Question

Oranges of \(2.5\)-in-diameter $\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\( and \)\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}$ ) initially at a uniform temperature of \(78^{\circ} \mathrm{F}\) are to be cooled by refrigerated air at $25^{\circ} \mathrm{F}\( flowing at a velocity of \)1 \mathrm{ft} / \mathrm{s}$. The average heat transfer coefficient between the oranges and the air is experimentally determined to be $4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Determine how long it will take for the center temperature of the oranges to drop to \(40^{\circ} \mathrm{F}\). Also, determine if any part of the oranges will freeze during this process. Solve this problem using the analytical one-term approximation method.

Short answer

Answer: It will take approximately 184.77 seconds for the center temperature of the oranges to drop to 40°F. No part of the oranges will freeze during this cooling process.

Step-by-step solution

Find the Biot Number (Bi)

The Biot number (Bi) can be calculated using the formula: $$ \text{Bi} = \frac{hL_c}{k} $$ where $$h$$ is the heat transfer coefficient, $$L_c$$ is the characteristic length of the object, and $$k$$ is the thermal conductivity. In this case, h = 4.6 Btu/h/ft²°F, L_c = diameter / 6 (for spheres) = 2.5 / 6, and k = 0.26 Btu/h/ft/°F. $${\rm Bi} = \frac {4.6 \, {\rm Btu}/h/ft^{2}/^{\circ}F \times \frac{2.5}{6} \, {\rm ft} }{0.26 \, {\rm Btu}/h/ \rm ft\cdot{ }^{\circ} \mathrm{F}} \approx 7.43$$

Find the Fourier Number (Fo)

The Fourier number (Fo) can be calculated using the formula: $$ \text{Fo} = \frac{\alpha t}{L_c^2}. $$ Here, α = \(1.4 \times 10^{-6} \, {\rm ft}^2/s\) and $$L_c = \frac{2.5}{6} \, {\rm ft}$$. We can rearrange the formula to give the time, t: $$ t = \frac{\text{Fo} \times L_c^2}{\alpha}. $$

Use the analytical one-term approximation method

We are given that we should use the analytical one-term approximation method. At the center of the sphere, r = 0. Hence, we will use Heisler charts or an analytical expression to find the value of the Fourier number (Fo) based on the values derived above. Heisler chart for the center of the sphere: $$ \frac {T-T_0}{T_i - T_0} = f({\rm Bi}, {\rm Fo}) $$ where T is the center temperature, T_0 is the ambient temperature, and T_i is the initial temperature. We will use the chart to find the value of Fo based on Bi = 7.43. For T = 40°F, T_0 = 25°F, and T_i = 78°F: $$ \frac {40-25}{78-25} = f(7.43, {\rm Fo}) $$ Using the chart, we find that Fo ≈ 0.35.

Calculate the time

Now that we have the Fo value, we can calculate the time using the formula we derived in Step 2: $$ t = \frac{\text{Fo} \times L_c^2}{\alpha} = \frac{0.35 \times (\frac{2.5}{6})^2 }{1.4 \times 10^{-6}} \approx 184.77\, \text{seconds}. $$ Thus, it will take approximately 184.77 seconds for the center temperature of the oranges to drop to 40°F.

Determine if any part of the oranges will freeze

Since the temperature at the center of the oranges drops to 40°F, which is above the freezing point of water (32°F), we can conclude that no part of the oranges will freeze during this cooling process. Alternatively, you can check the Heisler charts for the other values of r, and if those values are above 32°F as well, then no part of the oranges will freeze.