Question

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 \(3-\mathrm{m}\)-high and 10 -m-wide dry wall is \(12 \mathrm{~mm}\) thick with a solubility of water vapor in the wall material of approximately $0.007 \mathrm{kmol} / \mathrm{m}^{3}$. bar, and the 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.

Short answer

Answer: The mass diffusion rate of water vapor through the dry wall is approximately 1.889 g/s.

Step-by-step solution

List given information and identify the formula.

We are given the following information: - The vapor pressure in the air (\(P_{1}\)) is 3 kPa, - The vapor pressure in the insulation (\(P_{2}\)) is negligible, so we can consider it as 0 kPa, - The height (H) and width (W) of the drywall are 3 m and 10 m, respectively, - The thickness of the drywall (L) is 12 mm, - The solubility of water vapor in the wall material (S) is approximately 0.007 kmol/m³.bar, - The diffusion coefficient of water (D) vapor in the wall is \(0.2 * 10^{-9} \text{m}^2/\text{s}\). We will use Fick's law of diffusion to determine the mass diffusion rate of water vapor through the wall: \(J = -D \frac{\Delta C}{L}\) First, we will need to find the difference in concentration of the vapor on each side of the wall, which is related to the vapor pressures and the solubility.

Convert vapor pressures to concentrations using solubility.

To find the concentrations (\(C_1\) and \(C_2\)) of the vapor on each side of the wall, we use the solubility (S): \(C_1 = P_1 \times S\) and \(C_2 = P_2 \times S\) Since \(P_{2} = 0 \,\text{kPa}\), \(C_2\) is 0. For \(C_1\), we have: \(C_1 = 3\, \text{kPa} \times 0.007\, \text{kmol/m}^3\text{.bar}\) Note that we need to make sure the pressure is in bars so the units align: \(C_1 = (3/100)\, \text{bar} \times 0.007\, \text{kmol/m}^3\text{.bar} = 0.00021\, \text{kmol/m}^3\) The difference in concentration is \(\Delta C = C_1 - C_2 = 0.00021\, \text{kmol/m}^3\).

Apply Fick's Law.

Now that we have the difference in concentration (\(\Delta C\)), we can use Fick's law to find the molar diffusion rate of water vapor through the wall: \(J = -D \frac{\Delta C}{L}\) \(J = - (0.2 * 10^{-9}\, \text{m}^2/\text{s}) \frac{0.00021\, \text{kmol/m}^3}{0.012\, \text{m}}\) \(J = - (-3.5 * 10^{-6}\, \text{kmol/m}^2\text{s})\) As the signs were only to indicate direction, we can discard the negative sign, and we get: \(J = 3.5 * 10^{-6}\, \text{kmol/m}^2\text{s}\)

Calculate mass diffusion rate.

Finally, we will convert the molar diffusion rate into a mass diffusion rate using the molar mass of water (18.015 g/mol): \(Mass\_Diffusion\_Rate = J \times Area \times Molar\_Mass\_Water\) \(Area = Height \times Width = 3\, \text{m} \times 10\, \text{m} = 30\, \text{m}^2\) \(Mass\_Diffusion\_Rate = (3.5 * 10^{-6}\, \text{kmol/m}^2\text{s}) \times 30\, \text{m}^2 \times 18.015\, \text{g/mol}\) \(Mass\_Diffusion\_Rate = 1.889\, \text{g/s}\) Therefore, the mass diffusion rate of water vapor through the wall is approximately 1.889 g/s.