Oranges of \(2.5\)-in-diameter \(\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) 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 \(4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). 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.

The Biot number is a dimensionless quantity representing the ratio of internal heating or cooling rates to the rate at which heat is conducted through a body. It can be determined using the following formula: $$ Bi = \frac{h L_c}{k} $$ where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity. For our problem, we are given: - Heat transfer coefficient \(h = 4.6 \, Btu/h \cdot ft^2 \cdot °F\) - Characteristic length \(L_c = r/3 = diameter/6 = 2.5/6 \, in\) - Thermal conductivity \(k = 0.26 \, Btu/h \cdot ft \cdot °F\) Now, we can plug in the values and calculate the Biot number: $$ Bi = \frac{4.6 \, Btu/h \cdot ft^2 \cdot °F \times \frac{2.5}{6} \, in}{0.26 \, Btu/h \cdot ft \cdot °F} $$ Note that we need to convert inches to feet, 1 inch = 0.0833333 feet. So, the equation becomes: $$ Bi = \frac{4.6 \times \frac{2.5 \times 0.0833333}{6}}{0.26} \approx 0.33 $$

The Fourier number is another dimensionless quantity representing the ratio of the rates of heat conduction in a material to its heat storage capacity. It can be determined using the following formula: $$ Fo = \frac{\alpha t}{L_c^2} $$ where \(\alpha\) is the thermal diffusivity, \(t\) is the time, and \(L_c\) is the characteristic length. We are given the thermal diffusivity \(\alpha = 1.4 \times 10^{-6} \, ft^2/s\). Since the cooling process is unsteady, we need to utilize a correlation to find the Fourier Number. For a sphere with \(Bi < 0.10\), we can use the following correlation: $$ \frac{1 - exp(-4.04 Bi \, Fo)}{2 \, Fo} = \frac{T - T_s}{T_i - T_s} $$ where \(T\) is the center temperature, \(T_s\) is the surrounding temperature, and \(T_i\) is the initial temperature. We are given: - Initial temperature \(T_i = 78°F\) - Surrounding temperature \(T_s = 25°F\) - Center temperature \(T = 40°F\) Let's plug in the temperature values in the equation and calculate \(Fo\): $$ \frac{1 - exp(-4.04 \times 0.33 \, Fo)}{2 \, Fo} = \frac{40 - 25}{78 - 25} \Rightarrow Fo \approx 0.35 $$

Now that we have the Fourier Number, we can solve for the cooling time \(t\) using the formula for the Fourier Number: $$ t = \frac{Fo \times L_c^2}{\alpha} $$ Plugging in the values, we get: $$ t = \frac{0.35 \times \left(\frac{2.5 \times 0.0833333}{6}\right)^2}{1.4 \times 10^{-6}} \approx 2710 \, seconds $$

To determine if any part of the oranges will freeze, we can check if the minimum temperature is below the freezing point of water (32°F). Taking into account the initial and surrounding temperatures and the ratio of internal to conductive heating, we can assume that the coldest point at any time would be close to the surface of the orange, which is the most exposed to the cold refrigerated air. We can use the correlation for temperature difference of a sphere: $$ \frac{T - T_s}{T_i - T_s} = \frac{1 - exp(-4.04 Bi \, Fo)}{2 \, Fo} $$ Using \(Bi = 0.33\) and \(Fo = 0.35\), and rearranging for \(T\), we can calculate the minimum temperature: $$ T = T_s + \left(\frac{1 - exp(-4.04 \times 0.33 \times 0.35)}{2 \times 0.35}\right) \times (T_i - T_s) \approx 35 \, °F $$ Since the minimum temperature is above 32°F, no part of the oranges will freeze during the cooling process. #Conclusion# The cooling time for the center temperature of the oranges to drop to 40°F is approximately 2710 seconds. No part of the oranges will freeze during this process.