Question

Consider a brick house that is maintained at \(20^{\circ} \mathrm{C}\) and 60 percent relative humidity at a location where the atmospheric pressure is $85 \mathrm{kPa}$. The walls of the house are made of 20 -cm-thick brick whose permeance is \(23 \times 10^{-12} \mathrm{~kg} / \mathrm{s}\) - $\mathrm{m}^{2} \cdot \mathrm{Pa}$. 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 \(3-\mathrm{m} \times 5-\mathrm{m}\) section of a wall during a 24-h period.

Short answer

Answer: The maximum amount of water vapor that will diffuse through a 3 m x 5 m section of the brick wall during a 24-hour period is 1.29 kg.

Step-by-step solution

Find the vapor pressure inside the house

Since the atmospheric pressure P_atm in the house is given as 85 kPa, we need to find the vapor pressure P_v_inside by using the given values of temperature and relative humidity. We'll use the saturation pressure formula for water vapor at the given temperature and multiply it with the relative humidity (in decimal). First, find the saturation pressure P_sat at 20°C: \(P_\text{sat} = 0.0234 \times e^{(17.62 \times T) / (T + 243.12)}\) Where T is the temperature in °C. Substitute the knowns and solve for P_sat: \(P_\text{sat} = 0.0234 \times e^{(17.62 \times 20) / (20 + 243.12)} = 2.338\) Now, find the vapor pressure P_v_inside by multiplying P_sat with the relative humidity (60%): \(P_\text{v inside} = \text{relative humidity} \times P_\text{sat} = 0.6 \times 2.338 = 1.403 \ \text{kPa}\)

Find the pressure difference across the wall

Since the vapor pressure at the outer side of the wall is given as zero, the pressure difference across the wall will be equal to the vapor pressure inside the house: \(\Delta P = P_\text{v inside} - P_\text{v outside} = 1.403 \ \text{kPa}\)

Calculate the rate of diffusion using Fick's first law

Fick's first law of diffusion states that the rate of diffusion J is proportional to the pressure difference across the wall and inversely proportional to the thickness of the wall: \(J = \frac{P_\text{permeance} \times A \times \Delta P}{d}\) Where \(P_\text{permeance} = 23 \times 10^{-12} \ \frac{\text{kg}}{\text{s} \cdot \text{m}^2 \cdot \text{Pa}}\), A is the area of the wall section \((3\,\text{m} \times 5\,\text{m})\), and \(d=0.2\,\text{m}\) is the wall thickness. Substitute the known values and solve for J: \(J = \frac{(23 \times 10^{-12})(3 \times 5)(1.403 \times 10^{3})}{0.2} = 14.92 \times 10^{-6} \ \frac{\text{kg}}{\text{s}}\)

Calculate the amount of water vapor diffused in a 24-hour period

Now that we have the rate of diffusion, we can find the total amount of water vapor diffused in a 24-hour period by multiplying it with the length of time in seconds: \(\text{Amount of water vapor diffused} = J \times \text{time} = 14.92 \times 10^{-6} \ \frac{\text{kg}}{\text{s}} \times 24 \ \text{h} \times 3600\, \frac{\text{s}}{\text{h}} = 1.29 \ \text{kg}\) Therefore, the maximum amount of water vapor that will diffuse through a \(3\,\text{m} \times 5\,\text{m}\) section of a wall during a 24-hour period is 1.29 kg.