Moisture Loss During Blast Freezing

Reference ID: MET-FA21 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

Moisture loss during blast freezing is a critical parameter in food process engineering, directly impacting product yield, quality, and shelf life. This calculation estimates the mass transfer of water vapor from the surface of an unpackaged food product to the surrounding high‑velocity air stream during the initial freezing phase. By quantifying the moisture loss, engineers can optimize freezer residence times, air velocities, and humidity levels to minimize dehydration (often referred to as “freezer burn” or “shrinkage”). For a deeper understanding of how freezing rate influences liquid exudate, see our drip loss prediction from freezing rate methodology, which complements this moisture‑loss analysis. This methodology is typically applied in the design and operational analysis of IQF (Individually Quick Frozen) tunnels and blast freezing chambers where the product surface remains in an active drying state.

Methodology & Formulas

The calculation relies on the Chilton-Colburn analogy to relate heat and mass transfer, assuming a Lewis number of unity. The process follows a sequential determination of fluid properties, flow regimes, and mass flux.

First, determine the film temperature T_f and the Reynolds number Re_L to characterize the flow regime:

\[ T_{f} = \frac{T_{s} + T_{\infty}}{2} \] \[ Re_{L} = \frac{\rho \cdot U \cdot L}{\mu} \]

The Nusselt number Nu_L is selected based on the flow regime:

Regime	Condition	Formula
Laminar	\( Re_{L} < 5 \times 10^{5} \)	\( Nu_{L} = 0.664 \cdot Re_{L}^{0.5} \cdot Pr^{1/3} \)
Turbulent	\( Re_{L} \geq 5 \times 10^{5} \)	\( Nu_{L} = 0.037 \cdot Re_{L}^{0.8} \cdot Pr^{1/3} \)

The heat transfer coefficient h and the mass transfer coefficient h_m are calculated as follows:

\[ h = \frac{Nu_{L} \cdot k}{L} \] \[ h_{m} = \frac{h}{\rho \cdot c_{p}} \]

The driving force for mass transfer is the difference in vapor density between the product surface and the bulk air, calculated using the ideal gas law:

\[ \rho_{v,\infty} = \frac{\phi_{\infty} \cdot P_{\mathrm{sat,ice}}(T_{\infty})}{R_{v} \cdot T_{\infty}} \] \[ \rho_{v,s} = \frac{P_{\mathrm{sat,ice}}(T_{s})}{R_{v} \cdot T_{s}} \] \[ \Delta\rho_{v} = \max(0, \rho_{v,s} - \rho_{v,\infty}) \]

Finally, the total mass loss m_loss and the percentage weight loss are determined:

\[ \dot{m}'' = h_{m} \cdot \Delta\rho_{v} \] \[ m_{\mathrm{loss}} = \dot{m}'' \cdot A_{\mathrm{surf}} \cdot t_{\mathrm{active}} \] \[ \text{Loss}_{\%} = \left( \frac{m_{\mathrm{loss}}}{m_{\mathrm{initial}}} \right) \cdot 100 \]

Moisture loss, or dehydration, is primarily driven by the vapor pressure differential between the product surface and the surrounding air. Key process variables include:

Air velocity across the product surface.
Temperature delta between the product and the evaporator coils.
Relative humidity levels within the blast freezer environment.
Surface area to volume ratio of the product being processed.

To optimize yield while maintaining freezing efficiency, consider the following engineering controls:

Implement staged cooling to reduce the initial thermal shock.
Optimize airflow distribution to ensure uniform heat transfer without excessive surface scouring.
Utilize packaging materials with high moisture barrier properties where applicable.
Maintain evaporator coil temperatures as close to the product temperature as possible to minimize the driving force for sublimation.

The design of the evaporator coil directly impacts the moisture holding capacity of the air. A larger surface area on the coils allows for a smaller temperature difference between the refrigerant and the air, which significantly reduces the rate of frost formation and moisture migration from the product. Engineers should focus on:

Increasing coil surface area to allow for higher suction pressures.
Reducing the delta T between the air and the refrigerant.
Ensuring proper defrost cycles to prevent efficiency degradation.

Worked Example: Moisture Loss During Blast Freezing of a Fish Fillet

A 0.75 kg fish fillet (0.3 m × 0.1 m × 0.025 m, exposed surface area 0.06 m²) is blast frozen in a tunnel with air at –30 °C, 100 % relative humidity, flowing at 12 m/s. The product surface is assumed to remain at 0 °C during the first 30 min (1800 s) of crust freezing. Estimate the moisture loss (%) during this active drying period.

Knowns (Input Parameters)

Air bulk temperature, \(T_{\infty} = -30\;^{\circ}\mathrm{C}\) (243.15 K)
Product surface temperature, \(T_{s} = 0\;^{\circ}\mathrm{C}\) (273.15 K)
Air velocity, \(U = 12.0\;\mathrm{m/s}\)
Characteristic length (fillet length), \(L = 0.3\;\mathrm{m}\)
Exposed surface area, \(A_{\mathrm{surf}} = 0.06\;\mathrm{m^{2}}\)
Initial mass, \(m_{\mathrm{initial}} = 0.75\;\mathrm{kg}\)
Relative humidity, \(\phi_{\infty} = 1.0\)
Active drying time, \(t_{\mathrm{active}} = 1800\;\mathrm{s}\)
Air density (at film temperature), \(\rho_{\mathrm{air}} = 1.37\;\mathrm{kg/m^{3}}\)
Air specific heat, \(c_{p} = 1006\;\mathrm{J/(kg\cdot K)}\)
Air thermal conductivity, \(k = 0.023\;\mathrm{W/(m\cdot K)}\)
Air dynamic viscosity, \(\mu = 1.6\times10^{-5}\;\mathrm{Pa\cdot s}\)
Air Prandtl number, \(Pr = 0.72\)
Saturation pressure of ice at \(T_{\infty}\): \(P_{\mathrm{sat},\infty} = 38\;\mathrm{Pa}\)
Saturation pressure of ice at \(T_{s}\): \(P_{\mathrm{sat},s} = 611\;\mathrm{Pa}\)
Gas constant for water vapor, \(R_{v} = 461.5\;\mathrm{J/(kg\cdot K)}\)

Step-by-Step Calculation

Film temperature and air properties:
\[ T_{f} = \frac{T_{s} + T_{\infty}}{2} = 258.15\;\mathrm{K} \] The air properties were evaluated at this film temperature and are given in the Knowns.
Reynolds number:
\[ Re_{L} = \frac{\rho_{\mathrm{air}}\,U\,L}{\mu} = \frac{1.37\times 12.0\times 0.3}{1.6\times10^{-5}} = 308\,250 \] Since \(Re_{L} < 5\times10^{5}\), the flow is laminar.
Nusselt number (laminar correlation):
\[ Nu_{L} = 0.664\;Re_{L}^{1/2}\;Pr^{1/3} = 0.664\times (308\,250)^{0.5}\times (0.72)^{1/3} = 330.4 \]
Heat transfer coefficient:
\[ h = \frac{Nu_{L}\;k}{L} = \frac{330.4\times 0.023}{0.3} = 25.33\;\mathrm{W/(m^{2}\cdot K)} \]
Mass transfer coefficient (Chilton–Colburn analogy, \(Le \approx 1\)):
\[ h_{m} = \frac{h}{\rho_{\mathrm{air}}\,c_{p}} = \frac{25.33}{1.37\times 1006} = 0.01838\;\mathrm{m/s} \]
Vapor densities (ideal gas law):
\[ \rho_{v,\infty} = \frac{\phi_{\infty}\,P_{\mathrm{sat},\infty}}{R_{v}\;T_{\infty}} = \frac{1.0\times 38}{461.5\times 243.15} = 0.000339\;\mathrm{kg/m^{3}} \] \[ \rho_{v,s} = \frac{P_{\mathrm{sat},s}}{R_{v}\;T_{s}} = \frac{611}{461.5\times 273.15} = 0.004847\;\mathrm{kg/m^{3}} \]
Driving force for mass transfer:
\[ \Delta\rho_{v} = \max(0,\;\rho_{v,s}-\rho_{v,\infty}) = 0.004847 - 0.000339 = 0.004508\;\mathrm{kg/m^{3}} \]
Mass flux:
\[ \dot{m}'' = h_{m}\;\Delta\rho_{v} = 0.01838 \times 0.004508 = 8.286\times10^{-5}\;\mathrm{kg/(m^{2}\cdot s)} \]
Total moisture removed:
\[ m_{\mathrm{loss}} = \dot{m}''\;A_{\mathrm{surf}}\;t_{\mathrm{active}} = 8.286\times10^{-5}\times 0.06\times 1800 = 0.00895\;\mathrm{kg} \]
Weight loss percentage:
\[ \text{Loss}_{\%} = \frac{m_{\mathrm{loss}}}{m_{\mathrm{initial}}}\times 100 = \frac{0.00895}{0.75}\times 100 = 1.19\% \]

Final Answer

The estimated moisture loss during the first 30 minutes of blast freezing is 0.00895 kg, which corresponds to a weight loss of 1.19 % of the initial product mass.

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