Subtopic: Plank equation freezing time calculation
Question Statement:
A slab of meat of thickness $a = 5\text{ cm}$ is frozen in an air-blast freezer where ambient air is at $-19^\circ\text{C}$ and the initial freezing point of meat is $-1^\circ\text{C}$ ($\Delta T = 18^\circ\text{C}$). The density of frozen meat is $\rho_f = 950\text{ kg/m}^3$, latent heat of freezing is $\lambda = 250\text{ kJ/kg}$, frozen thermal conductivity is $k_f = 1.2\text{ W/(m}\,\text{K)}$, and surface convective coefficient is $h = 25\text{ W/(m}^2\,\text{K)}$. Using Plank's equation for an infinite slab ($P = 1/2$, $R = 1/8$), the freezing time is ________ $\text{hours}$ (round off to two decimal places).
Answer Key:
4.62
Step-by-Step Solution & Explanation
Plank's equation for freezing time of an infinite slab is:
$t_f = \frac{\rho_f \lambda}{\Delta T} \left[ \frac{a}{2 h} + \frac{a^2}{8 k_f} \right]$
$\frac{a}{2 h} = \frac{0.05}{2 \times 25.0} = 0.001000$
$\frac{a^2}{8 k_f} = \frac{0.0025000000000000005}{8 \times 1.2} = 0.000260$
$t_f = \frac{950 \times 250000}{18} \times (0.001000 + 0.000260) = 16630.5\text{ seconds}$
Converting to hours:
$t_f = \frac{16630.5}{3600} = 4.62\text{ hours}$