Heat and Mass Transfer Modeling

Reference ID: MET-7213 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The freeze drying (lyophilization) of a slab is a critical unit operation in pharmaceutical and food engineering, used to remove moisture from heat-sensitive products while maintaining structural integrity. This calculation models the primary drying phase, where a sublimation front moves through a frozen material. Understanding the drying time is essential for process scale-up, optimizing cycle times, and ensuring that the product does not exceed its collapse temperature. This model is typically applied in the design of vacuum shelf dryers where heat is supplied via conduction through the dry layer and vapor is removed via a condenser. The analysis assumes quasi-steady state conditions, one-dimensional heat and mass transport, and constant physical properties throughout the dry layer.

Methodology & Formulas

The drying process is governed by the movement of a sublimation front through a dry, porous layer. The total time required for drying is determined by the resistance to either heat transfer (energy required for sublimation) or mass transfer (vapor diffusion through the dry layer). Under quasi-steady conditions, the following formulas define the time required for each regime:

For heat-transfer limited processes, the time t_heat is calculated as:

\[ t_{\text{heat}} = \frac{Z^{2} \cdot \rho \cdot (w_{i} - w_{f}) \cdot \lambda_{s}}{2 \cdot k \cdot (T_{0} - T_{i})} \]

For mass-transfer limited processes, the time t_mass is calculated as:

\[ t_{\text{mass}} = \frac{Z^{2} \cdot \rho \cdot (w_{i} - w_{f})}{2 \cdot \Pi \cdot (p_{i} - p_{0})} \]

Where the variables are defined as:

Z — Slab thickness [m]
ρ — Bulk density of the frozen product [kg/m³]
w_i — Initial ice mass fraction (dimensionless)
w_f — Final ice mass fraction (dimensionless)
λ_s — Latent heat of sublimation [J/kg]
k — Effective thermal conductivity of the dry layer [W/(m·K)]
T₀ — Temperature at the slab surface (heating side) [K]
T_i — Temperature at the sublimation front [K]
Π — Vapor permeability of the dry layer [kg/(m·s·Pa)]
p_i — Saturation vapor pressure at the sublimation front [Pa]
p₀ — Partial pressure of water vapor at the condenser/surface [Pa]

The driving forces are expressed as:

\[ \Delta T = T_{0} - T_{i} \] \[ \Delta p = p_{i} - p_{0} \]

To determine the governing mechanism, equate the two time scales. Setting t_heat = t_mass and canceling common terms yields the regime criterion:

\[ \frac{\Delta p}{\Delta T} > \frac{k}{\Pi \cdot \lambda_{s}} \quad \Rightarrow \quad \text{Heat-Transfer Limited} \] \[ \frac{\Delta p}{\Delta T} < \frac{k}{\Pi \cdot \lambda_{s}} \quad \Rightarrow \quad \text{Mass-Transfer Limited} \]

Regime	Condition
Heat-Transfer Limited	\( \displaystyle \frac{\Delta p}{\Delta T} > \frac{k}{\Pi \cdot \lambda_{s}} \)
Mass-Transfer Limited	\( \displaystyle \frac{\Delta p}{\Delta T} < \frac{k}{\Pi \cdot \lambda_{s}} \)
Bottleneck Identification	\( t_{\text{final}} = \max(t_{\text{heat}}, t_{\text{mass}}) \)

The final drying time is selected as the maximum of the two calculated values, as the slower mechanism dictates the overall process duration. Engineers must ensure that the operating parameters remain within the empirical ranges for thermal conductivity k (typically 0.02–0.2 W/(m·K)), permeability Π (typically 10⁻¹⁰–10⁻⁷ kg/(m·s·Pa)), and temperature driving force ΔT (typically 10–30 K) to prevent product collapse or melting. These ranges are product-specific and should be validated experimentally.

To ensure numerical stability and accuracy in your heat and mass transfer models, follow these steps:

Perform a mesh independence study by systematically refining the grid until the temperature and concentration gradients converge within an acceptable tolerance (e.g., less than 1% change in results).
Focus mesh refinement on regions with high thermal gradients, such as boundary layers, near heat sources, or at moving phase-change interfaces like the sublimation front.
Verify that the dimensionless wall distance y₊ is appropriate for your chosen turbulence model if convective heat transfer is dominant.

Coupling reaction kinetics with mass transport often leads to stiff systems of equations. Avoid these common errors:

Neglecting the impact of local temperature variations on the reaction rate constant, which follows the Arrhenius law: k_r = A · exp(−E_a/(RT)).
Failing to account for the feedback loop where exothermic reactions alter the local fluid density and viscosity, which in turn affects convective mass transport.
Using an incorrect diffusion coefficient for species transport in non-ideal mixtures; always verify against experimental binary diffusion data or established correlations.

Validation is critical for process reliability. Use the following approach:

Compare predicted steady-state temperature profiles against thermocouple measurements at key process nodes. For freeze-drying applications, product temperature at the sublimation front is particularly critical.
Utilize dimensionless numbers, such as the Nusselt number (Nu) or Sherwood number (Sh), to benchmark your results against established empirical correlations from the literature.
Conduct a sensitivity analysis on input parameters like thermal conductivity and mass diffusivity to identify which variables most influence model uncertainty and warrant the tightest experimental control.

Worked Example: Freeze-Drying Time of a Slab

Scenario: A pharmaceutical company must determine the primary drying time for a slab of frozen product. The slab is heated from one side while the sublimation front recedes. All parameters are known from product characterization and process conditions.

Knowns

Slab thickness: \(Z = 0.01\ \text{m}\)
Bulk density of frozen product: \(\rho = 1000\ \text{kg/m}^3\)
Ice mass fraction removed: \(\Delta w = w_i - w_f = 0.8\) (dimensionless)
Latent heat of sublimation: \(\lambda_s = 2\,830\,000\ \text{J/kg}\)
Thermal conductivity of dry layer: \(k = 0.05\ \text{W/(m·K)}\)
Surface temperature: \(T_0 = 263\ \text{K}\)
Sublimation front temperature: \(T_i = 243\ \text{K}\)
Vapor permeability of dry layer: \(\Pi = 1 \times 10^{-7}\ \text{kg/(m·s·Pa)}\)
Saturation vapor pressure at sublimation front: \(p_i = 38\ \text{Pa}\)
Condenser pressure: \(p_0 = 10\ \text{Pa}\)

Step-by-Step Calculation

Compute temperature and pressure differences. \[ \Delta T = T_0 - T_i = 263 - 243 = 20\ \text{K} \] \[ \Delta p = p_i - p_0 = 38 - 10 = 28\ \text{Pa} \] Both are within the recommended empirical ranges: \(10 \le \Delta T \le 30\ \text{K}\) and \(\Delta p > 0\).
Calculate the ratio of driving forces and the ratio of transport resistances. \[ \frac{\Delta p}{\Delta T} = \frac{28}{20} = 1.4 \] \[ \frac{k}{\Pi \cdot \lambda_s} = \frac{0.05}{1 \times 10^{-7} \times 2\,830\,000} = \frac{0.05}{0.283} = 0.177 \] The empirical value of \(k\) (0.05) lies between 0.02 and 0.2, and \(\Pi\) (\(1 \times 10^{-7}\)) is at the upper end of the typical range \(10^{-10}\) to \(10^{-7}\).
Identify the controlling mechanism.
The ratio of driving forces (\(\Delta p / \Delta T = 1.4\)) is greater than the ratio of resistances (\(k/(\Pi \lambda_s) = 0.177\)). Since \(\Delta p / \Delta T > k/(\Pi \lambda_s)\), the heat transfer resistance is the bottleneck. This is confirmed by calculating both drying times explicitly: \[ t_{\text{mass}} = \frac{Z^{2} \cdot \rho \cdot \Delta w}{2 \cdot \Pi \cdot \Delta p} = \frac{(0.01)^{2} \times 1000 \times 0.8}{2 \times 1 \times 10^{-7} \times 28} = \frac{0.08}{5.6 \times 10^{-6}} = 14\,285.7\ \text{s} \] The heat-transfer time (computed below) is substantially longer; therefore, the process is heat-transfer limited.
Apply the heat-transfer controlled equation. \[ t_{\text{heat}} = \frac{Z^{2} \cdot \rho \cdot \Delta w \cdot \lambda_s}{2 \cdot k \cdot \Delta T} \] Substitute the known values: \[ t_{\text{heat}} = \frac{(0.01)^{2} \times 1000 \times 0.8 \times 2\,830\,000}{2 \times 0.05 \times 20} \] \[ t_{\text{heat}} = \frac{226\,400}{2} = 113\,200.0\ \text{s} \]
Convert to hours for practical interpretation. \[ t_{\text{hr}} = \frac{113\,200.0}{3600} = 31.44\ \text{h} \]

Final Answer
The primary drying time is approximately 113 200 s (31.44 h). This result assumes no melting or structural collapse and that the quasi-steady moving-front model applies. Note that this calculation addresses the primary drying phase only; secondary desorption drying may require additional time.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

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