Freezing Time with Packaging Resistance

Reference ID: MET-1AD8 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The calculation of freezing time for packaged food products is a critical task in food process engineering, particularly for the design and optimization of blast freezing systems; understanding the impact of package thickness on freezing time is essential, as detailed in our analysis of the effect of package thickness on freezing rates. In industrial settings, products are rarely exposed directly to the cooling medium; they are typically enclosed in packaging materials that introduce additional thermal resistance. This calculation utilizes Plank's Equation, modified to account for the series thermal resistance of the packaging layer and the external convective boundary layer. Accurate estimation of this time is essential for determining conveyor speeds, freezer residence times, and ensuring product quality by minimizing ice crystal growth through controlled freezing rates.

Methodology & Formulas

The methodology relies on a one-dimensional heat transfer model for an infinite slab. The total thermal resistance is determined by the sum of the convective resistance and the conductive resistance of the packaging. The freezing time is then calculated by accounting for both the external resistance and the internal conduction through the frozen product layer.

The latent heat factor, representing the energy required for phase change per unit temperature difference, is defined as:

\[ \lambda_{\text{factor}} = \frac{\rho \cdot L_{f}}{T_{f} - T_{\infty}} \]

The effective overall heat transfer coefficient, accounting for the packaging resistance, is calculated as:

\[ \frac{1}{U_{\text{eff}}} = \frac{1}{h} + \frac{z_{\text{pkg}}}{k_{\text{pkg}}} \]

The freezing time is determined by the sum of the external resistance term and the internal conduction term:

\[ t_{f} = \lambda_{\text{factor}} \cdot \left( \frac{P \cdot D}{U_{\text{eff}}} + \frac{R \cdot D^{2}}{k_{m}} \right) \]

To assess the validity of the model, the Biot number is calculated based on the characteristic length \(L_{c} = D / 2\):

\[ Bi = \frac{U_{\text{eff}} \cdot L_{c}}{k_{m}} \]

Regime / Condition	Criteria
Biot Number Validity	\(0.1 \leq Bi \leq 10.0\)
Frozen Food Conductivity	\(0.5 \leq k_{m} \leq 2.0 \text{ W/m}\cdot\text{K}\)
Packaging Conductivity	\(0.05 \leq k_{\text{pkg}} \leq 0.15 \text{ W/m}\cdot\text{K}\)
Convection Coefficient	\(10.0 \leq h \leq 200.0 \text{ W/m}^{2}\cdot\text{K}\)
Latent Heat	\(200 \leq L_{f} \leq 300 \text{ kJ/kg}\)

The packaging material acts as an additional thermal barrier that increases the total resistance to heat transfer. To calculate the impact on freezing time, process engineers must consider:

The thermal conductivity of the packaging material.
The thickness of the packaging layers.
The presence of air gaps between the product surface and the packaging.

When modeling freezing time for wrapped goods using Plank's equation, the key variables that govern the total thermal resistance and freezing duration include:

The convective heat transfer coefficient (\(h\)) of the cooling medium.
The thermal conductivity (\(k_{\text{pkg}}\)) and thickness (\(z_{\text{pkg}}\)) of the packaging material.
The effective overall heat transfer coefficient (\(U_{\text{eff}}\)), which combines the convective and packaging resistances in series: \(1/U_{\text{eff}} = 1/h + z_{\text{pkg}}/k_{\text{pkg}}\).
The frozen food thermal conductivity (\(k_{m}\)) and the product thickness (\(D\)).
The Biot number (\(Bi = U_{\text{eff}} \cdot L_{c} / k_{m}\)), which must remain within the valid range of 0.1 to 10 for this model.

Yes, you can optimize the process by modifying the external heat transfer conditions. Consider the following strategies:

Increasing the air velocity in the blast freezer to reduce the external boundary layer resistance.
Lowering the ambient temperature of the cooling medium.
Ensuring a tighter fit of the packaging to eliminate insulating air pockets.

Worked Example: Freezing Time with Packaging Resistance

Scenario: A thin slab of packaged fish fillet (thickness 4 cm) is frozen in a blast freezer. The packaging adds a significant conductive resistance compared to the convective air film. Plank's equation is used to calculate the freezing time with and without the packaging to quantify its impact.

Knowns:

Slab shape factors: \(P = 0.5\), \(R = 0.125\)
Slab thickness: \(D = 0.040\) m
Half-thickness (characteristic length for Biot): \(L_{c} = 0.020\) m
Product density: \(\rho = 1000.0\) kg/m³
Latent heat of freezing: \(L_{f} = 250000.0\) J/kg
Freezing point: \(T_{f} = -1.0\) °C
Coolant temperature: \(T_{\infty} = -12.5\) °C
Frozen food conductivity: \(k_{m} = 1.5\) W/(m·K)
Convection coefficient: \(h = 100.0\) W/(m²·K)
Packaging thickness: \(z_{\text{pkg}} = 0.003\) m
Packaging conductivity: \(k_{\text{pkg}} = 0.06\) W/(m·K)

Step-by-Step Calculation:

Temperature driving force:
\(\Delta T = T_{f} - T_{\infty} = -1.0 - (-12.5)\).
Numerical result: \(\Delta T = 11.5\) K.
Latent heat factor:
\(\lambda_{\text{factor}} = \rho \cdot L_{f} / \Delta T = 1000.0 \cdot 250000.0 / 11.5\).
Numerical result: \(\lambda_{\text{factor}} = 21,739,130.435\) J/m³·K.
Thermal resistances and effective coefficients:
Convection resistance: \(R_{\text{conv}} = 1 / h = 0.010\) m²·K/W.
Packaging resistance: \(R_{\text{pkg}} = z_{\text{pkg}} / k_{\text{pkg}} = 0.003 / 0.06 = 0.050\) m²·K/W.
Effective coefficient without packaging: \(U_{\text{no pkg}} = h = 100.0\) W/m²·K.
Effective coefficient with packaging: \(U_{\text{eff}} = 1 / (R_{\text{conv}} + R_{\text{pkg}}) = 1 / (0.010 + 0.050)\).
Numerical result: \(U_{\text{eff}} = 16.667\) W/m²·K.
Biot number validation:
Without packaging: \(Bi_{\text{no pkg}} = U_{\text{no pkg}} \cdot L_{c} / k_{m} = 100.0 \cdot 0.020 / 1.5\).
Numerical result: \(Bi_{\text{no pkg}} = 1.333\).
With packaging: \(Bi_{\text{with pkg}} = U_{\text{eff}} \cdot L_{c} / k_{m} = 16.667 \cdot 0.020 / 1.5\).
Numerical result: \(Bi_{\text{with pkg}} = 0.222\).
Both Biot numbers lie within the valid range (0.1 to 10) for Plank's equation.
Plank's equation shape terms:
External term without packaging: \(A_{1} = P \cdot D \cdot (1/h) = 0.5 \cdot 0.04 \cdot 0.01\).
Numerical result: \(A_{1} = 0.000200\).
External term with packaging: \(A_{2} = P \cdot D / U_{\text{eff}} = 0.5 \cdot 0.04 / 16.667\).
Numerical result: \(A_{2} = 0.001200\).
Internal conduction term (identical for both): \(B = R \cdot D^{2} / k_{m} = 0.125 \cdot 0.0016 / 1.5\).
Numerical result: \(B = 0.000133\).
Freezing time (seconds):
Without packaging: \(t_{f,\text{no pkg}} = \lambda_{\text{factor}} \cdot (A_{1} + B) = 21,739,130.435 \cdot (0.000200 + 0.000133)\).
Numerical result: \(t_{f,\text{no pkg}} = 7,246.377\) seconds.
With packaging: \(t_{f,\text{with pkg}} = \lambda_{\text{factor}} \cdot (A_{2} + B) = 21,739,130.435 \cdot (0.001200 + 0.000133)\).
Numerical result: \(t_{f,\text{with pkg}} = 28,985.507\) seconds.
Freezing time (hours):
Without packaging: \(t_{f,\text{no pkg}} = 7,246.377 / 3600\).
Numerical result: \(t_{f,\text{no pkg}} = 2.013\) hours.
With packaging: \(t_{f,\text{with pkg}} = 28,985.507 / 3600\).
Numerical result: \(t_{f,\text{with pkg}} = 8.052\) hours.

Final Answer:

The freezing time for the 4 cm slab in the blast freezer is calculated for two cases:

Without packaging: \(t_{f} = 2.013\) hours.
With 3 mm cardboard packaging: \(t_{f} = 8.052\) hours.

The presence of packaging introduces a thermal resistance (\(R_{\text{pkg}} = 0.05\) m²·K/W) that is five times greater than the convective air film resistance (\(R_{\text{conv}} = 0.01\) m²·K/W). This analysis demonstrates the significant impact of packaging on total freezing time.

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