Air Velocity Effect on Freezing Rate

Reference ID: MET-BFBB | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

The calculation of air velocity effects on freezing rates is a fundamental task in food process engineering and industrial refrigeration. In an air blast freezer, the rate at which a product reaches its target frozen state is governed by the convective heat transfer coefficient (h), which is highly sensitive to the velocity of the cooling medium and is closely linked to the effect of surface heat transfer on freezing.

This analysis is critical for optimizing energy consumption and throughput in cold chain logistics. By understanding the relationship between air velocity and heat flux, engineers can determine the optimal fan power required to achieve specific freezing times without incurring excessive operational costs. This model is typically applied to slab-shaped products where forced convection is the primary mechanism for heat removal.

Methodology & Formulas

The system assumes steady-state forced convection over an isothermal surface. The following equations define the relationship between fluid dynamics and thermal performance:

1. Mass Velocity: The mass flux of the air stream is defined by the product of air density and velocity:

\[ G = \rho \cdot V \]

2. Convective Heat Transfer Coefficient: The empirical correlation for turbulent flow in industrial blast freezers is given by:

\[ h = 20 \cdot G^{0.8} \]

3. Convective Heat Flux: The rate of heat removal per unit area is calculated using Newton's Law of Cooling:

\[ \dot{Q}'' = h \cdot (T_{s} - T_{\infty}) \]

4. Dimensionless Analysis: To validate the flow regime and determine if the process is internally or externally limited, we calculate the Reynolds number (Re) and the Biot number (Bi):

\[ Re = \frac{G \cdot L}{\mu} \] \[ Bi = \frac{h \cdot L_{char}}{k_{product}} \]

Parameter	Condition / Threshold	Engineering Significance
Flow Regime	\( Re > 5 \cdot 10^{5} \)	Required for the validity of the turbulent correlation.
Empirical Range	\( 5 \leq G \leq 25 \)	Valid range for the \( h \approx 20 \cdot G^{0.8} \) correlation.
Biot Number	\( Bi > 0.1 \)	Indicates internal conduction resistance is significant; freezing is internally limited.

Increasing air velocity significantly enhances the convective heat transfer coefficient, which accelerates the freezing rate. By reducing the thickness of the stagnant boundary layer surrounding the product, the system achieves more efficient thermal exchange. Key benefits include:

Reduced total freezing time for the product core.
Minimized ice crystal size, leading to better texture retention.
Increased throughput capacity for continuous freezing tunnels.

Yes, process engineers must account for the law of diminishing returns. While higher velocities improve heat transfer, the relationship is not linear. Beyond a certain threshold, the internal thermal resistance of the product becomes the limiting factor rather than the external convective resistance. Considerations include:

Exponential increases in fan power consumption and operational costs.
Potential for product dehydration or freezer burn if humidity is not strictly controlled.
Increased pressure drop across the evaporator coils.

High air velocity can lead to increased moisture loss through sublimation, which negatively impacts product yield. To mitigate this, engineers should focus on:

Maintaining a small temperature difference between the air and the evaporator surface.
Ensuring uniform airflow distribution to prevent localized high-velocity zones.
Optimizing the air velocity setpoint to balance rapid freezing with moisture retention requirements.

Worked Example: Air Velocity Effect on Freezing Rate in a Blast Freezer

Scenario: A food slab (minced meat, 5 cm thick, area 1 m²) is frozen in a tunnel freezer. Cold air at −20°C and 1 atm flows over the slab surface, which remains at 0°C during phase change. The air density is 1.4 kg/m³. We double the air velocity from 7.143 m/s to 14.286 m/s to assess the impact on convective heat transfer.

Knowns:

Air density: ρ = 1.4 kg/m³
Initial velocity: V₁ = 7.143 m/s
Final velocity: V₂ = 14.286 m/s
Surface temperature: T_s = 0.0°C
Air temperature: T_∞ = −20.0°C
Slab half-thickness: L_char = 0.025 m
Product thermal conductivity: k_product = 0.5 W/m·K
Air dynamic viscosity: μ = 1.5×10⁻⁵ Pa·s
Plate length (turbulence criterion): L_plate = 1.0 m
Empirical correlation: h = 20 · G⁰·⁸ for G ∈ [5, 25] kg/m²·s

Step-by-Step Calculation:

Convert velocity to mass velocity. \[ G_1 = \rho V_1 = 1.4 \times 7.143 = 10.0\ \text{kg/m}^2\!\cdot\!\text{s} \] \[ G_2 = \rho V_2 = 1.4 \times 14.286 = 20.0\ \text{kg/m}^2\!\cdot\!\text{s} \]
Apply the convective correlation. \[ h_1 = 20 \cdot (10.0)^{0.8} = 126.191\ \text{W/m}^2\!\cdot\!\text{K} \] \[ h_2 = 20 \cdot (20.0)^{0.8} = 219.712\ \text{W/m}^2\!\cdot\!\text{K} \]
Compute the percentage increase in h. \[ \frac{h_2 - h_1}{h_1} \times 100 = \frac{219.712 - 126.191}{126.191} \times 100 = 74.11\% \]
Determine convective heat flux (optional context). \[ \dot{Q}''_1 = h_1 (T_s - T_\infty) = 126.191 \times 20 = 2523.829\ \text{W/m}^2 \] \[ \dot{Q}''_2 = h_2 (T_s - T_\infty) = 219.712 \times 20 = 4394.242\ \text{W/m}^2 \] The heat flux also increases by 74.11%.
Check validity and flow regime.
Reynolds numbers: \[ Re_1 = \frac{G_1 L_{\text{plate}}}{\mu} = \frac{10.0 \times 1.0}{1.5\times10^{-5}} = 666666.667\ (> 5\times10^5,\ \text{turbulent}) \] \[ Re_2 = \frac{G_2 L_{\text{plate}}}{\mu} = \frac{20.0 \times 1.0}{1.5\times10^{-5}} = 1333333.333\ (> 5\times10^5,\ \text{turbulent}) \]
Mass velocity range: G₁ = 10.0 and G₂ = 20.0 lie within the valid empirical bounds [5, 25] kg/m²·s.
Biot number analysis (internal resistance check). \[ Bi_1 = \frac{h_1 L_{\text{char}}}{k_{\text{product}}} = \frac{126.191 \times 0.025}{0.5} = 6.31 \] \[ Bi_2 = \frac{h_2 L_{\text{char}}}{k_{\text{product}}} = \frac{219.712 \times 0.025}{0.5} = 10.986 \] Both Bi > 0.1 indicate that internal conduction resistance is significant; doubling the air velocity increases the convective coefficient, but the overall freezing time will be limited by the product’s thermal conductivity.

Final Answer: Doubling the air velocity from 7.143 m/s to 14.286 m/s increases the convective heat transfer coefficient by 74.11% (from 126.191 W/m²·K to 219.712 W/m²·K) and the convective heat flux by the same percentage (from 2523.829 W/m² to 4394.242 W/m²). The flow remains turbulent and within the correlation’s validity range. The Biot numbers (6.31 and 10.986) show that internal conduction resistance controls the freezing rate.

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