During cold weather periods, vapor in a room diffuses through the dry wall and condenses in the adjoining insulation. This process decreases the thermal resistance and degrades the insulation. Consider a condition at which the vapor pressure in the air at \(25^{\circ} \mathrm{C}\) inside a room is \(3 \mathrm{kPa}\), and the vapor pressure in the insulation is negligible. The 3 -m-high and 10 -m-wide dry wall is 12-mm thick with a solubility of water vapor in the wall material of approximately \(0.007 \mathrm{kmol} / \mathrm{m}^{3}\).bar, and diffusion coefficient of water vapor in the wall is \(0.2 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). Determine the mass diffusion rate of water vapor through the wall.

Given quantities: - Ambient vapor pressure in the room, \(P_A = 3 \mathrm{kPa}\) - Vapor pressure in insulation, \(P_I = 0 \mathrm{kPa}\) (negligible) - Drywall dimensions: height \(H = 3 \mathrm{m}\), width \(W = 10\mathrm{m}\), and thickness \(L = 12 \mathrm{mm} = 0.012 \mathrm{m}\) - Solubility of water vapor in the wall material, \(C = 0.007 \mathrm{kmol / m^3}.\mathrm{bar}\) - Diffusion coefficient of water vapor in the wall, \(D = 0.2 \times 10^{-9} \mathrm{m^2/s}\) Fick's law of diffusion states that the mass diffusion rate \(N\) is proportional to the concentration difference and the diffusion area divided by the diffusion distance: $$N = -D \times \frac{A \times (C_A - C_B)}{L}$$ where, \(N\) is the mass diffusion rate (kg/s) \(-D\) means that diffusion occurs from higher to lower concentration \(A\) is the diffusion area (m²) \(C_A\) and \(C_B\) are the concentrations at points A and B respectively \(L\) is the diffusion distance (m) We will use this formula to find the mass diffusion rate of water vapor through the drywall.

First, we need to find the concentrations at points A and B via the given solubility. $$C_A = C \times P_A$$ $$C_A = (0.007\, \mathrm{kmol/m^3.\, bar}) \times (3\, \mathrm{kPa})$$ Since 1 bar equals to 100 kPa, we can then convert kPa to bar: $$C_A = (0.007\, \mathrm{kmol/m^3.\, bar}) \times (0.03\, \mathrm{bar})$$ $$C_A = 0.00021\, \mathrm{kmol/m^3}$$ As the vapor pressure in the insulation is negligible, \(C_B\) will also be \(0\, \mathrm{kmol/m^3}\).

The diffusion area, \(A\), is equal to the product of the height \(H\) and width \(W\) of the wall. $$A = H \times W$$ $$A = (3\, \mathrm{m}) \times (10\, \mathrm{m})$$ $$A = 30\, \mathrm{m^2}$$

Now we have all the required values to find the mass diffusion rate using Fick's law. $$N = -D \times \frac{A\times (C_A - C_B)}{L}$$ $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00021\, \mathrm{kmol/m^3})}{0.012\, \mathrm{m}}$$ Now, we need to convert kmol to kg. Since 1 kmol of water is equal to \(18.015 \mathrm{kg}\): $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00021 \times 18.015\, \mathrm{kg/m^3})}{0.012\, \mathrm{m}}$$ $$N = - (0.2 \times 10^{-9}\, \mathrm{m^2/s}) \times \frac{(30\, \mathrm{m^2})\times (0.00378\, \mathrm{kg/m^3})}{0.012\, \mathrm{m}}$$ $$N \approx -2.975 \times 10^{-4}\, \mathrm{kg/s}$$ Since the negative sign indicates that the mass diffusion is in the direction from higher to lower concentration, and we only want the magnitude: $$N \approx 2.975 \times 10^{-4}\, \mathrm{kg/s}$$ Therefore, the mass diffusion rate of water vapor through the drywall is approximately \(2.975 \times 10^{-4} \mathrm{kg/s}\).