Question

The roof of a house is \(15 \mathrm{~m} \times 8 \mathrm{~m}\) and is made of a 30 -cm-thick concrete layer. The interior of the house is maintained at \(25^{\circ} \mathrm{C}\) and 50 percent relative humidity and the local atmospheric pressure is \(100 \mathrm{kPa}\). Determine the amount of water vapor that will migrate through the roof in \(24 \mathrm{~h}\) if the average outside conditions during that period are \(3^{\circ} \mathrm{C}\) and 30 percent relative humidity. The permeability of concrete to water vapor is $24.7 \times 10^{-12} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{Pa}$.

Short answer

Answer: Approximately 79.89 kg of water vapor will migrate through the roof in 24 hours.

Step-by-step solution

Find the saturation pressure of water vapor at both the inside and outside temperatures

We will use the Antoine equation to find the saturation pressure of water vapor at the inside temperature (\(25^{\circ}\mathrm C\)) and outside temperature (\(3^{\circ}\mathrm C\)). The Antoine equation is given as: $$p_{sat} = 10^{A - \frac{B}{T+C}}$$ For water, the coefficients are: \(A = 8.07131\), \(B = 1730.63\), and \(C = 233.426\). We can plug in the temperatures and calculate the saturation pressure of water vapor at both temperatures. $$p_{sat_{inside}} = 10^{8.07131 - \frac{1730.63}{25+233.426}} \approx 3.1696 \times 10^3 \mathrm{Pa}$$ $$p_{sat_{outside}} = 10^{8.07131 - \frac{1730.63}{3+233.426}} \approx 0.53823 \times 10^3 \mathrm{Pa}$$

Find the partial pressure of water vapor inside and outside the house

Now we will find the partial pressure of water vapor inside and outside the house by multiplying the saturation pressure by the relative humidity. $$p_{v_{inside}} = p_{sat_{inside}} \times \frac{50}{100} \approx 1.5848 \times 10^3 \mathrm{Pa}$$ $$p_{v_{outside}} = p_{sat_{outside}} \times \frac{30}{100} \approx 0.16147 \times 10^3 \mathrm{Pa}$$

Calculate the water vapor pressure difference

Subtract the outside partial pressure of water vapor from the inside partial pressure. $$\Delta p_v = p_{v_{inside}} - p_{v_{outside}} \approx (1.5848 - 0.16147) \times 10^3 \mathrm{Pa} \approx 1.4233 \times 10^3 \mathrm{Pa}$$

Calculate the water vapor flow rate through the roof

To find the water vapor flow rate, use the given permeability of concrete and the pressure difference calculated in Step 3. The area of the roof and the thickness of the concrete layer will also be needed. $$Flow\,rate = \frac{Permeability \times Area \times \Delta p_v}{Thickness}$$ Plug in the values to calculate the flow rate: $$Flow\,rate = \frac{24.7 \times 10^{-12} \mathrm{\frac{kg}{s\cdot m\cdot Pa}} \times 15 \mathrm{m} \times 8 \mathrm{m} \times 1.4233 \times 10^3 \mathrm{Pa}}{0.3 \mathrm{m}} \approx 9.2379 \times 10^{-7} \mathrm{\frac{kg}{s}}$$

Determine the amount of water vapor migrated through the roof in 24 hours

Multiply the flow rate by the time period (24 hours) to find the total amount of water vapor migrated. $$Water\,vapor\,amount = Flow\,rate \times 24 \mathrm{h} \times 3600 \mathrm{\frac{s}{h}} \approx 9.2379 \times 10^{-7} \mathrm{\frac{kg}{s}} \times 24 \times 3600 \mathrm{\frac{s}{h}} \approx 79.89 \mathrm{kg}$$ So, the amount of water vapor that will migrate through the roof in 24 hours is approximately \(79.89\,kg\).