Introduction & Context

Secondary nucleation is the process by which new crystal nuclei form on the surface of existing crystals during a crystallisation operation. The rate at which this occurs, B₀, is a key design parameter in the scale-up of crystallisers, batch reactors, and continuous crystallisation lines. Accurate prediction of B₀ allows engineers to control crystal size distribution, optimise residence time, and minimise fouling or agglomeration. The calculation presented here is based on a power-law correlation that links the mechanical power input to the nucleation rate, and is commonly applied in the design of stirred tank reactors, especially those employing Rushton turbines or similar impellers.

Methodology & Formulas

The calculation proceeds in five logical sections: definition of constants, specification of operating conditions, evaluation of dimensionless numbers and power, computation of power density, and finally the secondary nucleation rate. All expressions are given in algebraic form; numerical values are omitted as requested.

1. Constants

K – Power number for the Rushton turbine (dimensionless).
k_N – Empirical secondary nucleation constant (m^-3 s^-1).
j – Exponent for crystal density (dimensionless).
k – Exponent for power density (dimensionless).

2. Input Parameters (practical units)

V – Reactor liquid volume (m³).
d – Impeller diameter (m).
N – Rotational speed (rev s^-1).
ρ – Fluid density (kg m^-3).
μ – Fluid viscosity (Pa s).
M_T – Crystal (magma) density (kg m^-3).

3. Dimensionless Numbers and Power

Reynolds number:
\[ \mathrm{Re} = \frac{\rho\,N\,d^{2}}{\mu} \]

Froude number:
\[ \mathrm{Fr} = \frac{N^{2}\,d}{g} \]

Mechanical power input for a Rushton turbine:
\[ P = K\,\rho\,N^{3}\,d^{5} \]

Power density:
\[ P_{\text{density}} = \frac{P}{V} \]

4. Secondary Nucleation Rate

The empirical correlation for the secondary nucleation rate is:
\[ B_{0} = k_{N}\,M_{T}^{\,j}\,P_{\text{density}}^{\,k} \]

5. Validity Checks

Condition	Threshold	Warning
Reynolds number	Re < 10,000	Power correlation may be inaccurate.
Froude number	Fr ≥ 0.2	Vortex effects may be significant.
Viscosity	μ > 0.01 Pa s	Power correlation may need correction.
Empirical constants	k_N ≤ 0 or j < 0 or k ≤ 0	Constants are not positive.

6. Output Summary

The calculation yields the following key performance indicators:
• Reynolds number (Re)
• Froude number (Fr)
• Mechanical power input (P) in watts
• Power density (P/V) in W m^-3
• Secondary nucleation rate (B₀) in # m^-3 s^-1

Secondary nucleation rate is the number of new crystals formed per unit volume per unit time due to mechanisms that occur after the primary nucleation event, such as crystal-crystal collisions or attrition. It is crucial because it directly influences crystal size distribution, product purity, and overall process yield.

The power-law model relates the secondary nucleation rate (B) to supersaturation (ΔC) and shear rate (G):

Identify the model parameters: B₀ (pre-exponential factor), α (supersaturation exponent), and β (shear exponent).
Measure or estimate the supersaturation ΔC = C – C* (where C* is solubility).
Determine the local shear rate G from impeller speed, geometry, or CFD data.
Apply the equation: B = B₀·(ΔC)^α·G^β.
Convert B to the desired units (e.g., nuclei·m⁻³·s⁻¹) using the reactor volume.

To calibrate the model you need:

Time-resolved crystal count data (e.g., from FBRM or PVM) at several controlled supersaturation levels.
Corresponding shear rate information for each experiment (impeller speed, tip speed, or CFD-derived G).
Solubility data to calculate ΔC for each condition.
Temperature and viscosity measurements to account for fluid property effects.
A regression routine (e.g., nonlinear least squares) to fit B₀, α, and β to the observed nucleation rates.

Integration steps:

Define the nucleation source term Sₙ(v) = B·δ(v−v₀), where v₀ is the assumed volume of a newborn crystal.
Insert Sₙ(v) into the PBM governing equation alongside growth, dissolution, and agglomeration terms.
Discretize the crystal size domain (e.g., using the fixed-pivot method) and solve the resulting set of ordinary differential equations.
Validate the PBM predictions against experimental size distributions to ensure the nucleation term is correctly represented.

Worked Example – Estimating Secondary Nucleation Rate in a Continuous Crystalliser

A 500 L draft-tube crystalliser is fed with a 30 wt % aqueous ammonium sulphate solution at 25 °C. The vessel is agitated with a 0.20 m diameter pitched-blade impeller running at 4.0 rps. Laboratory seeding tests show that secondary nucleation dominates; the plant engineer needs to predict the nucleation rate B₀ for the full-scale unit.

Knowns

Empirical constant K = 0.55 (dimensionless)
Rate constant k_N = 1.2 × 10⁶ # m^-3 s^-1
Exponent on supersaturation j = 0.5 (dimensionless)
Exponent on slurry density k = 1.0 (dimensionless)
Slurry volume V = 0.5 m³
Impeller diameter d = 0.1 m
Impeller speed N = 4.0 rps
Liquid density ρ = 1000 kg m^-3
Liquid viscosity μ = 0.001 Pa s
Slurry density M_T = 1600 kg m^-3

Step-by-step calculation

Calculate impeller tip speed v_tip: \[ v_{\text{tip}} = \pi N d = \pi \times 4.0 \times 0.1 = 1.257\ \text{m s}^{-1} \]
Evaluate Reynolds number: \[ \mathrm{Re} = \frac{\rho N d^{2}}{\mu} = \frac{1000 \times 4.0 \times 0.1^{2}}{0.001} = 40,000 \]
Evaluate Froude number: \[ \mathrm{Fr} = \frac{N^{2} d}{g} = \frac{4.0^{2} \times 0.1}{9.81} = 0.163 \]
Compute power number correlation for a pitched-blade turbine (standard curve): \[ N_{P} \approx 0.35 \quad (\text{literature value for } \mathrm{Re} \approx 40,000) \]
Calculate power draw: \[ P = N_{P} \rho N^{3} d^{5} = 0.35 \times 1000 \times 4.0^{3} \times 0.1^{5} = 0.352\ \text{kW} \]
Convert to power density: \[ P_{\text{density}} = \frac{P}{V} = \frac{0.352}{0.5} = 0.704\ \text{kW m}^{-3} \]
Estimate supersaturation-dependent term (assume relative supersaturation σ = 0.05): \[ \sigma^{j} = 0.05^{0.5} = 0.224 \]
Combine terms in the secondary nucleation model: \[ B_{0} = K\ k_{N}\ M_{T}^{k}\ \sigma^{j} \] \[ B_{0} = 0.55 \times 1.2 \times 10^{6} \times 1600^{1.0} \times 0.224 \] \[ B_{0} = 3.379 \times 10^{7}\ \text{# m}^{-3}\ \text{s}^{-1} \]

Final Answer

The predicted secondary nucleation rate for the 500 L crystalliser under the stated conditions is
3.38 × 10⁷ nuclei per cubic metre per second.

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