In a manufacturing facility, \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\) wet brass plates coming out of a water bath are to be dried by passing them through a section where dry air at 1 atm and \(25^{\circ} \mathrm{C}\) is blown parallel to their surfaces at \(4 \mathrm{~m} / \mathrm{s}\). If the plates are at \(15^{\circ} \mathrm{C}\) and there are no dry spots, determine the rate of evaporation from both sides of a plate.

To calculate the mass transfer coefficient (k), we will use the following equation derived from heat and mass transfer principles: \[k = \frac{h_{fg}}{C_{pl}(T_s - T_\infty)}\] where \(h_{fg}\) is the latent heat of vaporization of water, \(C_{pl}\) is the specific heat of liquid water, \(T_s\) is the surface temperature of the plate, and \(T_\infty\) is the air temperature. We are given that \(T_s = 15^{\circ}\mathrm{C}\) and \(T_\infty = 25^{\circ}\mathrm{C}\). The latent heat of vaporization for water at room temperature is approximately \(h_{fg} = 2.4 \times 10^6 \mathrm{~J/kg}\). The specific heat of liquid water is \(C_{pl} = 4186 \mathrm{~J/(kg \cdot K)}\). Let's calculate k: \[k = \frac{2.4 \times 10^6 \mathrm{~J/kg}}{4186 \mathrm{~J/(kg \cdot K)} (15^{\circ}\mathrm{C} - 25^{\circ} \mathrm{C})} \approx 11.43 \mathrm{~m/s}\]

Now that we have the mass transfer coefficient, we can determine the rate of evaporation from the plate. The evaporation rate equation can be given as: \[q = A \times k \times \rho_{vapor}(T_\infty - T_s)\] where q is the evaporation rate, A is the area of the plate, and \(\rho_{vapor}\) is the vapor density. The given dimensions of the plate are \(40 \mathrm{~cm} \times 40 \mathrm{~cm}\), which is equal to \(0.4 \mathrm{~m} \times 0.4 \mathrm{~m}\), so the surface area is \(A = (0.4 \mathrm{~m} \times 0.4 \mathrm{~m}) = 0.16 \mathrm{~m^2}\). Assuming the vapor density at these conditions is \(\rho_{vapor} = 0.02 \mathrm{~kg/m^3}\), we can calculate the evaporation rate: \[q = 0.16 \mathrm{~m^2} \times 11.43 \mathrm{~m/s} \times 0.02 \mathrm{~kg/m^3} (25^{\circ}\mathrm{C} - 15^{\circ} \mathrm{C}) \approx 0.0366 \mathrm{~kg/s}\]

Since air is blown parallel to both sides of the plate, the evaporation rate from both sides should be the same. Therefore, the evaporation rate from both sides of the plate would be double the value we obtained in Step 2: \[q_{Total} = 2 \times 0.0366 \mathrm{~kg/s} = 0.0732 \mathrm{~kg/s}\] So, the rate of evaporation from both sides of the wet brass plate is approximately \(0.0732 \mathrm{~kg/s}\).