Question

White potatoes \(\left(k=0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(20^{\circ} \mathrm{C}\) and have an average diameter of \(6 \mathrm{~cm}\) are to be cooled by refrigerated air at \(2^{\circ} \mathrm{C}\) flowing at a velocity of $4 \mathrm{~m} / \mathrm{s}$. The average heat transfer coefficient between the potatoes and the air is experimentally determined to be $19 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Determine how long it will take for the center temperature of the potatoes to drop to \(6^{\circ} \mathrm{C}\). Also, determine if any part of the potatoes will experience chilling injury during this process. Solve this problem using the analytical one-term approximation method.

Short answer

Question: Determine the time it would take for the center temperature of the potatoes to drop down to \(6^{\circ} C\) using the analytical one-term approximation method. Answer: It will take approximately 7391 seconds (~2 hours) for the center temperature of the potatoes to drop to \(6^{\circ} C\).

Step-by-step solution

Calculation of Biot Number

First, let's calculate the Biot number using the following formula: $$Bi = \frac{hL_{C}}{k}$$ where, \(h\) is the heat transfer coefficient, \(L_{C}\) is the characteristic length or radius, and \(k\) is the thermal conductivity of the potato. The characteristic length, \(L_{C} = r = \frac{d}{2}\), is half of the diameter for a sphere. Hence, $$r = \frac{6\times10^{-2}}{2} ~m$$ Now, substituting the values into the Biot number formula: $$Bi = \frac{19 \frac{W}{m^2 \cdot K} \times 3\times10^{-2} m}{0.50 \frac{W}{m \cdot K}} $$

Evaluating the Biot Number

Evaluate the Biot number and make sure it is less than 0.1 to confirm the lumped system analysis validity: $$Bi = 1.14$$ Since Bi > 0.1, we cannot use lumped system analysis. We must use the analytical one-term approximation method.

Analytical One-Term Approximation Method

For the analytical one-term approximation method, we can use the following relationship: $$\frac{T - T_\infty}{T_i - T_\infty} = \sum_{n=1}^{\infty} C_n \exp \left( -\frac{n^2 \pi^2 \alpha t}{r^2} \right)$$ Consider only the first term (n=1) for the one-term approximation, $$\frac{T - T_\infty}{T_i - T_\infty} = C_1 \exp \left( -\frac{\pi^2 \alpha t}{r^2} \right)$$ Where \(T = 6^{\circ} C (final ~temperature)\), \(T_i = 20^{\circ} C(initial ~temperature)\), and \(T_\infty = 2^{\circ} C(air ~temperature)\). We need to find the value of \(t\) (time in seconds).

Calculating Time

First, let's solve for the constant \(C_1\), by substituting when \(x = 0\) and \(T(0) = T_i\): $$C_1 = \frac{T_i - T_\infty}{T_i - T_\infty} = 1$$ Now, let's solve for \(t\): $$\frac{6^{\circ} C - 2^{\circ} C}{20^{\circ} C - 2^{\circ} C} = \exp \left( -\frac{\pi^2 (0.13 \times 10^{-6} \frac{m^2}{s}) t}{(3\times10^{-2} m)^2} \right)$$ Take the natural logarithm of both sides: $$\ln \left( \frac{4}{18} \right) = -\frac{\pi^2 0.13 \times 10^{-6} t}{9\times10^{-4}} $$ Now solve for \(t\): $$t = \frac{-9\times10^{-4}\ln \left( \frac{2}{9} \right)}{\pi^2 0.13 \times 10^{-6}}$$

Time taken for cooling

Evaluate the time taken for the center temperature of the potatoes to drop to \(6^{\circ} C\): $$t = 7391 ~s$$ This reveals that it will take approximately 7391 seconds (~2 hours) for the center temperature of the potatoes to drop to \(6^{\circ} C\).

Chilling Injury

The chilling injury is not a part of the calculation since the question does not provide specific temperature limits or rates that cause chilling injury. However, the temperature gradient in the potatoes can cause uneven cooling, resulting in chilling injury at some parts of the potato.