Question

Layers of \(23-\mathrm{cm}\)-thick meat slabs $(k=0.47 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\( ) initially at a uniform temperature of \)7^{\circ} \mathrm{C}$ are to be frozen by refrigerated air at \(-30^{\circ} \mathrm{C}\) flowing at a velocity of \(1.4 \mathrm{~m} / \mathrm{s}\). The average heat transfer coefficient between the meat and the air is $20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming the size of the meat slabs to be large relative to their thickness, determine how long it will take for the center temperature of the slabs to drop to \(-18^{\circ} \mathrm{C}\). Also, determine the surface temperature of the meat slab at that time.

Short answer

Question: Determine the time it takes for a meat slab at an initial temperature of \(7^{\circ} \mathrm{C}\) to be frozen such that the temperature at the center becomes \(-18^{\circ} \mathrm{C}\), and calculate the surface temperature of the meat slab at that time.

Step-by-step solution

Determine if the lumped capacitance method is applicable

To determine the applicability of the lumped capacitance method, we need to calculate the Biot number: $$\mathrm{Bi} = \frac{hL_c}{k}$$ Where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity. For a slab with a thickness of \(23 \mathrm{~cm}\), the characteristic length is equal to half the thickness: $$L_c = \frac{23}{2} \times 10^{-2} \mathrm{~m}$$ Now, let's calculate the Biot number: $$\mathrm{Bi} = \frac{20 \mathrm{~W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-1}(11.5 \times 10^{-2} \mathrm{~m})}{0.47 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot \mathrm{K}^{-1}}$$ If the Biot number is less than 0.1, the lumped capacitance method can be applied.

Calculate the time required for the center temperature to drop to \(-18^{\circ} \mathrm{C}\)

Using the lumped capacitance method, the following formula relates the temperature at any time \(t\): $$\theta(t) = \theta_i - (\theta_i - \theta_\infty)(1 - e^{-t \frac{hA_s}{V\rho c})})$$ Here, \(\theta(t)\) is the temperature at time \(t\), \(\theta_i\) is the initial temperature, \(\theta_\infty\) is the surrounding temperature, and \(A_s\) is the surface area. From the given information, we know that the initial temperature (\(\theta_i\)) is \(7^{\circ} \mathrm{C}\) and the surrounding temperature (\(\theta_\infty\)) is \(-30^{\circ} \mathrm{C}\). We want to calculate the time it takes for the temperature at the center to be \(-18^{\circ} \mathrm{C}\). Hence, let \(\theta(t) = -18^{\circ} \mathrm{C}\), and we can solve for \(t\): $$-18 = 7 - (7 - (-30))(1 - e^{-t\frac{20A_s}{V\rho c)})$$ The surface area, \(A_s\), for a slab with a large area relative to its thickness can be considered as the area of one side. Assume \(1 \mathrm{~m}^2\) for simplicity: $$A_s = 1 \mathrm{~m}^2$$ We know that \(\rho = \frac{1}{\alpha}\) and since \(\alpha = 0.13 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{s}^{-1}\): $$\rho = \frac{1}{0.13 \times 10^{-6}}$$ Assuming constant volume, we can write the volume \(V\) as: $$V = A_s L_c$$ Since we have already calculated \(L_c\) in Step 1, substitute that value, and now we have all the necessary information to find the time, \(t\).

Calculate the surface temperature of the meat slab at that time

Once we have calculated the time \(t\), we can use the lumped capacitance method formula again to find the surface temperature at that time. In this case, we'll use the parameter \(x = 0\) to represent the surface temperature. The formula becomes: $$\theta_s(t) = \theta_i - (\theta_i - \theta_\infty)(1 - e^{-t\frac{hA_s}{V\rho c}})$$ Simply substitute the values for \(\theta_i\), \(\theta_\infty\), \(h\), \(A_s\), \(V\), \(\rho\), and \(c\) along with the time \(t\) we found in Step 2 to calculate the surface temperature of the meat slab at that time.