Introduction & Context

The Contact Freezing Time calculation is a fundamental process engineering assessment used to determine the duration required to solidify a food product, such as a fish block, when placed in direct contact with a refrigerated surface; incorporating proper glazing for frozen product protection further enhances product quality by minimizing moisture loss and surface dehydration during the freezing operation.

This methodology is typically applied in the design of plate freezers and cold‑chain logistics, where heat transfer is dominated by conduction through the product and the interface between the product packaging and the cooling medium. By predicting the freezing time, engineers can establish operational setpoints that balance energy consumption with production capacity, and a complementary brine freezing calculation can be consulted for processes involving immersion in chilled brine.

Methodology & Formulas

The calculation utilizes Plank’s equation, which assumes that the freezing process occurs at a constant temperature (the latent heat phase). The total time required is derived from the thermal resistance of the interface—see our plate freezer contact heat transfer optimization—and the internal thermal resistance of the product slab.

First, the overall heat transfer coefficient U is determined by the sum of the contact resistance and the freezing time with packaging resistance contribution.

\[ \frac{1}{U} = \frac{1}{h_{c}} + \frac{x_{p}}{k_{p}} \]

The freezing time t_f is then calculated using the following relationship, which accounts for the density of the product, the latent heat of fusion, the temperature gradient, and the geometric dimensions, and it directly ties into the product cooling load calculation for accurate process design:

\[ t_{f} = \frac{\rho_{f} \cdot L_{f}}{\Delta T} \cdot \left( \frac{L}{U} + \frac{L^{2}}{2 \cdot k_{f}} \right) \]

Where the temperature difference is defined as:

\[ \Delta T = T_{f} - T_{a} \]

To ensure the validity of the one-dimensional heat transfer assumption, the Biot number (Bi) is calculated to verify that internal resistance is significant relative to surface resistance:

\[ Bi = \frac{U \cdot L}{k_{f}} \]

Parameter	Condition / Threshold	Engineering Significance
Biot Number	\( Bi \geq 0.1 \)	Required for Plank's equation validity; indicates internal resistance is dominant.
Temperature Gradient	\( \Delta T > 0 \)	Ensures a positive heat flux from the product to the cooling medium.
Package Thickness	\( 0.05\ \text{mm} \leq x_{p} \leq 0.5\ \text{mm} \)	Empirical range for standard freezer films; values outside this may require adjusted thermal resistance models.

To accurately calculate the freezing time for contact systems, you must account for the thermal resistance of both the product and the contact interface. Key variables include:

Product thickness and initial temperature
Thermal conductivity and density of the food product
Surface heat transfer coefficient between the plate and the product
Temperature of the refrigerant circulating within the plates

The contact resistance represents the air gaps between the product surface and the freezing plate. If this variable is ignored, your model will significantly underestimate the total freezing time. Process engineers should consider:

Surface irregularities of the product packaging
The pressure applied by the freezing plates
The presence of frost buildup on the plate surfaces

The latent heat of fusion is the most energy-intensive phase of the freezing process. When calculating the time required, you must ensure the model accounts for:

The specific enthalpy change during the transition from liquid to solid state
The depression of the freezing point based on the solute concentration of the product
The non-linear progression of the freezing front through the product geometry

Worked Example: Contact Freezing Time of a Fish Block

A 6 cm thick fish block, insulated on top, is frozen by contact with a cold plate at \( T_{plate} = -35\,^{\circ}\text{C} \). The block is packaged in a 0.2 mm plastic film. The following known values are used:

Slab thickness: \( L = 0.06\ \text{m} \)
Plate temperature: \( T_{plate} = -35\ ^{\circ}\text{C} \)
Freezing point of fish: \( T_f = -2\ ^{\circ}\text{C} \)
Fish density (frozen): \( \rho_f = 1000\ \text{kg/m}^3 \)
Latent heat of freezing: \( L_f = 3.0 \times 10^5\ \text{J/kg} \)
Thermal conductivity of frozen fish: \( k_f = 1.5\ \text{W/(m·K)} \)
Package thickness: \( x_p = 0.0002\ \text{m} \)
Package thermal conductivity: \( k_p = 0.2\ \text{W/(m·K)} \)
Contact heat transfer coefficient: \( h_c = 1500\ \text{W/(m}^2\cdot\text{K)} \)

The freezing time is calculated using the overall heat transfer coefficient and Plank’s equation for a one-sided slab.

Temperature difference \[ \Delta T = T_f - T_{plate} = (-2) - (-35) = 33.0\ \text{K} \]
Contact resistance \[ R_{contact} = \frac{1}{h_c} = \frac{1}{1500} \approx 6.667 \times 10^{-4}\ \text{m}^2\cdot\text{K/W} \]
Package resistance \[ R_{package} = \frac{x_p}{k_p} = \frac{0.0002}{0.2} = 1.000 \times 10^{-3}\ \text{m}^2\cdot\text{K/W} \]
Total resistance \[ R_{total} = R_{contact} + R_{package} = 6.667 \times 10^{-4} + 1.000 \times 10^{-3} = 1.667 \times 10^{-3}\ \text{m}^2\cdot\text{K/W} \]
Overall heat transfer coefficient \[ U = \frac{1}{R_{total}} = \frac{1}{1.667 \times 10^{-3}} \approx 600.0\ \text{W/(m}^2\cdot\text{K)} \]
First Plank term \[ \frac{\rho_f L_f}{\Delta T} = \frac{1000 \cdot 3.0 \times 10^{5}}{33.0} = 9.090909091 \times 10^{6}\ \text{J/(m}^3\cdot\text{K)} \]
Second Plank term – conduction resistance \[ \frac{L}{U} = \frac{0.06}{600.0} = 1.0 \times 10^{-4}\ \text{m}^3\cdot\text{K/W} \]
Third Plank term – internal resistance \[ \frac{L^2}{2k_f} = \frac{0.06^2}{2 \cdot 1.5} = \frac{0.0036}{3.0} = 1.2 \times 10^{-3}\ \text{m}^3\cdot\text{K/W} \]
Sum of resistance terms \[ \frac{L}{U} + \frac{L^2}{2k_f} = 1.0 \times 10^{-4} + 1.2 \times 10^{-3} = 1.3 \times 10^{-3}\ \text{m}^3\cdot\text{K/W} \]
Freezing time (seconds) \[ t_f = \frac{\rho_f L_f}{\Delta T} \left( \frac{L}{U} + \frac{L^2}{2k_f} \right) = 9.090909091 \times 10^{6} \cdot 1.3 \times 10^{-3} = 11818.182\ \text{s} \]
Freezing time (hours) \[ t_f = \frac{11818.182}{3600} \approx 3.283\ \text{h} \]

Final Answer: The contact freezing time is approximately 3.283 hours.

Validity check: The Biot number \( \text{Bi} = U L / k_f = (600.0 \cdot 0.06) / 1.5 = 24.0 \gg 0.1 \), confirming that Plank’s equation is applicable and internal resistance is not limiting.

"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

Contact Freezing Time Calculation

Introduction & Context

Methodology & Formulas

Worked Example: Contact Freezing Time of a Fish Block

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