i = \frac{m_{i,\text{kg}}}{\rho_{\text{p}}\,V_i} \]
where \(\rho_{\text{p}}\) is the true density of the material (kg m−3).
Accumulate the total number of particles and the total “number-length” product:
\[ N = \sum n_i \quad\text{and}\quad \sum(n_i\,d_i) \]
Calculate the number-average diameter (returned in µm for convenience):
\[ d_{\text{n}} = \frac{\sum(n_i\,d_i)}{N} \]
Parameter
Typical Range / Criterion
True density \(\rho_{\text{p}}\)
800–3000 kg m−3 (food powders)
Particle diameter \(d_i\)
\(d_i > 0\) (strictly positive)
Number Average Diameter (Dn) is the arithmetic mean of all particle diameters weighted by the number of particles in each size class. It is critical for process engineers because:
It directly influences pressure drop and residence-time distribution in fixed-bed reactors.
It is the primary input for calculating specific surface area and, therefore, reaction or dissolution rates.
Regulatory filings often require Dn to demonstrate batch-to-batch consistency in catalyst or excipient specifications.
Instruments that natively count particles give the most accurate Dn:
Electrical sensing zone (Coulter principle) measures volume of each particle but reports number if calibrated with NIST spheres.
Particle counters using light obscuration or light scattering with single-particle counting mode.
Dynamic image analysis systems that bin every detected particle by its Feret diameter.
To avoid conversion errors, disable any built-in Mie or Fraunhofer volume-weighting algorithms and export the raw number count table before performing the arithmetic mean.
Laser diffraction yields volume-weighted data; convert to number as follows:
Export the cumulative volume percent in fixed size bins.
For each bin i, calculate the number of particles ni = (Vi / Vsphere,i) where Vsphere,i = (π/6)di3.
Compute Dn = Σ(ni·di) / Σni.
Note: this conversion amplifies noise in the smallest bins; include only bins with >0.5 % of total volume to keep uncertainty below ±5 %.
Follow these guidelines:
Count at least 10,000 particles in the finest size class of interest; for sub-micron powders this usually means >0.5 g dispersed.
Use wet dispersion with surfactant concentration 0.01–0.1 % w/w to break agglomerates without causing particle dissolution.
Verify dispersion by measuring Dn at three sonication times; plateau indicates optimum.
Run five repeat measurements on independently drawn samples; if RSD >3 %, increase counting statistics or improve dispersion.
Worked Example – Number-Average Diameter of a Latex Product
A process engineer needs to verify the average particle size of a poly-vinyl-acetate latex batch before it is fed to the spray dryer. A 0.15 g sample is taken from the reactor, dispersed in 1 L of water, and analysed with an optical particle counter. The instrument reports size-channel data that must be converted to a number-average diameter.
Compute the volume of one average particle:
\[
V = \frac{\pi}{6} d_i^3 = \frac{\pi}{6}(302.5\ \mathrm{\mu m})^3 = 1.449 \times 10^{-11}\ \mathrm{m^3}
\]
Convert this volume to mass per particle:
\[
m_\text{particle} = \rho_p V = 1050\ \mathrm{kg\ m^{-3}} \times 1.449 \times 10^{-11}\ \mathrm{m^3} = 1.521 \times 10^{-8}\ \mathrm{kg}
\]
Estimate the total number of particles in the 0.15 g sample:
\[
\sum n_i = \frac{m}{m_\text{particle}} = \frac{0.00015\ \mathrm{kg}}{1.521 \times 10^{-8}\ \mathrm{kg}} \approx 9856.6
\]
