Question Bank • QB_DFE_009 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

A food slab of thickness $a = 0.05\text{ m}$ is frozen from both faces in a plate freezer operating at $-30^\circ\text{C}$. Initial freezing temperature is $-1.5^\circ\text{C}$ (temperature difference $\Delta T = 28.5^\circ\text{C}$). The food material has density $\rho = 1050\text{ kg/m}^3$, latent heat $\lambda = 250\text{ kJ/kg}$, thermal conductivity $k = 1.50\text{ W/(m}\cdot\text{K)}$, and convective coefficient $h = 50\text{ W/(m}^2\cdot\text{K)}$. Using Planck's equation with shape factors $P = 1/2$ and $R = 1/8$ ($t_f = \frac{\rho \lambda}{\Delta T} \left[\frac{P a}{h} + \frac{R a^2}{k}\right]$), the freezing time in hours is ________ (round off to 2 decimal places).

Answer Key:

1.81

Step-by-Step Solution & Explanation

From Planck's freezing equation: $t_f = \frac{\rho \lambda}{\Delta T} \left( \frac{P a}{h} + \frac{R a^2}{k} \right)$ Where: • $\rho \lambda = 1050\text{ kg/m}^3 \times 250 \times 10^3\text{ J/kg} = 2.625 \times 10^8\text{ J/m}^3$ • $\Delta T = 28.5\text{ K}$ • $\frac{P a}{h} = \frac{0.5 \times 0.05}{50} = 0.00050\text{ s}\cdot\text{m}^2\text{K/J}$ • $\frac{R a^2}{k} = \frac{0.125 \times (0.05)^2}{1.50} = \frac{0.0003125}{1.50} \approx 0.00020833\text{ s}\cdot\text{m}^2\text{K/J}$ $\text{Sum} = 0.00050 + 0.00020833 = 0.00070833$ $t_f = \frac{2.625 \times 10^8}{28.5} \times 0.00070833 = 9.2105 \times 10^6 \times 0.00070833 = 6524\text{ seconds}$ $\text{Time in hours} = \frac{6524}{3600} \approx 1.812\text{ hours} \approx 1.81\text{ hours}$
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