Question 7 of 9: Particle Size Distribution — Spray-Dried Milk Droplets and Powder Screen Analysis

Question 7: Particle Size Distribution — Spray-Dried Milk Droplets and Powder Screen Analysis (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Given. Cumulative mass-percent undersize varies linearly with droplet diameter D_p, from X = 0% at D_p = 50 µm to X = 100% at D_p = 250 µm — i.e. the mass-fraction density is uniform, dX/dD_p = 1/(D_max−D_min), over [50, 250] µm.

Find. The mass mean diameter D̄_mass = ∫D_pdX and the mean volume-surface (Sauter) diameter D_vs = 1/∫(dX/D_p).

Approach. Integrate each definition directly over the uniform mass-density distribution on [D_min, D_max].

1. Part (a) — mass mean diameter. $$\overline{D}_{mass}=\int_{D_{min}}^{D_{max}} D_p\,\frac{dX}{dD_p}\,dD_p=\frac{1}{D_{max}-D_{min}}\int_{D_{min}}^{D_{max}} D_p\,dD_p=\frac{D_{max}^2-D_{min}^2}{2(D_{max}-D_{min})}$$ Substituting D_min = 50 µm, D_max = 250 µm: $$\overline{D}_{mass}=\frac{250^2-50^2}{2(250-50)}=\frac{62500-2500}{400}=\boxed{150\ \mu\text{m}}$$ (the arithmetic mean of the two end points, as expected for a uniform mass-density distribution).
2. Part (a) — mean volume-surface (Sauter) diameter. $$D_{vs}=\frac{1}{\displaystyle\int_{D_{min}}^{D_{max}}\frac{1}{D_p}\frac{dX}{dD_p}dD_p}=\frac{D_{max}-D_{min}}{\ln(D_{max}/D_{min})}$$ Substituting the same limits: $$D_{vs}=\frac{250-50}{\ln(250/50)}=\frac{200}{\ln 5}=\frac{200}{1.6094}=\boxed{124.3\ \mu\text{m}}$$
3. Part (b) — volume and surface shape factors. Earle's screen-analysis formulae express a particle's volume as v_p = bD_p³ and its surface area as s_p = aD_p², where a and b are related through the sphericity Φ_s (the ratio of the surface area of a sphere of equal volume to the particle's own surface area) by a/b = 6/Φ_s: $$b=\frac{a\,\Phi_s}{6}=\frac{2\times0.7}{6}=0.2333$$
4. Part (b) — specific surface, A_W. Summing the surface area contributed by every size fraction per unit mass, using each fraction's particle count (∝ X_i/D_pi³) times its individual surface area (aD_pi²), collapses to $$A_W=\frac{a}{\rho_p\,b}\sum\frac{X_i}{D_{pi}}$$ $$A_W=\frac{2}{(0.0018)(0.2333)}\times0.8749=\frac{2}{4.200\times10^{-4}}\times0.8749=4761.9\times0.8749$$ $$A_W=\boxed{4166\ \text{mm}^2/\text{g}}$$
5. Part (b) — specific number of particles, N_W. Similarly, summing the particle count contributed by every size fraction per unit mass gives $$N_W=\frac{1}{\rho_p\,b}\sum\frac{X_i}{D_{pi}^3}$$ $$N_W=\frac{1}{4.200\times10^{-4}}\times4.676=2380.95\times4.676$$ $$N_W=\boxed{11{,}133\ \text{particles/g}}$$