The diffusion of water vapor through plaster boards and its condensation in the wall insulation in cold weather are of concern since they reduce the effectiveness of insulation. Consider a house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is \(97 \mathrm{kPa}\). The inside of the walls is finished with \(9.5\)-mm-thick gypsum wallboard. Taking the vapor pressure at the outer side of the wallboard to be zero, determine the maximum amount of water vapor that will diffuse through a \(3-\mathrm{m} \times 8-\mathrm{m}\) section of a wall during a 24-h period. The permeance of the \(9.5\)-mm-thick gypsum wallboard to water vapor is \(2.86 \times 10^{-9} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{Pa}\).

To determine the partial pressure of vapor (Pi) inside the house, we need to use the given information: the temperature inside the house is \(20^{\circ} \mathrm{C}\), the relative humidity is 60% and the atmospheric pressure is \(97\ \mathrm{kPa}\). First, we need to find the saturation pressure (\(P_{sat}\)) of water vapor at the inside temperature (20°C). We can use the Antoine equation for this: \(P_{sat}=610.5\times{10^{\frac{7.5 \times T}{T + 237.3}}}\) where \(T\) is the temperature in Celsius. Substituting the temperature into the equation, we have: \(P_{sat}=610.5\times{10^{\frac{7.5\times20}{20+237.3}}}=2.338\ \mathrm{kPa}\) Now, we apply the definition of relative humidity to find the partial pressure of vapor (Pi) inside the house: \(Relative\ Humidity=\frac{P_i}{P_{sat}}\) \(P_i=Relative\ Humidity\times P_{sat}=0.6\times2.338 = 1.403\ \mathrm{kPa}\)

Now that we have calculated the partial pressure of water vapor inside the house (Pi), we can find the vapor pressure difference (ΔP) between the two sides of the wall. The vapor pressure outside the wall is given as zero, so: \(\Delta P=P_i-0=1.403\ \mathrm{kPa}\)

To calculate the mass flow rate (m) of water vapor through the wall, we need to use the permeance of the wall and the vapor pressure difference. We are given that the permeance of the wall is \(2.86\times10^{-9}\ \mathrm{kg/s\cdot m^2\cdot Pa}\). To use this value, we need to convert the vapor pressure difference from kPa to Pa: \(\Delta P=1.403\ \mathrm{kPa}\times1000\ \mathrm{Pa/kPa}=1403\ \mathrm{Pa}\) Now, we can find the mass flow rate (m) per unit area of the wall: \(m=\mathrm{Permeance}\times\Delta P=2.86\times10^{-9}\ \mathrm{kg/s\cdot m^2\cdot Pa}\times1403\ \mathrm{Pa}=4.010\times10^{-6}\ \mathrm{kg/s\cdot m^2}\)

We now have the mass flow rate of water vapor through the wall per unit area. To find the maximum amount of water vapor that will diffuse through the specified area (\(3\ \mathrm{m}\times8\ \mathrm{m}\)) of the wall during a 24-hour period, we need to multiply the mass flow rate by the area of the wall and the duration (in seconds) of the 24-hour period: \(Mass\ Flow\ Rate\times Area\times Time=4.010\times10^{-6}\ \mathrm{kg/s\cdot m^2}\times(3\ \mathrm{m}\times8\ \mathrm{m})\times(24\ \mathrm{h\cdot3600\ s/h})=2.639\ \mathrm{kg}\) So, the maximum amount of water vapor that will diffuse through the \(3-\mathrm{m} \times 8-\mathrm{m}\) section of a wall during a 24-hour period under these conditions is approximately 2.639 kg.