Bin-Based Dissolution Calculations for Polydisperse Powders

Why Calculate Dissolution by Size Bin?

A powder dose can contain particles with a broad range of sizes. Under the same medium conditions, a finer fraction generally presents more exposed surface area per unit mass than a coarser fraction, so its contribution to dissolution can be disproportionately large. A single representative particle size can conceal this difference. A bin-based calculation preserves the separate contributions of the measured size fractions while keeping the arithmetic transparent.

The calculation here uses the bin masses and surface areas associated with converting particle-size data to dissolving surface area. For a spherical-particle approximation, a bin’s mass and representative diameter determine its initial surface area. Each bin then receives its own interfacial mass-transfer rate, and the predicted dissolved mass is the sum across bins. This is a deliberately simple, fixed-area calculation: it is useful for short-time estimates and comparisons, but it does not track particle geometry as particles shrink.

Model and Assumptions

For spherical particles of common density \(\rho\), the surface area corresponding to a bin’s mass and representative diameter is

This formula treats the particles in a bin as if they all had diameter \(d_i\). When the input is a mass fraction \(w_i\) of a total dose \(D\), first calculate \(m_{i,0}=Dw_i\). Fractions should be nonnegative and sum to one, subject to rounding.

The interfacial mass-transfer expression for the dissolution rate is \(kA_i(C_s-C_{\mathrm{bulk}})\), where \(k\) is the mass-transfer coefficient, \(C_s\) is the concentration at the particle surface, and \(C_{\mathrm{bulk}}\) is the bulk dissolved concentration. In this tutorial, a sink-condition approximation is used: \(C_{\mathrm{bulk}}\) is negligible relative to \(C_s\), so the concentration driving force is approximately \(C_s\).

With these assumptions, the initial rate for bin \(i\) is \(R_i=kA_iC_s\). The rate cannot continue once the bin’s available mass has dissolved. Thus the bin’s cumulative dissolved mass is the smaller of its initial mass and the amount predicted by its constant rate over time:

When \(m_{i,0}\), \(A_i\), and \(R_i\) are positive, the bin’s nominal exhaustion time is \(t_i^*=m_{i,0}/R_i\). Before this time its predicted dissolved mass increases linearly; after this time its contribution stays at \(m_{i,0}\). The combined profile is therefore piecewise linear, with a change in slope whenever a bin is exhausted.

Why the Bin Contributions Add

For each bin that has not yet exhausted, the fixed-area sink model gives a constant rate \(R_i\). Over a time interval of length \(t\), its dissolved mass is therefore \(R_it\), provided this amount does not exceed the bin’s starting mass. If it would exceed that mass, the bin can contribute only \(m_{i,0}\). These two cases give \(M_i(t)=\min(m_{i,0},R_it)\). Since the bins partition the starting powder mass and their dissolved contributions are measured in the same units, total dissolved mass is their sum.

This justification also gives a useful conservation check. For every \(t\geq 0\), each bin satisfies \(0\leq M_i(t)\leq m_{i,0}\), so

A prediction above the initial dose signals an implementation or unit error. This bound does not make the fixed-area assumption exact; it only ensures that the bin-wise calculation does not dissolve more mass than was initially assigned to the bins.

Using Several Bins and Checking the Calculation

For multiple bins, the same sequence applies: calculate each bin mass, convert it to area, calculate its rate, apply the mass cap at the requested time, and then sum. Keep all quantities in compatible units before multiplying. In particular, a diameter in micrometres must be converted if \(\rho\), \(k\), or area are expressed using millimetres.

Diagnostics, Alternatives, and Limitations

A practical check is to tabulate \(m_{i,0}\), \(A_i\), \(R_i\), and \(t_i^*\) for every bin. This makes unit mistakes and unexpectedly dominant bins easier to spot. Check that fractions sum to one, that all bin masses and diameters are positive, that rate units reduce to mass per time, and that the summed dissolved mass never exceeds the dose. If the smallest bin appears to dominate, verify that its mass fraction and representative diameter were entered correctly: a small diameter can produce a large area per unit mass.

A single effective diameter is simpler, but it can obscure the high area per unit mass of fine fractions. Explicit bins retain this heterogeneity with modest computational effort. At the other extreme, approaches that update particle sizes and surface areas continuously can represent changing geometry more directly, but require additional assumptions and computation. The fixed-area bin calculation is most defensible over intervals where treating the area as approximately constant is adequate for the intended estimate.

The principal limitation is that surface area is not actually constant as material dissolves. Consequently, a constant bin rate can overpredict dissolution over longer periods, even when the mass cap is applied. Also, sink conditions may fail as dissolved concentration rises; then the driving force is not simply \(C_s\), and the bins can no longer be treated as having independent rates driven by a negligible bulk concentration. The calculation is therefore a transparent approximation, not a universal dissolution law. Its value is greatest when its assumptions match the question being asked and the result is checked against observed dissolution data.

Common Pitfall

Wrong reasoning: “Each bin has a rate, so total dissolved mass at time \(t\) is \(t\) times the sum of all bin rates, even if that value exceeds the dose.” This ignores the finite mass available in each bin. Correction: cap each bin’s contribution at \(m_{i,0}\) first, using \(M_i(t)=\min(m_{i,0},R_it)\), and only then sum across bins. A total-dose cap applied after summing may hide which bin was incorrectly allowed to dissolve beyond its own available mass.