Question

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=0.45\) $\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at a uniform temperature of \)15^{\circ} \mathrm{C}$ are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be $0.95 \mathrm{~g} / \mathrm{cm}^{3}$ and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and $45 \mathrm{~min}$. Also, determine if any part of the chicken will freeze during this process. Solve this problem using the analytical one-term approximation method.

Short answer

Based on the given problem, calculate the center and surface temperatures of the chicken after being in the brine for 1 hour, and determine if any part of the chicken will freeze during this time.

Step-by-step solution

Calculate the Biot number (Bi) and Fourier number (Fo)

The Biot number is given by the formula: \[Bi = \frac{hR_c}{k}\] and the Fourier number is given by: \[Fo = \frac{\alpha t}{R_c^2}\] where - \(h\) is the heat transfer coefficient (\(440 \mathrm{W/m^2 K}\)) - \(R_c\) is the characteristic radius of the chicken - \(k\) is the thermal conductivity of the chicken (\(0.45 \mathrm{W/mK}\)) - \(\alpha\) is the thermal diffusivity of the chicken (\(0.13 \times 10^{-6} \mathrm{m^2/s}\)) - \(t\) is the time in seconds First, let's find \(R_c\). The volume of a sphere is given by the formula: \[V = \frac{4}{3}\pi R_c^3\] From the given average density and mass of the chicken, we can find the volume: \[V = \frac{m}{\rho} = \frac{1.7 \mathrm{kg}}{0.95 \times 10^3 \mathrm{kg/m^3}}\] Now, we can find \(R_c\) using the volume formula:

Calculate the dimensionless temperature

Now we need to calculate the dimensionless temperature using the one-term approximation method: \[Θ^* = \frac{1}{3\sqrt{Bi\cdot Fo}} = \frac{1}{3\sqrt{(hR_ck)\cdot(\frac{\alpha t}{R_c^2})}}\] This will give us the dimensionless temperature at the center of the sphere. Next, let's find the dimensionless temperature at the surface, which is given by \(Θ^*_s = 0\) (since the surface is in contact with the brine at constant temperature).

Calculate the actual temperatures at the center and the surface

Now we can find the actual temperatures at the center and the surface of the chicken using the dimensionless temperatures and the formula: \[T = T_\infty + (T_i - T_\infty) Θ^*\] where - \(T_\infty\) is the environmental temperature, which is the brine temperature (\(-7^\circ\mathrm{C}\)) - \(T_i\) is the initial temperature of the chicken (\(15^\circ\mathrm{C}\))

Determine if any part of the chicken will freeze

To determine if any part of the chicken will freeze, we need to check if the temperature at any point falls below the freezing point of the chicken (\(0^\circ\mathrm{C}\)). If the center or surface temperature calculated in Step 3 is less than \(0^\circ\mathrm{C}\), then that part of the chicken will freeze. Otherwise, the chicken will not freeze.