Question Bank • QB_DFE_EXP_041 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

Plank's equation for freezing time of a slab of thickness $a = 0.05\text{ m}$ frozen from both sides is $t_f = \frac{\rho \lambda}{\Delta T} \left(\frac{P a}{h} + \frac{R a^2}{k}\right)$ with geometric factors $P = \frac{1}{2}$ and $R = \frac{1}{8}$. Given density of frozen food $\rho = 1000\text{ kg/m}^3$, latent heat $\lambda = 250\text{ kJ/kg} = 2.5 \times 10^5\text{ J/kg}$, temperature difference $\Delta T = 20.0\text{ K}$, surface heat transfer coefficient $h = 25.0\text{ W/(m}^2\cdot\text{K)}$, and frozen thermal conductivity $k = 1.25\text{ W/(m}\cdot\text{K)}$. The freezing time $t_f$ in seconds is ________ (answer in integer).

Answer Key:

15625

Step-by-Step Solution & Explanation

1. Term 1: $\frac{\rho \lambda}{\Delta T}$: $\frac{1000 \times 2.5 \times 10^5}{20.0} = 1.25 \times 10^7\text{ J/(m}^3\cdot\text{K)}$ 2. Term 2: $\frac{P a}{h}$: $\frac{0.50 \times 0.05}{25.0} = \frac{0.025}{25.0} = 0.0010\text{ m}^2\cdot\text{K/W}$ 3. Term 3: $\frac{R a^2}{k}$: $\frac{0.125 \times (0.05)^2}{1.25} = \frac{0.125 \times 0.0025}{1.25} = 0.10 \times 0.0025 = 0.00025\text{ m}^2\cdot\text{K/W}$ 4. Sum of resistances: $0.0010 + 0.00025 = 0.00125\text{ m}^2\cdot\text{K/W}$. 5. Freezing time $t_f$: $t_f = (1.25 \times 10^7) \times 0.00125 = 15,625\text{ seconds} \approx 4.34\text{ hours}$
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