Introduction & Context

The thermal processing of food products in flexible pouches is a critical operation in the food industry, ensuring commercial sterility while maintaining product quality. This calculation estimates the time required for the cold point—the geometric center of the pouch—to reach a target sterilization temperature. By modeling the pouch as an infinite slab subjected to external convective heating, process engineers can optimize retort cycle times, minimize over-processing, and ensure food safety compliance.

Methodology & Formulas

The calculation follows a systematic approach to determine the transient heat conduction within the pouch. The process begins by characterizing the external fluid dynamics to determine the convective heat transfer coefficient, followed by an evaluation of the internal thermal resistance to select the appropriate heat transfer model.

1. External Convection Analysis

The Reynolds number (Re_L) is calculated to determine the flow regime over the pouch surface:

\[ \mathrm{Re}_{L} = \frac{V \cdot L_{\mathrm{char}}}{\nu} \]

The Nusselt number (Nu) for turbulent flow over a flat plate is determined by:

\[ \mathrm{Nu} = 0.037 \cdot \mathrm{Re}_{L}^{0.8} \cdot \mathrm{Pr}^{1/3} \]

The convective heat transfer coefficient (h) is then derived:

\[ h = \frac{\mathrm{Nu} \cdot k_{f}}{L_{\mathrm{char}}} \]

2. Internal Thermal Resistance

The Biot number (Bi) dictates the internal temperature distribution regime:

\[ \mathrm{Bi} = \frac{h \cdot L}{k_{\mathrm{food}}} \]

3. Transient Temperature Analysis

For systems where \( \mathrm{Bi} \geq 0.1 \) and \( \mathrm{Fo} \geq 0.2 \), the one-term approximation of the Heisler solution is employed. First, the eigenvalue (λ₁) is solved via the transcendental equation:

\[ \lambda_{1} \cdot \tan(\lambda_{1}) = \mathrm{Bi} \]

The coefficient (A₁) is calculated as:

\[ A_{1} = \frac{4 \cdot \sin(\lambda_{1})}{2 \cdot \lambda_{1} + \sin(2 \cdot \lambda_{1})} \]

The dimensionless temperature ratio (θ) is defined as:

\[ \theta = \frac{T_{c} - T_{\infty}}{T_{i} - T_{\infty}} \]

The Fourier number (Fo) is then isolated to solve for the process time:

\[ \mathrm{Fo} = -\frac{\ln(\theta / A_{1})}{\lambda_{1}^{2}} \]

\[ t = \frac{\mathrm{Fo} \cdot L^{2}}{\alpha} \]

Regime	Condition	Model Approach
Lumped Capacitance	\( \mathrm{Bi} < 0.1 \)	\( \mathrm{Fo} = -\dfrac{\ln(\theta)}{\mathrm{Bi}} \)
One-Term Approximation	\( \mathrm{Bi} \geq 0.1 \) and \( \mathrm{Fo} \geq 0.2 \)	Heisler transcendental solution
Turbulent Flow	\( 5 \cdot 10^{5} \leq \mathrm{Re}_{L} \leq 10^{7} \)	\( \mathrm{Nu} = 0.037 \cdot \mathrm{Re}_{L}^{0.8} \cdot \mathrm{Pr}^{1/3} \)

To enhance thermal efficiency while maintaining package safety, consider the following process adjustments:

Implement mechanical agitation or rotary processing to disrupt the boundary layer on the external surface of the pouch, increasing the convective heat transfer coefficient.
Optimize the fill volume to ensure consistent contact between the product and the pouch surface.
Utilize high-conductivity laminate materials that allow for faster heat penetration without requiring higher sealing temperatures.

The heat penetration rate is governed by several critical process variables:

The viscosity of the product, which dictates the mode of heat transfer (conduction vs. convection).
The surface area to volume ratio of the pouch geometry.
The thermal resistance of the packaging film layers.
The temperature differential between the heating medium and the product core.

Pouch orientation is critical for optimizing convective currents. When pouches are oriented vertically, internal product movement is facilitated, which significantly reduces the time required to reach the target lethality compared to horizontal stacking. Ensure that your racking system allows for maximum exposure of the pouch faces to the heating medium.

Worked Example: Flexible Pouch Heat Transfer Enhancement

Consider a flexible pouch containing conduction-heating food subjected to pressurized hot-water agitation at 121°C. The pouch half-thickness is 0.01 m, and the food properties are thermal conductivity \(k = 0.6\) W/(m·K) and thermal diffusivity \(\alpha = 1.5\times10^{-7}\) m²/s. The initial food temperature is 25°C, and the target center temperature is 120°C. Water at 121°C has Pr = 1.44, kinematic viscosity \(\nu = 2.6\times10^{-7}\) m²/s, and thermal conductivity \(k_f = 0.685\) W/(m·K). The characteristic length for flow is 0.3 m at velocity 0.5 m/s.

Knowns

\(L = 0.01\) m
\(\alpha = 1.5\times10^{-7}\) m²/s
\(k = 0.6\) W/(m·K)
\(T_{\infty} = 121\) °C
\(T_i = 25\) °C
\(T_c = 120\) °C
\(V = 0.5\) m/s
\(L_{\mathrm{char}} = 0.3\) m
\(\mathrm{Pr} = 1.44\)
\(\nu = 2.6\times10^{-7}\) m²/s
\(k_f = 0.685\) W/(m·K)

Step-by-Step Calculation

Reynolds number: \(\mathrm{Re}_L = V L_{\mathrm{char}} / \nu = 576923.077\).
Nusselt number (turbulent flat plate): \(\mathrm{Nu} = 0.037\, \mathrm{Re}_L^{0.8} \mathrm{Pr}^{1/3} = 1697.788\).
Convective coefficient: \(h = \mathrm{Nu}\, k_f / L_{\mathrm{char}} = 3876.617\) W/(m²·K).
Biot number: \(\mathrm{Bi} = h L / k = 64.61\). Since \(\mathrm{Bi} \gg 0.1\), a lumped analysis is invalid; use the Heisler one-term approximation.
Eigenvalue \(\lambda_1\) solved from \(\lambda_1 \tan \lambda_1 = \mathrm{Bi}\): \(\lambda_1 = 1.547\).
Coefficient \(A_1 = \dfrac{4\sin\lambda_1}{2\lambda_1 + \sin(2\lambda_1)} = 1.273\).
Temperature ratio: \(\theta = \dfrac{T_c - T_{\infty}}{T_i - T_{\infty}} = 0.01042\).
Fourier number: \(\mathrm{Fo} = -\dfrac{\ln(\theta / A_1)}{\lambda_1^2} = 2.008\).
Time: \(t = \dfrac{\mathrm{Fo}\, L^2}{\alpha} = 1338.667\) s = 22.311 min.

Final Answer: The cold point reaches 120°C after approximately 22.3 minutes of heating under the given conditions.

"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