Reference ID: MET-DF8B | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The calculation of freezing time is a fundamental requirement in Process Engineering, particularly within the food processing, cryogenics, and chemical storage industries. Predicting the time required for a substance to undergo a phase change from liquid to solid is critical for ensuring product quality, optimizing refrigeration energy consumption, and designing batch processing cycles. This specific model utilizes Plank's equation for a slab geometry, which provides a robust analytical approach to estimate the time required for a slab of material to freeze completely from one side under convective cooling conditions.
Methodology & Formulas
The freezing process is modeled by considering the latent heat removal and the thermal resistance encountered during the phase change. The total time required for freezing is determined by the temperature gradient between the freezing point of the substance and the surrounding medium, as well as the thermal properties of the material being frozen. This formulation assumes one-sided freezing where the characteristic length equals the full slab thickness.
First, the temperature difference is defined as:
\[ \Delta T = T_{\text{freeze}} - T_{\infty} \]
The total freezing time \( t \) is calculated by combining the conductive resistance of the ice layer and the convective resistance at the surface. The governing equation is:
\( \Delta T \): Temperature difference between freezing point and medium (K)
\( k_{\text{ice}} \): Thermal conductivity of ice (W/m·K)
\( h_{\text{conv}} \): Convective heat transfer coefficient (W/m2·K)
\( T_{\text{freeze}} \): Freezing point temperature (°C or K)
\( T_{\infty} \): Bulk temperature of the cooling medium (°C or K)
To ensure the physical validity of the model, the following operational constraints must be satisfied:
Parameter
Constraint Condition
Reasoning
Thickness
\( x > 0 \)
Physical dimension must be positive.
Temperature Gradient
\( \Delta T > 0 \)
Medium must be colder than the freezing point to initiate phase change.
Convection Coefficient
\( h_{\text{conv}} > 0 \)
Heat transfer must occur from the slab to the medium.
Thermal Conductivity
\( k_{\text{ice}} > 0 \)
Material must possess thermal conductive properties.
The relationship between thickness and freezing time is non-linear, typically following a square-law relationship when conduction dominates. As the thickness of the product increases, the time required to reach the target core temperature increases significantly because:
Heat must travel a longer distance from the center to the surface.
The growing frozen layer adds thermal resistance, progressively slowing the rate of heat extraction.
The overall thermal conductance (\(k_{\text{ice}}/x\)) decreases as the ice front progresses inward.
While the square-law suggests that doubling the thickness quadruples the freezing time, real-world process conditions introduce variables that deviate from this ideal model:
Variations in surface heat transfer coefficients.
Changes in the latent heat of fusion based on product composition.
The presence of air gaps or packaging materials that alter the effective thermal resistance.
Fluctuations in the temperature gradient between the refrigerant and the product surface.
Geometry dictates the heat flow path and the surface-area-to-volume ratio, which are critical for process engineers to consider:
Infinite slabs freeze primarily in one dimension, following the standard thickness-time relationship.
Cylinders and spheres experience faster freezing times than slabs of the same characteristic thickness due to multi-dimensional heat flow.
Irregular shapes create localized hot spots where the center point is further from the surface, leading to uneven freezing rates.
Worked Example: Freezing Time vs Thickness Relationship
Scenario: A 0.05 m thick slab of pure water is frozen from one side in a freezing medium at −15.0 °C. The convection heat transfer coefficient is 120.0 W/m2·K. Using Plank's equation, determine the freezing time. The slab is considered to freeze from the exposed surface inward (one-sided freezing).
Knowns (Input Parameters and Units):
Latent heat of fusion: \(L_{f} = 333.7\ \text{kJ/kg}\)
Thermal conductivity of ice: \(k_{\text{ice}} = 2.22\ \text{W/(m·K)}\)
Density of ice: \(\rho_{\text{ice}} = 917.0\ \text{kg/m}^3\)
Slab thickness: \(x = 0.05\ \text{m}\)
Freezing medium temperature: \(T_{\infty} = -15.0\ °\text{C}\)
Freezing point of water: \(T_{\text{freeze}} = 0.0\ °\text{C}\)
Temperature difference:
\[
\Delta T = T_{\text{freeze}} - T_{\infty} = 0.0 - (-15.0) = 15.0\ \text{K}
\]
Characteristic length for one-sided freezing:
For freezing from one side, the characteristic length equals the full slab thickness:
\[
L = x = 0.05\ \text{m}
\]
First term of Plank’s equation (latent heat, density, length):
Convert \(L_{f}\) to J/kg (multiply by 1000) and combine with density and thickness, divided by \(\Delta T\):
\[
\text{TERM}_{1} = \frac{L_{f} \cdot 1000 \cdot \rho_{\text{ice}} \cdot x}{\Delta T} = \frac{333.7 \times 1000 \times 917.0 \times 0.05}{15.0} = 1\,020\,009.667\ \text{J/(m}^2·\text{K)}
\]
Freezing time in seconds:
Multiply the two terms (using the unrounded value of TERM2 = 0.0195946…):
\[
t = \text{TERM}_{1} \times \text{TERM}_{2} = 1\,020\,009.667 \times 0.0195946 = 19\,986.7\ \text{s}
\]
Final Answer:
The freezing time for a one-sided, 0.05 m thick water slab under the given conditions is approximately 19,987 seconds (or 333.1 minutes).
"Un projet n'est jamais trop grand s'il est bien conçu."— André Citroën
"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise."— Charles de Gaulle
