Introduction & Context

In thermal food processing, the heating rate index, denoted as f h, represents the time required for the temperature difference between the product and the heating medium to change by one log cycle. In static retorts, heat transfer is limited by natural convection or conduction, resulting in longer process times. Agitation, specifically end‑over‑end rotation, induces forced convection within the container, significantly accelerating the heating rate; the effect is closely tied to the immersion retort basket rotation speed, which governs the intensity of the induced flow.

This calculation is critical in Process Engineering for optimizing thermal sterilization cycles. By quantifying the reduction in f_h, engineers can reduce the total process time required to achieve a target lethality (F₀), thereby improving product quality, reducing energy consumption, and increasing throughput in industrial canning operations.

Methodology & Formulas

The agitation effect is modeled using an empirical reduction factor (RF) that relates the static heating rate to the agitated heating rate. The following steps outline the physical derivation based on rotational dynamics:

1. Calculate the angular velocity of the retort rotation:

\[ \omega = N_{rpm} \cdot \left( \frac{2 \cdot \pi}{60} \right) \]

2. Determine the rotational Reynolds number to ensure the flow regime is sufficiently turbulent for the empirical correlation:

\[ Re_{rot} = \frac{\rho \cdot \omega \cdot D_{can}^{2}}{\mu} \]

3. Calculate the reduction factor, which accounts for the increased heat transfer coefficient due to forced convection:

\[ RF = 1 + 0.1 \cdot (N_{rpm})^{0.8} \]

4. Compute the agitated heating rate index:

\[ f_{h,agitated} = \frac{f_{h,static}}{RF} \]

5. Adjust the total process time to maintain the required thermal lethality:

\[ t_{adjusted} = t_{static} \cdot \left( \frac{f_{h,agitated}}{f_{h,static}} \right) \]

Parameter	Condition / Constraint
Rotation Speed	\( 8 \leq N_{rpm} \leq 20 \)
Flow Regime	\( Re_{rot} > 10,000 \)
Fill Volume Fraction	\( \phi > 0.90 \)
Reduction Factor	\( 1.5 \leq RF \leq 5.0 \)

Increasing the agitation speed typically results in a reduction of the f_h value, which represents the time required for the heating curve to traverse one log cycle. This occurs because:

Enhanced convective heat transfer coefficients are achieved at the product-container interface.
The boundary layer thickness at the wall is minimized, facilitating faster thermal energy penetration.
Internal product mixing reduces the temperature gradient between the center and the surface of the container.

While reducing f_h improves throughput, process engineers must monitor the following risks:

Mechanical damage to delicate particulates or shear-sensitive ingredients.
Potential for excessive headspace gas incorporation, which may alter heat transfer characteristics.
Inconsistent agitation patterns leading to cold spots if the rotation speed is not uniform across all containers.

Validation requires a systematic approach to ensure the thermal process remains lethal and safe:

Perform heat penetration studies using calibrated thermocouples placed at the slowest heating point.
Calculate the new f_h and j values under the specific agitation speed and product viscosity conditions.
Verify that the lethality (F₀ value) remains within the validated safety margins for the target pathogen.
Conduct a sensitivity analysis to account for variations in product fill weight or initial temperature.

Worked Example: Agitation Effect on Heating Rate (f_h Reduction)

Scenario: A liquid food product (e.g., cream soup) is processed in a sealed cylindrical can using an end-over-end rotary retort. The retort operates at a constant steam temperature. The goal is to determine the reduced heating rate index under agitation and the corresponding adjustment to the sterilization process time.

Known Parameters

Can diameter, \( D = 0.3 \, \text{m} \)
Fill volume fraction, \( \phi = 0.95 \)
Product density, \( \rho = 1000 \, \text{kg/m}^3 \)
Dynamic viscosity, \( \mu = 0.001 \, \text{Pa·s} \)
Rotational speed, \( N_{rpm} = 20 \, \text{RPM} \)
Static heating rate index, \( f_{h,\text{static}} = 25.0 \, \text{min} \)
Static process time (to achieve target \( F_0 \)), \( t_{\text{static}} = 60.0 \, \text{min} \)

Step-by-Step Calculation

Compute angular velocity: \[ \omega = \frac{2\pi}{60} \cdot N_{rpm} = \frac{2\pi}{60} \cdot 20 = 2.094 \, \text{rad/s} \]
Calculate rotational Reynolds number: \[ \text{Re}_{\text{rot}} = \frac{\rho \, \omega \, D^{2}}{\mu} = \frac{1000 \cdot 2.094 \cdot (0.3)^{2}}{0.001} = 188{,}495.6 \] Since \( \text{Re}_{\text{rot}} > 10,000 \), the correlation is valid for turbulent flow.
Determine reduction factor: \[ \text{RF} = 1 + 0.1 \cdot N_{rpm}^{0.8} = 1 + 0.1 \cdot (20)^{0.8} = 2.099 \] Check empirical range: \( 1.5 \leq \text{RF} \leq 5.0 \) – satisfied.
Compute agitated heating rate index: \[ f_{h,\text{agitated}} = \frac{f_{h,\text{static}}}{\text{RF}} = \frac{25.0}{2.099} = 11.91 \, \text{min} \]
Adjust process time: \[ t_{\text{agitated}} = t_{\text{static}} \cdot \frac{f_{h,\text{agitated}}}{f_{h,\text{static}}} = 60.0 \cdot \frac{11.91}{25.0} = 28.59 \, \text{min} \]

Final Answer

The agitated heating rate index is \( f_{h,\text{agitated}} = 11.91 \, \text{min} \). The adjusted process time to achieve the same sterilization effect is \( t_{\text{agitated}} = 28.59 \, \text{min} \).

"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

Agitation Effect on Heating Rate (f_h reduction)