Question Bank • QB_DFE_EXP_103 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

A plate freezer freezes fish fillets packed in flat boxes of thickness $a = 0.05\text{ m}$. In Plank's equation $t_f = \frac{\rho \lambda}{T_f - T_a} \left[ \frac{P a}{h} + \frac{R a^2}{k} \right]$, the density is $\rho = 1000\text{ kg/m}^3$, latent heat is $\lambda = 250\text{ kJ/kg} = 250000\text{ J/kg}$, temperature difference is $\Delta T = 25^\circ\text{C}$, $h = 50\text{ W/(m}^2\cdot\text{K)}$, $k = 1.2\text{ W/(m}\cdot\text{K)}$, $P = 1/2$, and $R = 1/8$. Calculate the freezing time in $\text{seconds}$.

Answer Key:

7604.17

Step-by-Step Solution & Explanation

Plank's terms: $\frac{P a}{h} = \frac{0.5 \times 0.05}{50} = \frac{0.025}{50} = 0.0005\text{ m}^2\cdot\text{K/W}$ $\frac{R a^2}{k} = \frac{0.125 \times (0.05)^2}{1.2} = \frac{0.125 \times 0.0025}{1.2} = \frac{0.0003125}{1.2} \approx 0.0002604\text{ m}^2\cdot\text{K/W}$ Sum of terms $= 0.0005 + 0.0002604 = 0.0007604$. Constant factor: $\frac{\rho \lambda}{\Delta T} = \frac{1000 \times 250000}{25} = 10000000 = 10^7$ $t_f = 10^7 \times 0.000760417 \approx 7604.17\text{ seconds}$
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