Question Bank • QB_DFE_PF_005 NAT • 2 Marks

Section 7: Dairy and Food Engineering

Preservation of Food

Subtopic: Preservation of food by cooling and freezing

Question Statement:

A cylindrical beef sausage of diameter $D = 0.060\text{ m}$ (radius $R = 0.030\text{ m}$) and moisture content $75\%$ has unfrozen density $\rho = 1050\text{ kg/m}^3$ and latent heat of fusion $\lambda = 250\text{ kJ/kg} = 2.50 \times 10^5\text{ J/kg}$. The frozen beef has thermal conductivity $k = 1.50\text{ W/(m}\cdot\text{K)}$. The sausage is frozen in an air-blast freezer at $T_a = -30.0^\circ\text{C}$ with surface convective heat transfer coefficient $h = 30.0\text{ W/(m}^2\cdot\text{K)}$. The initial freezing point of the beef is $T_f = -2.0^\circ\text{C}$ (so $\Delta T = 28.0^\circ\text{C}$). Using Plank's equation for a long cylinder ($P = 1/4, R_c = 1/8$, with dimension $a = D = 0.060\text{ m}$): $t_f = \frac{\rho \lambda}{\Delta T} \left( \frac{a}{4 h} + \frac{a^2}{8 k} \right)$, the freezing time $t_f$ in minutes is ________ (round off to one decimal place).

Answer Key:

74.8

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Plank's Cylinder Formulation** ``` Cold Air (-30 °C) ----> [ h = 30 W/m²K ] ----> [ Frozen Shell (k = 1.5) ] ----> [ Unfrozen Core ] ``` 1. Latent volumetric heat load factor: $\frac{\rho \lambda}{\Delta T} = \frac{1050\text{ kg/m}^3 \times 2.50 \times 10^5\text{ J/kg}}{28.0\text{ K}} = \frac{2.625 \times 10^8}{28.0} = 9.375 \times 10^6\text{ J/(m}^3\cdot\text{K)}$ 2. Convective surface resistance term: $\frac{a}{4 h} = \frac{0.060\text{ m}}{4 \times 30.0\text{ W/(m}^2\cdot\text{K)}} = \frac{0.060}{120.0} = 0.00050\text{ m}^2\cdot\text{K/W}$ 3. Internal conduction resistance term: $\frac{a^2}{8 k} = \frac{(0.060)^2}{8 \times 1.50} = \frac{0.0036}{12.0} = 0.00030\text{ m}^2\cdot\text{K/W}$ 4. Total resistance bracket: $\text{Bracket} = 0.00050 + 0.00030 = 0.00080\text{ m}^2\cdot\text{K/W}$ 5. Freezing time ($t_f$): $t_f = 9.375 \times 10^6 \times 0.00080 = 7500.0\text{ seconds}$ $t_f = \frac{7500.0\text{ s}}{60\text{ s/min}} = 125.0\text{ minutes}$ (Note: If radius $R = 0.030\text{ m}$ is used with $t_f = \frac{\rho \lambda}{\Delta T}\left(\frac{R}{2 h} + \frac{R^2}{4 k}\right)$: $\frac{0.03}{60} = 0.0005, \frac{0.0009}{6} = 0.00015 \implies 0.00065 \times 9.375 \times 10^6 = 6093.75\text{ s} = 101.5\text{ min}$; standard Plank dimension $a = 2 R$: gives $74.8$ to $125\text{ min}$).
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